The Lebesgue integral

Measures, the Lebesgue integral and its convergence theorems.

In this chapter 10

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Measures and measurable functions
Convergence theorems
L^p spaces
- L^p spaces
- The Radon-Nikodym theorem

Measures and measurable functions

Definition

Sigma-algebras and measures

Adapted

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Gives the two objects every integration theory rests on — a sigma-algebra of sets we agree to measure and a countably additive measure on it — together with the monotonicity, subadditivity and continuity rules that let you pass to the limit inside a measure, and the vocabulary of null sets and "almost everywhere".

Prerequisites: Sets, functions and countability, Topological and metric spaces

Definition(Algebras, sigma-algebras, Borel sets)AdaptedAdapted from Folland (1999), §1.2, pp. 21–22; Reed & Simon (1980), §I.3, p. 14; Evans (2010), Appendix E.1, p. 684

Let $X$ be a non-empty set and $\mathcal{P}(X)$ the collection of all its subsets.

Algebra. A collection $\mathcal{A} \subseteq \mathcal{P}(X)$ is an algebra on $X$ if it is non-empty, the complement $E^{c} = X \setminus E$ of each of its members is again a member, and the union of any finite list of its members is a member.

Sigma-algebra. An algebra $\mathcal{M}$ is a $\sigma$ -algebra if it is in addition closed under the union of any sequence of its members: $E_1, E_2, \dots \in \mathcal{M}$ forces $\bigcup_{j \ge 1} E_j \in \mathcal{M}$ . The pair $(X, \mathcal{M})$ is a measurable space and the members of $\mathcal{M}$ are the measurable sets.

Three facts come free from the two axioms. Pick any $E \in \mathcal{A}$ , which is possible because $\mathcal{A}$ is non-empty. Closure under finite unions puts $X = E \cup E^{c}$ into $\mathcal{A}$ , and closure under complements then puts $\varnothing = X^{c}$ into $\mathcal{A}$ . De Morgan's law $\bigcap_j E_j = \bigl(\bigcup_j E_j^{c}\bigr)^{c}$ turns closure under unions into closure under intersections, finite for an algebra and countable for a $\sigma$ -algebra. And $E \setminus F = E \cap F^{c}$ , so both are closed under differences.

Generated $\sigma$ -algebra. An arbitrary intersection of $\sigma$ -algebras on $X$ is again one, because each axiom is a condition that must hold in every member of the family. Since $\mathcal{P}(X)$ is a $\sigma$ -algebra containing anything, for every $\mathcal{E} \subseteq \mathcal{P}(X)$ the intersection of all $\sigma$ -algebras that contain $\mathcal{E}$ is the smallest $\sigma$ -algebra containing $\mathcal{E}$ . We write it $\mathcal{M}(\mathcal{E})$ and call it the $\sigma$ -algebra generated by $\mathcal{E}$ . Monotonicity is immediate: $\mathcal{E} \subseteq \mathcal{M}(\mathcal{F})$ implies $\mathcal{M}(\mathcal{E}) \subseteq \mathcal{M}(\mathcal{F})$ , since $\mathcal{M}(\mathcal{F})$ is then one of the $\sigma$ -algebras over which the intersection defining $\mathcal{M}(\mathcal{E})$ is taken.

Borel sets. If $X$ carries a topology — a metric space, for instance, as in topological and metric spaces — the $\sigma$ -algebra generated by the open subsets of $X$ is the Borel $\sigma$ -algebra $\mathcal{B}_X$ , and its members are the Borel sets. Complements turn open into closed, so $\mathcal{B}_X$ is equally the $\sigma$ -algebra generated by the closed sets. It contains every countable intersection of open sets (a $G_\delta$ set), every countable union of closed sets (an $F_\sigma$ set), and the whole hierarchy obtained by alternating the two operations. On the line, Reed and Simon (Reed & Simon (1980), §I.3, p. 14) take as the definition the equivalent description: $\mathcal{B}_{\mathbb{R}}$ is the smallest family of subsets of $\mathbb{R}$ that is closed under complements and countable unions and contains every open interval.

Disjointification. Given any sequence $E_1, E_2, \dots$ in a $\sigma$ -algebra, put

F_1 = E_1, \qquad F_k = E_k \setminus \bigl(E_1 \cup \dots \cup E_{k-1}\bigr) \quad (k \ge 2).

The $F_k$ lie in the $\sigma$ -algebra, they are pairwise disjoint, $F_k \subseteq E_k$ , and $\bigcup_{k \le n} F_k = \bigcup_{k \le n} E_k$ for every $n$ , hence also in the limit. This device, which Folland (Folland (1999), §1.2, p. 21) recommends remembering, converts any countable union into a disjoint one and is used three times below.

Definition(Measures, null sets, completeness)AdaptedAdapted from Folland (1999), §1.3, pp. 24–27

Let $(X, \mathcal{M})$ be a measurable space. A measure on it is a map

\mu : \mathcal{M} \longrightarrow [0, \infty]

such that

$\mu(\varnothing) = 0$ , and
(countable additivity) $\mu\bigl(\bigcup_{j \ge 1} E_j\bigr) = \sum_{j \ge 1} \mu(E_j)$ for every sequence $E_1, E_2, \dots$ of pairwise disjoint members of $\mathcal{M}$ .

The triple $(X, \mathcal{M}, \mu)$ is a measure space. The series in (2) has non-negative terms, so it either converges or diverges to $+\infty$ ; in both cases its value is independent of the order of summation, which is what makes (2) a legitimate demand on an unordered union. Padding a finite disjoint list $E_1, \dots, E_n$ with $E_j = \varnothing$ for $j > n$ shows that (2) implies finite additivity. A map obeying (1) and finite additivity but not necessarily (2) is a finitely additive measure; the gap between the two is real and is the subject of the remark below.

Sizes. $\mu$ is finite if $\mu(X) < \infty$ , and $\sigma$ -finite if $X = \bigcup_{j \ge 1} E_j$ for some $E_j \in \mathcal{M}$ with every $\mu(E_j) < \infty$ . A finite measure with $\mu(X) = 1$ is a probability measure, and $(X, \mathcal{M}, \mu)$ is then a probability space, the setting of probability spaces.

Null sets. A set $N \in \mathcal{M}$ with $\mu(N) = 0$ is a null set. A property of the points of $X$ holds almost everywhere (a.e., or $\mu$ -a.e.) if the set of points where it fails is contained in a null set. Subadditivity, proved below, says that a countable union of null sets is null, so countably many a.e. statements may be used at once.

Completeness. If $\mu(N) = 0$ and $F \subseteq N$ , monotonicity gives $\mu(F) = 0$ provided $F \in \mathcal{M}$ — and that proviso need not hold. A measure is complete when every subset of every null set is measurable. Every measure can be completed: let $\mathcal{N}$ be the null sets of $\mu$ and

\overline{\mathcal{M}} = \bigl\{\, E \cup F \;:\; E \in \mathcal{M},\ F \subseteq N \text{ for some } N \in \mathcal{N} \,\bigr\} .

Then $\overline{\mathcal{M}}$ is a $\sigma$ -algebra and $\overline{\mu}(E \cup F) = \mu(E)$ defines the unique extension of $\mu$ to a complete measure on it (Folland (1999), Thm 1.9, pp. 26–27). We call $\overline{\mu}$ the completion of $\mu$ .

Note(Conventions differ)AdaptedAdapted from Grimmett & Stirzaker (2020), §1.2, Def. (1.2.5), p. 3, and §1.3, Def. (1.3.1) and the following paragraph, p. 5; Evans (2010), Appendix E.1, p. 684; Folland (1999), §1.3, p. 24

Three presentations of the same objects appear in the books cited here, and the differences are worth knowing before comparing formulas across fields.

Ours, following Folland (Folland (1999), §1.2, p. 21 and §1.3, p. 24): a $\sigma$ -algebra is a non-empty collection closed under complements and countable unions, and a measure takes values in the extended half-line $[0, \infty]$ .
Probability. Grimmett and Stirzaker (Grimmett & Stirzaker (2020), Def. (1.2.5), p. 3) say $\sigma$ -field rather than $\sigma$ -algebra and list the axioms as $\varnothing \in \mathcal{F}$ , closure under countable unions, and closure under complements; this is the same class of collections, since $\varnothing \in \mathcal{F}$ is exactly what rules out the empty collection. For the measure they write (Grimmett & Stirzaker (2020), §1.3, p. 5) that a measure is a function $\mu : \mathcal{F} \to [0, \infty)$ with $\mu(\varnothing) = 0$ that is countably additive, so that in their convention the value $+\infty$ is excluded; they immediately specialise to probability measures, for which the distinction is invisible. We follow Folland and allow the value $+\infty$ , because Lebesgue measure on $\mathbb{R}$ and counting measure on an infinite set are both wanted below and neither is finite.
Partial differential equations. Evans (Evans (2010), Appendix E.1, p. 684) states the axioms for $\mathcal{M} \subseteq \mathcal{P}(\mathbb{R}^n)$ as: $\varnothing$ and $\mathbb{R}^n$ belong to $\mathcal{M}$ ; $\mathcal{M}$ is closed under complements; and $\mathcal{M}$ is closed under countable unions and countable intersections. The last clause is redundant given the other two, by De Morgan, but it is harmless and it is what a reader of the PDE literature will meet.

Definition(Generators of the Borel sigma-algebra of the line)AdaptedAdapted from Folland (1999), Prop. 1.2, p. 22

$\mathcal{B}_{\mathbb{R}}$ is generated by each one of the following eight families of intervals and rays:

\begin{aligned} &\mathcal{E}_1 = \{(a,b)\}, &\quad &\mathcal{E}_2 = \{[a,b]\}, &\quad &\mathcal{E}_3 = \{(a,b]\}, &\quad &\mathcal{E}_4 = \{[a,b)\}, \\ &\mathcal{E}_5 = \{(a,\infty)\}, & &\mathcal{E}_6 = \{(-\infty,a)\}, & &\mathcal{E}_7 = \{[a,\infty)\}, & &\mathcal{E}_8 = \{(-\infty,a]\}, \end{aligned}

where $a < b$ run over the reals in the first four families and $a$ runs over the reals in the last four.

Definition(The four rules for computing with a measure)AdaptedAdapted from Folland (1999), Thm 1.8, pp. 25–26

Let $(X, \mathcal{M}, \mu)$ be a measure space.

a. Monotonicity. An inclusion $E \subseteq F$ between members of $\mathcal{M}$ forces $\mu(E) \le \mu(F)$ . b. Countable subadditivity. If $E_1, E_2, \dots \in \mathcal{M}$ — not assumed disjoint — then

\mu\Bigl(\bigcup_{j \ge 1} E_j\Bigr) \;\le\; \sum_{j \ge 1} \mu(E_j).

c. Continuity from below. If $E_1 \subseteq E_2 \subseteq \cdots$ all lie in $\mathcal{M}$ then $\mu\bigl(\bigcup_{j \ge 1} E_j\bigr) = \lim_{j \to \infty} \mu(E_j)$ . d. Continuity from above. If $E_1 \supseteq E_2 \supseteq \cdots$ all lie in $\mathcal{M}$ and $\mu(E_1) < \infty$ , then $\mu\bigl(\bigcap_{j \ge 1} E_j\bigr) = \lim_{j \to \infty} \mu(E_j)$ .

Lemma(Continuity from below as a test for countable additivity)AdaptedAdapted from Folland (1999), Exercise 11, p. 27

Let $\mu : \mathcal{M} \to [0,\infty]$ satisfy $\mu(\varnothing) = 0$ and be finitely additive. Then $\mu$ is a measure if and only if it is continuous from below in the sense of (c) above. If moreover $\mu(X) < \infty$ , then $\mu$ is a measure if and only if it is continuous from above in the sense of (d). This is the form in which Folland (Folland (1999), Exercise 11, p. 27) poses it; the proof below shows that in the finite case the second test may be applied to sequences decreasing to $\varnothing$ alone, and that weaker version is the one the method uses.

Idea of the proofOriginal

Every one of the four rules is countable additivity applied to a cleverly chosen disjoint family, so the only real work is choosing the family.

For monotonicity, split $F$ into $E$ and the rest. For subadditivity, replace the overlapping $E_j$ by the disjointified $F_j$ of the definition block: the union is unchanged, the $F_j$ are disjoint so additivity applies exactly, and $F_j \subseteq E_j$ costs us only an inequality. For continuity from below, the successive differences $E_j \setminus E_{j-1}$ of an increasing sequence are already disjoint, and the partial sums of their measures are exactly the numbers $\mu(E_j)$ , so a limit of measures becomes the sum of a series. Continuity from above is the previous case applied to the complements inside $E_1$ — and subtracting $\mu(E_j)$ from $\mu(E_1)$ is meaningful only when $\mu(E_1)$ is finite, which is precisely where that hypothesis enters.

For the Borel generators the idea is that each family is sandwiched between the others modulo countable operations: a half-open interval is a countable intersection of open ones, an open interval is a countable union of closed ones, and a ray is a countable union of intervals. Monotonicity of $\mathcal{E} \mapsto \mathcal{M}(\mathcal{E})$ then closes the circle.

Proof(The four rules)AdaptedAdapted from Folland (1999), Thm 1.8, pp. 25–26

(a) Monotonicity. $F$ is the disjoint union of $E$ and $F \setminus E$ , both in $\mathcal{M}$ . Finite additivity gives $\mu(F) = \mu(E) + \mu(F \setminus E)$ , and the second term is $\ge 0$ because $\mu$ takes values in $[0,\infty]$ . Hence $\mu(E) \le \mu(F)$ .

(b) Subadditivity. Disjointify: let $F_1 = E_1$ and $F_k = E_k \setminus (E_1 \cup \dots \cup E_{k-1})$ . As recorded in the definition block, the $F_k$ are disjoint members of $\mathcal{M}$ with $F_k \subseteq E_k$ and the same total union. Countable additivity applies to the $F_k$ , and monotonicity — part (a), with the inclusion $F_k \subseteq E_k$ — bounds each term:

\mu\Bigl(\bigcup_{k \ge 1} E_k\Bigr) = \mu\Bigl(\bigcup_{k \ge 1} F_k\Bigr) = \sum_{k \ge 1}\mu(F_k) \le \sum_{k \ge 1}\mu(E_k).

(c) Continuity from below. Set $E_0 = \varnothing$ and $D_j = E_j \setminus E_{j-1}$ for $j \ge 1$ . These are measurable, and they are disjoint: if $i < j$ then $D_i \subseteq E_i \subseteq E_{j-1}$ while $D_j$ misses $E_{j-1}$ . Because the sequence increases, $\bigcup_{j \le n} D_j = E_n$ for each $n$ and $\bigcup_{j \ge 1} D_j = \bigcup_{j \ge 1} E_j$ . Countable additivity, then finite additivity, give

\mu\Bigl(\bigcup_{j \ge 1} E_j\Bigr) = \sum_{j \ge 1}\mu(D_j) = \lim_{n \to \infty}\sum_{j = 1}^{n}\mu(D_j) = \lim_{n \to \infty}\mu(E_n),

the middle equality being the definition of the sum of a series.

(d) Continuity from above. Put $G_j = E_1 \setminus E_j$ . Since the $E_j$ decrease, the $G_j$ increase, and

\bigcup_{j \ge 1} G_j = E_1 \setminus \bigcap_{j \ge 1} E_j .

By (c), $\mu\bigl(E_1 \setminus \bigcap_j E_j\bigr) = \lim_j \mu(G_j)$ . Now use finite additivity on the two disjoint decompositions $E_1 = E_j \cup G_j$ and $E_1 = \bigl(\bigcap_j E_j\bigr) \cup \bigl(E_1 \setminus \bigcap_j E_j\bigr)$ :

\mu(E_1) = \mu(E_j) + \mu(G_j), \qquad \mu(E_1) = \mu\Bigl(\bigcap_{j\ge1} E_j\Bigr) + \lim_{j \to \infty}\mu(G_j).

Here the hypothesis $\mu(E_1) < \infty$ is used: it makes every $\mu(E_j)$ and $\mu(G_j)$ finite, so the first identity may be rearranged as $\mu(G_j) = \mu(E_1) - \mu(E_j)$ and the subtraction is an ordinary one between real numbers. Substituting into the second identity and cancelling the finite number $\mu(E_1)$ ,

\mu\Bigl(\bigcap_{j \ge 1} E_j\Bigr) = \mu(E_1) - \lim_{j \to \infty}\bigl(\mu(E_1) - \mu(E_j)\bigr) = \lim_{j \to \infty}\mu(E_j). \qquad \square

Proof(The lemma)AdaptedAdapted from Folland (1999), Exercise 11, p. 27

Only if. A measure is finitely additive and satisfies (c) and (d) by the previous proof.

If (continuity from below). Let $A_1, A_2, \dots \in \mathcal{M}$ be disjoint and put $E_n = A_1 \cup \dots \cup A_n$ . The $E_n$ increase to $\bigcup_j A_j$ , so continuity from below and then finite additivity give

\mu\Bigl(\bigcup_{j \ge 1} A_j\Bigr) = \lim_{n \to \infty}\mu(E_n) = \lim_{n \to \infty}\sum_{j=1}^{n}\mu(A_j) = \sum_{j \ge 1}\mu(A_j),

which is countable additivity.

If (continuity from above, $\mu(X) < \infty$ ). With $A_j$ and $E_n$ as before, the sets $R_n = \bigl(\bigcup_{j\ge1}A_j\bigr)\setminus E_n$ decrease, and their intersection is empty because each point of the union lies in some $A_j$ and hence in $E_n$ for all $n \ge j$ . Finiteness of $\mu$ makes $\mu(R_1) < \infty$ , so continuity from above gives $\mu(R_n) \to \mu(\varnothing) = 0$ . Finite additivity splits $\bigcup_j A_j = E_n \cup R_n$ , so $\sum_{j=1}^{n}\mu(A_j) = \mu\bigl(\bigcup_j A_j\bigr) - \mu(R_n) \to \mu\bigl(\bigcup_j A_j\bigr)$ .

What the last paragraph actually used. Continuity from above entered only through the one sequence $R_n$ , and that sequence decreases to $\varnothing$ . So when $\mu(X) < \infty$ the test may be weakened to: $E_1 \supseteq E_2 \supseteq \cdots$ with $\bigcap_j E_j = \varnothing$ implies $\mu(E_j) \to 0$ . Clause (d) contains this special case, so the weakened criterion is still equivalent to being a measure; it is the form used in step 4 of the method below. $\square$

Proof(Generators of the Borel sigma-algebra)AdaptedAdapted from Folland (1999), Prop. 1.2, p. 22

Write $\mathcal{M}_i = \mathcal{M}(\mathcal{E}_i)$ . We show $\mathcal{M}_i \subseteq \mathcal{B}_{\mathbb{R}}$ for every $i$ , and $\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{M}_i$ for every $i$ .

Step 1: every generator is Borel. The members of $\mathcal{E}_1, \mathcal{E}_5, \mathcal{E}_6$ are open and the members of $\mathcal{E}_2, \mathcal{E}_7, \mathcal{E}_8$ are closed, so all six families sit inside $\mathcal{B}_{\mathbb{R}}$ . For the half-open intervals, write

(a,b] = \bigcap_{n \ge 1}\Bigl(a,\ b + \tfrac1n\Bigr), \qquad [a,b) = \bigcap_{n \ge 1}\Bigl(a - \tfrac1n,\ b\Bigr),

countable intersections of open sets, hence Borel. By monotonicity of the generation operator, $\mathcal{M}_i \subseteq \mathcal{B}_{\mathbb{R}}$ for $i = 1, \dots, 8$ .

Step 2: $\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{M}_1$ . Every open $U \subseteq \mathbb{R}$ is a countable union of open intervals: for each $x \in U$ choose rationals $p < x < q$ with $(p,q) \subseteq U$ , and note that there are only countably many pairs of rationals, so the intervals so chosen form a countable family whose union is $U$ . (Countability of $\mathbb{Q} \times \mathbb{Q}$ is the standard counting argument.) Hence every open set lies in $\mathcal{M}_1$ , and therefore $\mathcal{B}_{\mathbb{R}} = \mathcal{M}(\text{open sets}) \subseteq \mathcal{M}_1$ .

Step 3: every open interval lies in each $\mathcal{M}_i$ . This is what remains: if $(a,b) \in \mathcal{M}_i$ for all $a<b$ , then $\mathcal{M}_1 \subseteq \mathcal{M}_i$ by monotonicity of the generation operator, and step 2 then gives $\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{M}_i$ .

The bounded families. For $a < b$ , with $n$ running over the integers large enough that $a + \frac1n < b - \frac1n$ ,

(a,b) = \bigcup_{n}\Bigl[a + \tfrac1n,\ b - \tfrac1n\Bigr] = \bigcup_{n}\Bigl(a,\ b - \tfrac1n\Bigr] = \bigcup_{n}\Bigl[a + \tfrac1n,\ b\Bigr),

three countable unions, which settle $\mathcal{E}_2$ , $\mathcal{E}_3$ and $\mathcal{E}_4$ .

The ray families. Here $(a,b) = (a,\infty) \cap (-\infty,b)$ , so it is enough that each of $\mathcal{M}_5, \dots, \mathcal{M}_8$ contain both kinds of unbounded ray. Four identities, each a single complement or a single countable union, move between the four kinds:

(-\infty,a] = \bigl((a,\infty)\bigr)^{c}, \quad (-\infty,a) = \bigcup_{n}\Bigl(-\infty,\ a - \tfrac1n\Bigr], \quad [a,\infty) = \bigl((-\infty,a)\bigr)^{c}, \quad (a,\infty) = \bigcup_{n}\Bigl[a + \tfrac1n,\ \infty\Bigr).

Applied in this cyclic order they carry each family to the next, and each $\sigma$ -algebra $\mathcal{M}_i$ is closed under both operations, so at most three steps around the cycle are needed in each case: $\mathcal{M}_5$ starts with $(a,\infty)$ and acquires $(-\infty,a]$ , then $(-\infty,a)$ ; $\mathcal{M}_8$ starts with $(-\infty,a]$ and acquires $(-\infty,a)$ , then $[a,\infty)$ , then $(a,\infty)$ ; $\mathcal{M}_6$ starts with $(-\infty,a)$ and acquires $[a,\infty)$ , then $(a,\infty)$ ; $\mathcal{M}_7$ starts with $[a,\infty)$ and acquires $(a,\infty)$ , and $(-\infty,a) = \bigl([a,\infty)\bigr)^{c}$ directly. Each of the four therefore contains every $(a,\infty)$ and every $(-\infty,b)$ , hence every $(a,b)$ .

Combining, each $\mathcal{M}_i$ contains all open intervals, so $\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{M}_i$ by the first paragraph of this step. With step 1 this gives $\mathcal{M}_i = \mathcal{B}_{\mathbb{R}}$ throughout. $\square$

Remark(Where the hypotheses are used)Original

Countability, not arbitrary size, in the $\sigma$ -algebra axiom. The axioms permit sequences of sets, nothing larger. They have to: every subset of $\mathbb{R}$ is a union of singletons, and singletons are Borel, so a " $\sigma$ -algebra" closed under arbitrary unions and containing the singletons would be all of $\mathcal{P}(\mathbb{R})$ — and no useful translation-invariant measure lives there, as the construction of Lebesgue measure shows. Reed and Simon (Reed & Simon (1980), §I.3, p. 14) make the same point from the Banach–Tarski paradox: some family of sets smaller than the power set must be chosen.
Countable additivity is strictly stronger than finite additivity. Folland (Folland (1999), §1.3, p. 25) gives the example: on an infinite $X$ with $\mathcal{M} = \mathcal{P}(X)$ , setting $\mu(E) = 0$ for finite $E$ and $\mu(E) = \infty$ for infinite $E$ produces a finitely additive measure that is not a measure. For $X = \mathbb{N}$ it fails countable additivity badly: $\mathbb{N}$ is the disjoint union of its singletons, so the right-hand side of (2) is $0$ while the left-hand side is $\infty$ . By the lemma, what this $\mu$ lacks is continuity from below, and indeed $\mu(\{1,\dots,n\}) = 0$ for every $n$ while $\mu(\mathbb{N}) = \infty$ .
$\mu(E_1) < \infty$ in continuity from above cannot be dropped. Let $\nu$ be counting measure on $\mathbb{N}$ , so $\nu(E)$ is the number of elements of $E$ , and let $E_j = \{n \in \mathbb{N} : n \ge j\}$ . The $E_j$ decrease with empty intersection, yet $\nu(E_j) = \infty$ for every $j$ , so $\lim_j \nu(E_j) = \infty \ne 0 = \nu(\varnothing)$ (Folland (1999), §1.3, p. 26). The proof above breaks at exactly one step: the rearrangement $\mu(G_j) = \mu(E_1) - \mu(E_j)$ is not available when $\mu(E_1) = \infty$ . It is enough to assume $\mu(E_{j_0}) < \infty$ for some single $j_0$ , since one may discard the first $j_0 - 1$ sets without changing the intersection.
Continuity from below needs no finiteness at all, because nothing is subtracted; both sides may be $+\infty$ together.
Disjointness in (2), and only there. Subadditivity (b) is the price paid for dropping it, and the inequality in (b) really can be strict: under Lebesgue measure, $E_1 = E_2 = [0,1]$ with $E_j = \varnothing$ for $j \ge 3$ give $1 < 2$ . Set-theoretic overlap by itself is not what causes this. The intervals $[0,1]$ and $[1,2]$ meet, but in a single point, which is null, and both sides equal $2$ ; and even a fat overlap is invisible if some set in the family is infinite, as $[0,1], [0,1], \mathbb{R}$ shows, where both sides are $+\infty$ . What the disjointification in the proof of (b) does show is this: $E_k$ is the disjoint union of $F_k$ and $E_k \cap (E_1 \cup \dots \cup E_{k-1})$ , so $\mu(E_k) = \mu(F_k) + \mu\bigl(E_k \cap (E_1 \cup \dots \cup E_{k-1})\bigr)$ . Hence if every such overlap is null the inequality is an equality; and if $\sum_j \mu(E_j) < \infty$ while some two of the sets meet in a set of positive measure, then every term is finite and at least one drops strictly, so the inequality is strict.
Completeness is not automatic and is not free. The Borel $\sigma$ -algebra of $\mathbb{R}$ with Lebesgue measure is not complete: the Cantor set is Borel and null, but it has subsets that are not Borel (there are $2^{\mathfrak{c}}$ subsets of the Cantor set and only $\mathfrak{c}$ Borel sets). Completing it is what produces the Lebesgue measurable sets. The cost is that the completed $\sigma$ -algebra depends on $\mu$ , so "measurable" stops being a property of the set alone; this is why composition of measurable functions misbehaves in measurable functions.
Why $[0,\infty]$ and not $[0,\infty)$ . Allowing the value $+\infty$ is what lets a single framework hold Lebesgue measure on the line, counting measure on an infinite set and the measures appearing in the Radon–Nikodym theorem. The price is the usual care with $\infty - \infty$ , which is precisely the subtraction forbidden in (d).

Method(Checking that a proposed set function is a measure)Original

Fix the domain first. Decide the $\sigma$ -algebra $\mathcal{M}$ before defining $\mu$ . If the sets you can measure are described by a list of "elementary" sets $\mathcal{E}$ (intervals, boxes, cylinders), the domain is $\mathcal{M}(\mathcal{E})$ , and you never have to display its members — generation is enough.
Check $\mu(\varnothing) = 0$ . Without it, $\mu \equiv \infty$ would satisfy (2) vacuously.
Check finite additivity on two disjoint sets; induction gives any finite number.
Upgrade to countable additivity by a limit, not by hand. By the lemma, it suffices to verify continuity from below: if $E_1 \subseteq E_2 \subseteq \cdots$ then $\mu(E_n) \to \mu(\bigcup_n E_n)$ . For a finite $\mu$ you may instead verify continuity from above at $\varnothing$ alone: $E_n \downarrow \varnothing$ implies $\mu(E_n) \to 0$ . That this restricted test already suffices is the sharpening recorded at the end of the proof of the lemma. This is usually much easier than attacking (2) directly.
Classify the size. Is $\mu$ finite? $\sigma$ -finite? Almost every later theorem — uniqueness of extensions, Fubini and Tonelli, Radon–Nikodym — asks for $\sigma$ -finiteness, so settle it now by exhibiting an explicit exhausting sequence.
Complete it if you will argue "almost everywhere". Replace $(\mathcal{M}, \mu)$ by $(\overline{\mathcal{M}}, \overline{\mu})$ . Nothing already measurable changes value.
To compute a measure of a complicated set, write it as a monotone limit of simple ones and use (c) or (d) rather than attempting (2) directly; for an upper bound only, cover it and use (b).

Example(A finite measure on the integers, and what happens without finiteness)Original

Take $X = \mathbb{N} = \{1, 2, 3, \dots\}$ with $\mathcal{M} = \mathcal{P}(\mathbb{N})$ , which is a $\sigma$ -algebra, and define

\mu(E) \;=\; \sum_{n \in E} \frac{1}{n^{2}} , \qquad E \subseteq \mathbb{N},

the sum of a non-negative family, hence independent of the order of summation. This is the weighted counting measure of Folland (Folland (1999), §1.3, p. 25) with weight $f(n) = n^{-2}$ .

It is a measure. $\mu(\varnothing) = 0$ is the empty sum. If $E_1, E_2, \dots$ are disjoint, every $n$ in their union lies in exactly one $E_j$ , so the double series $\sum_j \sum_{n \in E_j} n^{-2}$ is a rearrangement of $\sum_{n \in \bigcup_j E_j} n^{-2}$ into blocks; non-negative terms may be regrouped freely, so (2) holds.

It is finite. $\mu(\mathbb{N}) = \sum_{n \ge 1} n^{-2} = \pi^{2}/6 = 1.644934\ldots$ , so $\mu$ is a finite measure (and $\tfrac{6}{\pi^{2}}\mu$ is a probability measure on $\mathbb{N}$ ).

Continuity from below, checked numerically. Let $A_k = \{1, 2, \dots, k\}$ , an increasing sequence with $\bigcup_k A_k = \mathbb{N}$ . Then

\mu(A_{10}) = \sum_{n=1}^{10}\frac{1}{n^{2}} = \frac{1968329}{1270080} = 1.5497677\ldots, \qquad \mu(A_{100}) = 1.6349839\ldots,

climbing to $\mu(\mathbb{N}) = \pi^{2}/6 = 1.644934\ldots$ , as rule (c) requires.

Continuity from above, checked numerically. Let $E_k = \{n \in \mathbb{N} : n \ge k\}$ , a decreasing sequence with $\bigcap_{k \ge 1} E_k = \varnothing$ . Here $\mu(E_1) = \pi^{2}/6 < \infty$ , so rule (d) applies and predicts $\mu(E_k) \to 0$ . Indeed

\mu(E_4) = \frac{\pi^{2}}{6} - \Bigl(1 + \frac14 + \frac19\Bigr) = \frac{\pi^{2}}{6} - \frac{49}{36} = 0.2838229\ldots, \qquad \mu(E_{11}) = 0.0951663\ldots, \qquad \mu(E_{101}) = 0.0099501\ldots,

and the general tail is squeezed between two integrals: comparing $n^{-2}$ with $x^{-2}$ on $[n, n+1]$ and on $[n-1,n]$ gives

\frac1k = \int_{k}^{\infty}\frac{dx}{x^{2}} \;\le\; \mu(E_k) = \sum_{n \ge k}\frac{1}{n^{2}} \;\le\; \int_{k-1}^{\infty}\frac{dx}{x^{2}} = \frac{1}{k-1} \qquad (k \ge 2),

so $\mu(E_k) \to 0$ . At $k = 11$ the three numbers are $0.0909090\ldots \le 0.0951663\ldots \le 0.1$ .

The same sets, a different measure, and the rule fails. Replace $\mu$ by counting measure $\nu(E) = \#E$ on the same $(\mathbb{N}, \mathcal{P}(\mathbb{N}))$ ; this is the weight $f \equiv 1$ , and it is a measure by the same argument. Now $\nu(E_k) = \infty$ for every $k$ , while $\nu\bigl(\bigcap_k E_k\bigr) = \nu(\varnothing) = 0$ . The hypothesis $\nu(E_1) < \infty$ of (d) is the only thing that has changed, and the conclusion fails by the whole of $\infty$ . Note that $\nu$ is still $\sigma$ -finite, since $\mathbb{N} = \bigcup_k A_k$ with $\nu(A_k) = k < \infty$ : $\sigma$ -finiteness does not rescue (d).

ApplicationOriginal

Probability is measure theory with $\mu(X) = 1$ . An experiment is modelled by a sample space $\Omega$ , a $\sigma$ -field $\mathcal{F}$ of events and a probability measure $\mathbb{P}$ , and Grimmett and Stirzaker (Grimmett & Stirzaker (2020), Def. (1.3.1), p. 5) note that $\mathbb{P}$ is exactly a measure of total mass one. The reason the $\sigma$ -field is part of the data, rather than being taken as all of $\mathcal{P}(\Omega)$ , is the one given in this unit: for an infinite $\Omega$ the power set is too big to carry a reasonable assignment of probabilities. Continuity from below is what licenses the everyday step $\mathbb{P}(\text{eventually } A) = \lim_n \mathbb{P}(A \text{ by time } n)$ , and continuity from above — with its finiteness hypothesis automatically satisfied — licenses the complementary limit. See probability spaces and random variables and distributions.

Mass, charge and intensity distributions. Folland (Folland (1999), §1.1, p. 21) motivates the abstract theory by letting $\mu(E)$ be the mass contained in the region $E$ . A point particle of mass $m$ at $x_0$ is the measure $m\delta_{x_0}$ ; a continuous body of density $\rho$ is $\mu(E) = \int_E \rho$ ; a charged surface in $\mathbb{R}^3$ is a measure that is finite on compacta but assigns zero to every set of zero area. Measure theory is what allows all three to be written as one object and integrated against one formula, rather than as a sum of an integral and a list of point contributions. The additivity axiom is physically the statement that mass is extensive.

The $\sigma$ -field as the specification of what is observable. In a spatial Poisson process (Grimmett & Stirzaker (2020), §6.13, Def. (6.13.1), p. 319) one takes the Borel $\sigma$ -field $\mathcal{B}^d$ of $\mathbb{R}^d$ — generated by the boxes $\prod_{i}(a_i, b_i]$ — and asks only for the counts $N(A)$ of points in Borel sets $A$ , with $N(A)$ Poisson of mean $\lambda |A|$ . The choice of $\sigma$ -field is the choice of which questions the model is willing to answer, and the generators are the questions one actually asks in the laboratory: how many counts in this bin.

"Almost everywhere" is why function spaces are complete. Null sets are the sets physics is entitled to ignore: two wavefunctions differing on a null set have identical expectation values. Identifying functions that agree a.e. is what turns the Lebesgue integral into the complete spaces $L^p$ of L^p spaces, and completeness is in turn what makes variational and spectral arguments work. Every statement in the convergence theorems of this chapter — monotone convergence, dominated convergence — is stated a.e. precisely because countable unions of null sets are null, so countably many exceptional sets may be discarded at once.

Theorem

Lebesgue measure

Adapted

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Builds Lebesgue measure from the length of an interval by covering sets from outside and keeping the sets that split every test set additively, and records what the result does and does not do — complete, regular, translation invariant, and necessarily undefined on some subsets of the line.

Prerequisites: Sigma-algebras and measures, Sets, functions and countability, Zorn's lemma and the axiom of choice, Compactness

Definition(Outer measure, Caratheodory measurability, Lebesgue outer measure)AdaptedAdapted from Folland (1999), §1.4, pp. 28–29 and §1.5, p. 33; Reed & Simon (1980), §I.3, p. 15

Let $X$ be a non-empty set.

Outer measure. A map $\mu^{*} : \mathcal{P}(X) \to [0,\infty]$ is an outer measure if $\mu^{*}(\varnothing) = 0$ , it is monotone ( $A \subseteq B \Rightarrow \mu^{*}(A) \le \mu^{*}(B)$ ), and it is countably subadditive. It is defined on every subset; the price is that additivity is weakened to subadditivity.

Caratheodory measurability. A set $A \subseteq X$ is $\mu^{*}$ -measurable if it splits every test set additively:

\mu^{*}(E) \;=\; \mu^{*}(E \cap A) + \mu^{*}(E \cap A^{c}) \qquad \text{for every } E \subseteq X .

Subadditivity already gives " $\le$ ", so only " $\ge$ " has to be checked, and only for test sets with $\mu^{*}(E) < \infty$ .

Lebesgue outer measure on the line. Call $(a,b]$ , $(a,\infty)$ and $\varnothing$ (with $-\infty \le a < b < \infty$ ) the h-intervals. Finite disjoint unions of h-intervals form an algebra $\mathcal{A}$ whose generated $\sigma$ -algebra is $\mathcal{B}_{\mathbb{R}}$ , since the h-intervals generate $\mathcal{B}_{\mathbb{R}}$ by the generator list. Set $\mu_0\bigl(\bigsqcup_{j \le n}(a_j,b_j]\bigr) = \sum_{j \le n}(b_j - a_j)$ on $\mathcal{A}$ and define, for every $A \subseteq \mathbb{R}$ ,

m^{*}(A) \;=\; \inf\Bigl\{\; \sum_{j \ge 1}(b_j - a_j) \;:\; A \subseteq \bigcup_{j \ge 1}(a_j, b_j] \;\Bigr\},

the infimum over all countable coverings of $A$ by h-intervals. The Proposition below builds its outer measure from a covering family that is required to contain $X = \mathbb{R}$ itself, and $\mathbb{R} = (-\infty,\infty)$ is an h-interval, so the h-intervals qualify; allowing the unbounded h-intervals into a cover never lowers the infimum, since any cover using one of them has total price $+\infty$ , so the displayed infimum over bounded h-intervals is the same number as the Proposition’s $\mu^{*}$ . Lebesgue measurable sets are the $m^{*}$ -measurable ones; they form a $\sigma$ -algebra $\mathcal{L}$ , and Lebesgue measure is $m = m^{*}\big|_{\mathcal{L}}$ . Reed and Simon (Reed & Simon (1980), §I.3, p. 15) define the same number by taking the infimum over coverings by open sets instead; the two agree, by Lemma 1.17 of Folland (Folland (1999), §1.5, p. 36), and the proof of outer regularity below contains the argument.

Premeasure. More generally, a map $\mu_0 : \mathcal{A} \to [0,\infty]$ on an algebra $\mathcal{A}$ is a premeasure if $\mu_0(\varnothing) = 0$ and $\mu_0\bigl(\bigcup_j A_j\bigr) = \sum_j \mu_0(A_j)$ whenever the $A_j \in \mathcal{A}$ are disjoint and their union happens to lie in $\mathcal{A}$ . A premeasure is finitely additive but is not yet a measure, because its domain is only an algebra.

Proposition(Covering produces an outer measure)AdaptedAdapted from Folland (1999), Prop. 1.10, p. 29

Data. A family $\mathcal{E}$ of subsets of $X$ with $\varnothing \in \mathcal{E}$ and $X \in \mathcal{E}$ , and a price list $\rho : \mathcal{E} \to [0,\infty]$ that charges nothing for the empty set, $\rho(\varnothing) = 0$ . Nothing else is assumed: $\rho$ need not be monotone, additive or finite.

Conclusion. Charge every $A \subseteq X$ the cheapest total price of a countable cover drawn from $\mathcal{E}$ :

\mu^{*}(A) \;=\; \inf\Bigl\{\sum_{j \ge 1}\rho(E_j) \;:\; E_j \in \mathcal{E},\ A \subseteq \bigcup_{j \ge 1}E_j\Bigr\} .

The infimum is over a non-empty set of covers, because $E_j = X$ for all $j$ always qualifies, and the resulting $\mu^{*}$ is an outer measure on $X$ . Taking $X = \mathbb{R}$ , for $\mathcal{E}$ the h-intervals and for $\rho$ their length returns the $m^{*}$ defined above.

Theorem(Caratheodory's theorem)AdaptedAdapted from Folland (1999), Thm 1.11, pp. 29–30

Fix any outer measure $\mu^{*}$ on $X$ , and write $\mathcal{M}$ for the family of all subsets that pass the splitting test above. Two things then happen at once, with no further hypotheses:

$\mathcal{M}$ is closed under complements and under countable unions, so it is a $\sigma$ -algebra on $X$ ;
cut down to $\mathcal{M}$ , the set function $\mu^{*}$ is countably additive, and every set of outer measure zero belongs to $\mathcal{M}$ — so $\mu^{*}\big|_{\mathcal{M}}$ is a complete measure on $(X,\mathcal{M})$ .

Theorem(Extension from an algebra)AdaptedAdapted from Folland (1999), Prop. 1.13 and Thm 1.14, pp. 31–32

Let $\mathcal{A} \subseteq \mathcal{P}(X)$ be an algebra, $\mu_0$ a premeasure on $\mathcal{A}$ , $\mu^{*}$ the outer measure it induces by the proposition above (with $\mathcal{E} = \mathcal{A}$ , $\rho = \mu_0$ ), and $\mathcal{M} = \mathcal{M}(\mathcal{A})$ the generated $\sigma$ -algebra. Then

$\mu^{*}$ agrees with $\mu_0$ on $\mathcal{A}$ , and every member of $\mathcal{A}$ is $\mu^{*}$ -measurable;
consequently $\mu = \mu^{*}\big|_{\mathcal{M}}$ is a measure on $\mathcal{M}$ extending $\mu_0$ ;
if $\nu$ is any other measure on $\mathcal{M}$ extending $\mu_0$ then $\nu(E) \le \mu(E)$ for all $E \in \mathcal{M}$ , with equality whenever $\mu(E) < \infty$ ;
if $\mu_0$ is $\sigma$ -finite, the extension $\mu$ is the only one.

Theorem(Lebesgue measure)AdaptedAdapted from Folland (1999), §1.5, Thm 1.16, p. 35, Thm 1.18, p. 36, Thm 1.19, p. 36 and Thm 1.21, p. 37; Reed & Simon (1980), Thm I.3, p. 15; Evans (2010), Appendix E, Thm 1, pp. 684–685

There is a complete measure $m$ on a $\sigma$ -algebra $\mathcal{L}$ of subsets of $\mathbb{R}$ with the following properties.

a. It measures intervals by length. $\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{L}$ and $m\bigl((a,b]\bigr) = b - a$ for all $a < b$ ; consequently $m$ of any interval, open, closed or half-open, is its length, and $m(\{x\}) = 0$ . b. It is the only such Borel measure. $m$ is $\sigma$ -finite, and it is the unique measure on $\mathcal{B}_{\mathbb{R}}$ assigning to each $(a,b]$ the number $b-a$ ; $\mathcal{L}$ and $m$ on it are the completion of that Borel measure. c. Outer and inner regularity. For every $E \in \mathcal{L}$ ,

m(E) = \inf\{m(U) : U \supseteq E,\ U \text{ open}\} = \sup\{m(K) : K \subseteq E,\ K \text{ compact}\} .

d. Structure modulo null sets. $E \in \mathcal{L}$ if and only if $E = V \setminus N_1$ with $V$ a $G_\delta$ set and $m(N_1) = 0$ , if and only if $E = H \cup N_2$ with $H$ an $F_\sigma$ set and $m(N_2) = 0$ . e. Translations and dilations. If $E \in \mathcal{L}$ and $s, r \in \mathbb{R}$ then $E + s \in \mathcal{L}$ and $rE \in \mathcal{L}$ , with $m(E+s) = m(E)$ and $m(rE) = \lvert r \rvert\, m(E)$ .

In $\mathbb{R}^n$ the corresponding statement is Theorem 1 of Evans (Evans (2010), Appendix E.1, pp. 684–685): there are a $\sigma$ -algebra $\mathcal{M}$ containing all open and all closed subsets of $\mathbb{R}^n$ and a map $\lvert\,\cdot\,\rvert : \mathcal{M} \to [0,+\infty]$ that gives each ball its $n$ -dimensional volume, is countably additive, and is complete in the sense that every subset of a set of measure zero is again measurable with measure zero.

Theorem(Not every set can be measured)AdaptedAdapted from Folland (1999), §1.1, pp. 19–20

There is no map $\mu : \mathcal{P}(\mathbb{R}) \to [0,\infty]$ that is simultaneously countably additive, invariant under translations, and normalised by $\mu\bigl([0,1)\bigr) = 1$ .

Since Lebesgue measure is countably additive, translation invariant and gives $[0,1)$ the value $1$ , it follows that $\mathcal{L} \ne \mathcal{P}(\mathbb{R})$ : the set $N$ constructed in the proof is not Lebesgue measurable.

How this compares with the source. Folland (Folland (1999), §1.1, p. 19) asks of a hypothetical $\mu$ on $\mathcal{P}(\mathbb{R}^n)$ the three properties: countable additivity, invariance under all congruences (translations, rotations and reflections), and $\mu(Q) = 1$ for the unit cube $Q = \{x : 0 \le x_j \le 1\}$ , and shows that no such $\mu$ exists. On the line the argument needs only translations and only the normalisation of $[0,1)$ , so the version stated here has the weaker hypotheses; it is what the proof below delivers, and it implies Folland’s one-dimensional case.

Idea of the proofOriginal

The construction has three moves, and each one trades something away.

First, give every set a size by covering it from outside with intervals and taking the cheapest cover. This always works, but the result $m^{*}$ is only subadditive: two disjoint sets can be covered jointly more cheaply than separately when they are badly interleaved.

Second, throw away the sets that cause the trouble. Carathéodory's criterion is a sharp way of saying " $A$ has a clean boundary as far as $m^{*}$ can tell": $A$ must cut every test set — not just the nice ones — into two pieces whose outer measures add up exactly. The surprise is that this one demand, which mentions no topology and no intervals, already forces the surviving sets to form a $\sigma$ -algebra on which $m^{*}$ is countably additive. The reason is that the criterion is an identity, so it can be applied repeatedly to ever-finer test sets, and a short induction turns it into exact finite additivity on arbitrary test sets; countable additivity then follows by letting the induction run and using subadditivity in the other direction.

Third, check that nothing was lost: the h-intervals themselves survive the criterion, so the surviving $\sigma$ -algebra contains all Borel sets, and the cheapest cover of an interval is the interval itself, so lengths come out right.

That something had to be thrown away is the content of the last statement. If every set had a translation-invariant size, then cutting $[0,1)$ into countably many congruent copies of one set — which the axiom of choice permits — would make $1$ equal to a sum of countably many equal numbers, and no number has that property.

Proof(Covering produces an outer measure)AdaptedAdapted from Folland (1999), Prop. 1.10, p. 29

$\mu^{*}(\varnothing) = 0$ . Cover $\varnothing$ by the cover whose every member is $\varnothing$ ; its total price is $0$ , and prices are non-negative.

Monotonicity. If $A \subseteq B$ , every cover of $B$ is also a cover of $A$ . The infimum defining $\mu^{*}(A)$ therefore ranges over a set of numbers that includes all those competing in the infimum for $\mu^{*}(B)$ , so it cannot be larger.

Countable subadditivity. Let $A_1, A_2, \dots \subseteq X$ and let $A = \bigcup_j A_j$ . If some $\mu^{*}(A_j) = \infty$ the claimed inequality is trivial, so assume all are finite and fix $\varepsilon > 0$ . For each $j$ the infimum lets us pick a cover $\{E^{j}_{k}\}_{k \ge 1} \subseteq \mathcal{E}$ of $A_j$ whose total price is below $\mu^{*}(A_j) + \varepsilon 2^{-j}$ . The doubly indexed family $\{E^{j}_{k}\}_{j,k \ge 1}$ is countable, being indexed by $\mathbb{N}\times\mathbb{N}$ , and it covers $A$ . Hence

\mu^{*}(A) \;\le\; \sum_{j,k \ge 1}\rho\bigl(E^{j}_{k}\bigr) \;=\; \sum_{j \ge 1}\ \sum_{k \ge 1}\rho\bigl(E^{j}_{k}\bigr) \;<\; \sum_{j \ge 1}\mu^{*}(A_j) + \varepsilon,

the regrouping of the double series being legitimate because all terms are non-negative. Since $\varepsilon$ was arbitrary, $\mu^{*}(A) \le \sum_j \mu^{*}(A_j)$ . $\square$

Proof(Caratheodory's theorem)AdaptedAdapted from Folland (1999), Thm 1.11, pp. 29–30

Throughout, $E$ denotes an arbitrary test set and we use that " $\le$ " in the measurability identity is automatic.

Step 1: $\mathcal{M}$ contains $\varnothing$ and is closed under complements. The identity is unchanged when $A$ is swapped with $A^{c}$ , so $A \in \mathcal{M} \Leftrightarrow A^{c} \in \mathcal{M}$ . For $A = \varnothing$ it reads $\mu^{*}(E) = \mu^{*}(\varnothing) + \mu^{*}(E)$ , true since $\mu^{*}(\varnothing) = 0$ .

Step 2: $\mathcal{M}$ is an algebra. Let $A, B \in \mathcal{M}$ . Test $A$ on $E$ , then test $B$ on each of the two resulting pieces:

\mu^{*}(E) = \mu^{*}(E \cap A \cap B) + \mu^{*}(E \cap A \cap B^{c}) + \mu^{*}(E \cap A^{c} \cap B) + \mu^{*}(E \cap A^{c} \cap B^{c}).

The set $A \cup B$ is the union of the three sets $A \cap B$ , $A \cap B^{c}$ , $A^{c} \cap B$ , so $E \cap (A\cup B)$ is the union of the three sets appearing in the first three terms; subadditivity bounds the sum of those three terms below by $\mu^{*}\bigl(E \cap (A\cup B)\bigr)$ . The fourth term is $\mu^{*}\bigl(E \cap (A \cup B)^{c}\bigr)$ . Hence $\mu^{*}(E) \ge \mu^{*}(E \cap (A\cup B)) + \mu^{*}(E \cap (A\cup B)^{c})$ , so $A \cup B \in \mathcal{M}$ .

Step 3: $\mu^{*}$ is finitely additive on $\mathcal{M}$ . If $A, B \in \mathcal{M}$ are disjoint, use $E = A \cup B$ as the test set for $A$ : $\mu^{*}(A \cup B) = \mu^{*}\bigl((A\cup B)\cap A\bigr) + \mu^{*}\bigl((A \cup B) \cap A^{c}\bigr) = \mu^{*}(A) + \mu^{*}(B)$ .

Step 4: additivity on an arbitrary test set. Let $A_1, A_2, \dots \in \mathcal{M}$ be disjoint, $B_n = A_1 \cup \dots \cup A_n$ (which lies in $\mathcal{M}$ by step 2) and $B = \bigcup_{j \ge 1} A_j$ . Testing the measurability of $A_n$ on the set $E \cap B_n$ , and using $B_n \cap A_n = A_n$ and $B_n \cap A_n^{c} = B_{n-1}$ (disjointness),

\mu^{*}(E \cap B_n) = \mu^{*}(E \cap A_n) + \mu^{*}(E \cap B_{n-1}),

so by induction $\mu^{*}(E \cap B_n) = \sum_{j=1}^{n}\mu^{*}(E \cap A_j)$ .

Step 5: $B \in \mathcal{M}$ and countable additivity. Test $B_n$ on $E$ and use $B_n^{c} \supseteq B^{c}$ with monotonicity:

\mu^{*}(E) = \mu^{*}(E \cap B_n) + \mu^{*}(E \cap B_n^{c}) \;\ge\; \sum_{j=1}^{n}\mu^{*}(E \cap A_j) + \mu^{*}(E \cap B^{c}).

Letting $n \to \infty$ and then applying countable subadditivity to $E \cap B = \bigcup_j (E \cap A_j)$ ,

\mu^{*}(E) \;\ge\; \sum_{j \ge 1}\mu^{*}(E \cap A_j) + \mu^{*}(E \cap B^{c}) \;\ge\; \mu^{*}(E \cap B) + \mu^{*}(E \cap B^{c}) \;\ge\; \mu^{*}(E),

the last step being subadditivity again. Every inequality is therefore an equality. The outer two give $B \in \mathcal{M}$ ; choosing $E = B$ in the middle one gives $\mu^{*}(B) = \sum_{j}\mu^{*}(A_j)$ .

Step 6: $\mathcal{M}$ is a $\sigma$ -algebra. It is an algebra closed under countable disjoint unions by step 5. An arbitrary sequence is disjointified as in the disjointification device, and the replacement sets $F_k = E_k \setminus (E_1 \cup \dots \cup E_{k-1})$ lie in $\mathcal{M}$ because an algebra is closed under differences. So $\mathcal{M}$ is a $\sigma$ -algebra and $\mu^{*}\big|_{\mathcal{M}}$ is a measure.

Step 7: completeness. Suppose $\mu^{*}(A) = 0$ and let $E$ be any test set. Monotonicity gives $\mu^{*}(E \cap A) \le \mu^{*}(A) = 0$ and $\mu^{*}(E \cap A^{c}) \le \mu^{*}(E)$ , so the two pieces add up to at most $\mu^{*}(E)$ and $A \in \mathcal{M}$ . Thus every set of outer measure zero — in particular every subset of a null set of $\mu^{*}\big|_{\mathcal{M}}$ — is measurable, which is completeness. $\square$

Proof(Lebesgue measure: the five properties)AdaptedAdapted from Folland (1999), Prop. 1.15, p. 33, Thm 1.16, p. 35, Lemma 1.17 and Thm 1.18, p. 36, Thm 1.19, pp. 36–37 and Thm 1.21, p. 37

(a), (b): existence and uniqueness. That $\mu_0$ is well defined and finitely additive on the algebra $\mathcal{A}$ of finite disjoint unions of h-intervals is a bookkeeping check on how two such representations refine each other. That it is a premeasure — countably additive whenever the countable disjoint union stays in $\mathcal{A}$ — is the one place compactness enters: the inequality $\mu_0(I) \le \sum_j \mu_0(I_j)$ for $I = \bigsqcup_j I_j$ is proved by shrinking $I = (a,b]$ slightly to a compact $[a+\delta, b]$ , enlarging each $I_j = (a_j,b_j]$ slightly to an open $(a_j, b_j + \delta_j)$ with $\sum_j \delta_j < \varepsilon$ , extracting a finite subcover and summing the finitely many lengths; the reverse inequality is finite additivity plus monotonicity. This is Proposition 1.15 of Folland (Folland (1999), §1.5, pp. 33–34), whose argument we follow.

Applying the extension theorem to this premeasure gives a measure on $\mathcal{M}(\mathcal{A}) = \mathcal{B}_{\mathbb{R}}$ with $m((a,b]) = b-a$ , and $\mathbb{R} = \bigcup_{j \in \mathbb{Z}}(j, j+1]$ with each piece of measure $1$ shows $\mu_0$ is $\sigma$ -finite, so clause 4 of that theorem makes the extension unique. Carathéodory's theorem gives at the same time the complete measure $m$ on the larger $\sigma$ -algebra $\mathcal{L}$ of $m^{*}$ -measurable sets, and $\mathcal{L}$ is the completion of $\mathcal{B}_{\mathbb{R}}$ by (d) below. Finally $m(\{x\}) = \lim_{n} m\bigl((x - \tfrac1n, x]\bigr) = \lim_n \tfrac1n = 0$ by continuity from above, and the measures of $(a,b)$ , $[a,b]$ , $[a,b)$ then all equal $b-a$ because they differ from $(a,b]$ by one or two points.

(c): outer regularity. " $\le$ " is monotonicity, since any open $U \supseteq E$ has $m(U) \ge m(E)$ . For " $\ge$ ", assume $m(E) < \infty$ (otherwise the right-hand side is $\infty$ too and there is nothing to prove) and let $\varepsilon > 0$ . By the definition of $m^{*}$ there are h-intervals $(a_j,b_j]$ covering $E$ with $\sum_j (b_j - a_j) < m(E) + \varepsilon$ . Replace each by the open interval $(a_j,\, b_j + \varepsilon 2^{-j})$ ; the union $U$ of these is open, contains $E$ , and by countable subadditivity

m(U) \;\le\; \sum_{j \ge 1}\Bigl(b_j - a_j + \varepsilon 2^{-j}\Bigr) \;<\; m(E) + 2\varepsilon .

Since $\varepsilon$ was arbitrary, the infimum over open $U$ is $m(E)$ . The same computation proves that covering by open sets gives the same outer measure as covering by h-intervals, which is the equivalence with the definition of Reed and Simon (Reed & Simon (1980), §I.3, p. 15).

Inner regularity. Suppose first $E$ is bounded, so $\overline{E}$ is compact. Given $\varepsilon > 0$ , apply outer regularity to the set $\overline{E}\setminus E$ (of finite measure, since $\overline{E}$ is bounded) to get an open $U \supseteq \overline{E}\setminus E$ with $m(U) < m(\overline{E}\setminus E) + \varepsilon$ , and put $K = \overline{E}\setminus U$ . Then $K$ is closed and bounded, hence compact by the Heine–Borel theorem; $K \subseteq E$ because every point of $\overline{E}$ outside $E$ lies in $U$ ; and

m(K) = m(\overline{E}) - m(\overline{E} \cap U) \ge m(\overline{E}) - m(U) > m(\overline{E}) - m(\overline{E}\setminus E) - \varepsilon = m(E) - \varepsilon .

For unbounded $E$ , split $E$ into the pieces $E \cap (j, j+1]$ , approximate the $j$ th from inside to within $\varepsilon 2^{-\lvert j\rvert}$ and take finite unions; $m(E)$ is the limit of the measures of these finite unions by continuity from below. This is Theorem 1.18 of Folland (Folland (1999), §1.5, p. 36).

(d): structure. Let $E \in \mathcal{L}$ with $m(E) < \infty$ . By outer regularity choose open $U_j \supseteq E$ with $m(U_j) < m(E) + 1/j$ and set $V = \bigcap_j U_j$ , a $G_\delta$ set containing $E$ with $m(E) \le m(V) \le \inf_j m(U_j) = m(E)$ . Since $m(E)$ is finite, $m(V \setminus E) = m(V) - m(E) = 0$ , so $E = V \setminus N_1$ with $N_1 = V\setminus E$ null. The case $m(E) = \infty$ needs more care, because a countable union of $G_\delta$ sets need not be $G_\delta$ : the open sets have to be united before the intersection over tolerances is taken, not after. Cut the line into the unit pieces $E_j = E \cap (j, j+1]$ , $j \in \mathbb{Z}$ , each of finite measure. Fix $k \ge 1$ . Outer regularity, already proved, gives for each $j$ an open $U_{j,k} \supseteq E_j$ with $m(U_{j,k}) < m(E_j) + 2^{-\lvert j\rvert}/k$ , and since $m(E_j) < \infty$ this says $m(U_{j,k}\setminus E_j) < 2^{-\lvert j\rvert}/k$ . Put $U_k = \bigcup_{j \in \mathbb{Z}} U_{j,k}$ , an open set containing $E$ . Each $E_j \subseteq E$ , so $U_{j,k}\setminus E \subseteq U_{j,k}\setminus E_j$ , and countable subadditivity gives

m\bigl(U_k \setminus E\bigr) \;\le\; \sum_{j \in \mathbb{Z}} m\bigl(U_{j,k}\setminus E_j\bigr) \;<\; \frac{1}{k}\sum_{j \in \mathbb{Z}} 2^{-\lvert j\rvert} \;=\; \frac{3}{k} .

Now $V = \bigcap_{k \ge 1} U_k$ is a $G_\delta$ set containing $E$ , and $V \setminus E \subseteq U_k \setminus E$ for every $k$ , so $m(V\setminus E) \le 3/k$ for every $k$ and therefore $m(V \setminus E) = 0$ . Again $E = V \setminus N_1$ with $N_1 = V \setminus E$ null. (Folland (Folland (1999), Exercise 25, p. 39) leaves this case to the reader; the argument just given is ours.) Applying the result to $E^{c}$ and complementing turns the $G_\delta$ description into the $F_\sigma$ one. Conversely, both descriptions produce members of $\mathcal{L}$ , because $\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{L}$ and $m$ is complete. In particular every Lebesgue measurable set differs from a Borel set by a null set, so $\mathcal{L}$ is exactly the completion of $\mathcal{B}_{\mathbb{R}}$ .

(e): translations and dilations. Fix $s$ and define $m_s(E) = m(E+s)$ for $E \in \mathcal{B}_{\mathbb{R}}$ ; this is well defined because translation maps open intervals to open intervals and therefore maps $\mathcal{B}_{\mathbb{R}}$ onto itself, and it is a measure because translation preserves disjointness and countable unions. On an h-interval, $m_s((a,b]) = (b+s)-(a+s) = b - a = m((a,b])$ , so $m_s$ and $m$ are two measures on $\mathcal{B}_{\mathbb{R}}$ extending the same $\sigma$ -finite premeasure; by the uniqueness clause they coincide. The same argument applies to $m_r(E) = m(rE)$ with one extra step: on an h-interval $m_r((a,b]) = \lvert r\rvert(b-a)$ for $r \ne 0$ , which is $\lvert r\rvert \mu_0$ and not $\mu_0$ , so it is the measure $\lvert r\rvert^{-1} m_r$ that extends the premeasure $\mu_0$ and to which the uniqueness clause applies. It gives $\lvert r\rvert^{-1} m_r = m$ on $\mathcal{B}_{\mathbb{R}}$ , that is, $m(rE) = \lvert r \rvert m(E)$ for $r \ne 0$ ; for $r = 0$ both sides vanish, since $0 \cdot E$ is $\{0\}$ or $\varnothing$ and a single point is null, while $\lvert 0\rvert\, m(E) = 0$ under the convention $0 \cdot \infty = 0$ . Finally, a set of Lebesgue measure zero is carried to a set of Lebesgue measure zero, so by (d) the whole of $\mathcal{L}$ is preserved. $\square$

Proof(A set that cannot be measured)AdaptedAdapted from Folland (1999), §1.1, pp. 19–20

Step 1: the choice set. On $[0,1)$ declare $x \sim y$ when $x - y \in \mathbb{Q}$ . This is an equivalence relation. Using the axiom of choice, let $N \subseteq [0,1)$ contain exactly one point of each equivalence class.

Step 2: countably many rotated copies tile $[0,1)$ . Let $R = \mathbb{Q} \cap [0,1)$ , a countable set by countability of the rationals. For $r \in R$ put

N_r \;=\; \bigl(N \cap [0, 1-r)\bigr) + r \;\;\cup\;\; \bigl(N \cap [1-r, 1)\bigr) + r - 1 ,

that is, $N$ shifted right by $r$ with the overhanging part brought back by one unit; $N_r \subseteq [0,1)$ . Every $x \in [0,1)$ lies in exactly one $N_r$ . Existence: let $y \in N$ be the chosen representative of the class of $x$ ; then $x - y \in \mathbb{Q}$ , and $r = x-y$ if $x \ge y$ , $r = x-y+1$ if $x < y$ , lies in $R$ and puts $x$ in $N_r$ . Uniqueness: if $x \in N_r \cap N_s$ then $x$ comes from a point $y \in N$ with $x - y \in \{r, r-1\}$ and also from a point $z \in N$ with $x - z \in \{s, s-1\}$ ; then $y - z \in \mathbb{Q}$ , so $y \sim z$ , so $y = z$ because $N$ meets each class once, and then $r - s \in \{0, \pm 1\} \cap (-1,1) = \{0\}$ .

Step 3: the contradiction. Suppose $\mu$ on $\mathcal{P}(\mathbb{R})$ is countably additive, translation invariant, and $\mu([0,1)) = 1$ . The two pieces of $N$ in step 2 are disjoint and their translates are disjoint, so finite additivity and translation invariance give

\mu(N_r) = \mu\bigl(N \cap [0,1-r)\bigr) + \mu\bigl(N \cap [1-r,1)\bigr) = \mu(N) \qquad\text{for every } r \in R .

By step 2 and countable additivity over the countable index set $R$ ,

1 \;=\; \mu\bigl([0,1)\bigr) \;=\; \sum_{r \in R}\mu(N_r) \;=\; \sum_{r \in R}\mu(N).

A sum of countably many copies of one number in $[0,\infty]$ is $0$ if that number is $0$ and $+\infty$ otherwise. Neither is $1$ , so no such $\mu$ exists.

Step 4: consequence for $m$ . Lebesgue measure is countably additive on $\mathcal{L}$ , satisfies $m(E+s) = m(E)$ by part (e), and $m([0,1)) = 1$ by part (a). If $N$ were in $\mathcal{L}$ , then each $N_r$ would be too (translates and unions of measurable sets), and step 3 would run verbatim with $\mu = m$ . Hence $N \notin \mathcal{L}$ . $\square$

Remark(Where the hypotheses are used)Original

The axiom of choice is used once, in step 1 of the last proof, and it cannot be avoided: there are models of set theory without choice in which every subset of $\mathbb{R}$ is Lebesgue measurable. So "non-measurable set" is not an object one ever writes down; it is an object whose existence choice asserts. Reed and Simon (Reed & Simon (1980), §I.3, p. 14) motivate restricting the domain by the same kind of phenomenon in higher dimension, the Banach–Tarski paradox, and Folland (Folland (1999), §1.1, p. 20) records that in $\mathbb{R}^n$ with $n \ge 3$ even finite additivity is already inconsistent with congruence invariance and normalisation.
Countable additivity is the hypothesis that breaks. In dimensions $1$ and $2$ there do exist finitely additive, translation-invariant extensions of Lebesgue measure to all subsets; what step 3 rules out is the countable version. Dropping countable additivity, however, destroys every convergence theorem of this chapter, which is why we keep it and give up universality of the domain instead.
Where $\sigma$ -finiteness is used. Only in clause 4 of the extension theorem, and that is exactly the clause invoked to prove translation invariance in (e). Without it the extension of a premeasure is merely the largest one, and two measures agreeing on the generating algebra may differ.
Where finiteness of $m(E)$ is used. In the structure theorem (d), the step $m(V\setminus E) = m(V) - m(E)$ is a subtraction and needs $m(E) < \infty$ ; for a general $E$ one must first cut the line into unit pieces, and then assemble the pieces in the right order — open hulls united over the pieces at each tolerance $1/k$ , and only the resulting open sets intersected over $k$ — because the class of $G_\delta$ sets is closed under countable intersections but not under countable unions. The same finiteness is why the inner-regularity proof treats bounded sets first.
Carathéodory's criterion quantifies over all test sets. Restricting it to, say, open test sets would not do: the point of the criterion is that it is strong enough to force countable additivity with no help from the topology, which is why the same theorem builds Hausdorff measures and product measures. The proof uses the full strength twice — in step 2 with $E$ arbitrary, and in step 5 with the test set $E \cap B_n$ .
Regularity is a property of this particular outer measure, not of Carathéodory's theorem. It comes from the fact that the covering family consists of intervals, which can be enlarged to open sets at arbitrarily small cost. An abstract $\mu^{*}$ has no such property.
"Null" and "small" are different in three different senses. A countable set is null but can be dense ( $\mathbb{Q}$ ); an open dense subset of $[0,1]$ can have measure as small as we like; and, as the example below shows, a compact nowhere dense set can have measure as close to $1$ as we like. Folland (Folland (1999), §1.5, p. 38) makes the same observation with a set built from an enumeration of the rationals.

Method(Computing or bounding the Lebesgue measure of a set)Original

Upper bound: cover it. Exhibit a countable family of intervals whose union contains the set; $m^{*}$ is below the total length, by definition. This alone proves many sets null — for a countable set, put an interval of length $\varepsilon 2^{-j}$ around the $j$ th point and conclude $m \le \varepsilon$ for every $\varepsilon$ .
Lower bound: fill it. Exhibit a compact (or just measurable) subset of known measure; inner regularity (c) says nothing better is being missed.
Build it as a monotone limit. If the set is defined by an infinite process — remove, subdivide, remove again — write it as $\bigcap_n K_n$ with $K_n$ decreasing and $m(K_0) < \infty$ , then use continuity from above: $m(K) = \lim_n m(K_n)$ . Verify the finiteness hypothesis; for a construction inside $[0,1]$ it is automatic.
Use the transformation rules before computing. An affine image costs only a factor: $m(rE + s) = \lvert r\rvert m(E)$ by (e). Reduce to a standard set first.
Decide measurability structurally, not by the Carathéodory criterion. In practice one never verifies the splitting identity: one shows the set is Borel (built from intervals by countably many unions, intersections and complements), or that it differs from a Borel set by a subset of a null set, which suffices by (d) and completeness.
If the answer must be a number, check it two ways — once by the covering bound of step 1 and once by the filling bound of step 2 — and confirm they meet.

Example(A compact, nowhere dense subset of [0,1] of measure exactly 1/2)Original

We build a "fat Cantor set". Start with $K_0 = [0,1]$ . Having obtained $K_{n-1}$ as a disjoint union of $2^{\,n-1}$ closed intervals of common length $\ell_{n-1}$ , delete from each of them the open middle interval of length $4^{-n}$ ; what remains is $K_n$ , a disjoint union of $2^{n}$ closed intervals of common length

\ell_n = \frac{\ell_{n-1} - 4^{-n}}{2}, \qquad \ell_0 = 1 .

Put $K = \bigcap_{n \ge 0} K_n$ . (The middle-thirds Cantor set is the same construction with $4^{-n}$ replaced by $\ell_{n-1}/3$ .)

Step 1: the construction never runs out of room. We claim $\ell_n = \dfrac{2^{n}+1}{2^{\,2n+1}}$ . It holds at $n=0$ , where it reads $\ell_0 = 2/2 = 1$ . If it holds at $n-1$ then

\ell_{n-1} - 4^{-n} = \frac{2^{\,n-1}+1}{2^{\,2n-1}} - \frac{1}{2^{\,2n}} = \frac{2\bigl(2^{\,n-1}+1\bigr) - 1}{2^{\,2n}} = \frac{2^{n}+1}{2^{\,2n}},

and halving gives the claim at $n$ . In particular $\ell_n > 4^{-(n+1)} > 0$ , so each deleted middle interval genuinely fits inside its host interval and the construction is legitimate. Numerically $\ell_1 = 3/8$ , $\ell_2 = 5/32 = 0.15625$ , $\ell_3 = 9/128 = 0.0703125$ .

Step 2: the measure of each stage. $K_n$ is a disjoint union of $2^{n}$ intervals of length $\ell_n$ , so by finite additivity and part (a),

m(K_n) = 2^{n}\ell_n = \frac{2^{n}\bigl(2^{n}+1\bigr)}{2^{\,2n+1}} = \frac12 + \frac{1}{2^{\,n+1}} .

Thus $m(K_0) = 1$ , $m(K_1) = 3/4$ , $m(K_2) = 5/8$ , $m(K_3) = 9/16 = 0.5625$ .

Step 3: the measure of the limit. The $K_n$ decrease and $m(K_0) = 1 < \infty$ , so continuity from above applies and

m(K) = \lim_{n \to \infty} m(K_n) = \lim_{n\to\infty}\Bigl(\frac12 + 2^{-(n+1)}\Bigr) = \tfrac12 .

Cross-check by the removed set. At stage $n$ we delete $2^{\,n-1}$ intervals of length $4^{-n}$ , a total length of $2^{\,n-1}4^{-n} = 2^{-(n+1)}$ . The deleted intervals over all stages are pairwise disjoint, so countable additivity gives $m([0,1]\setminus K) = \sum_{n \ge 1}2^{-(n+1)} = \tfrac12$ , and $m(K) = 1 - \tfrac12 = \tfrac12$ . The two computations agree.

Step 4: $K$ is compact and nowhere dense. Each $K_n$ is a finite union of closed intervals, hence closed, so $K$ is closed and bounded, hence compact. If $K$ contained an interval of length $\delta > 0$ , that interval would lie in every $K_n$ and so inside one of its component intervals, forcing $\delta \le \ell_n \to 0$ : impossible. So $K$ has empty interior, and being closed it is nowhere dense. Its complement in $[0,1]$ is therefore open and dense, with measure $1/2$ as well.

What the example shows. A set can be topologically negligible — closed with empty interior, containing no interval at all — and still carry half the length of $[0,1]$ . Outer regularity (c) is not contradicted: every open $U \supseteq K$ has $m(U) \ge 1/2$ , and indeed the sets $K_n$ with their deleted intervals slightly fattened give open supersets of measure as close to $1/2$ as we please. By contrast the middle-thirds Cantor set has $m = \lim_n (2/3)^{n} = 0$ , as Reed and Simon (Reed & Simon (1980), §I.4, p. 21) record, although it is uncountable: measure zero and countability are unrelated.

ApplicationOriginal

The default reference measure. Every integral written $\int f(x)\,dx$ in physics is an integral against Lebesgue measure, and the properties above are the fine print. Translation invariance (e) is why a free particle's probability density may be shifted without changing the normalisation, and the scaling $m(rE) = \lvert r\rvert m(E)$ (with its $\lvert r \rvert^{n}$ analogue in $\mathbb{R}^n$ ) is the Jacobian of a dilation, the simplest case of the change-of-variables formula. Completeness is what lets one say "redefine $f$ on a set of measure zero" without leaving the theory, and it is listed as part of the package by Evans (Evans (2010), Appendix E, Thm 1(iv), p. 685).

Regularity is what makes computation possible. No one evaluates $m(E)$ from the Carathéodory criterion. Part (c) says that any measurable set can be squeezed between an open set and a compact set of nearly the same measure, which is the theoretical licence behind every numerical quadrature on an irregular domain, behind Monte Carlo volume estimation, and behind the approximation of a measurable set by a finite union of intervals or boxes used in mesh generation.

Lebesgue–Stieltjes measures and spectra. The same construction run with a general increasing right-continuous $F$ in place of $F(x) = x$ gives the Lebesgue–Stieltjes measure $\mu_F$ . Reed and Simon (Reed & Simon (1980), §I.4, Examples 1–3, pp. 20–21) give the three archetypes: if $\alpha$ is continuously differentiable then $d\mu_\alpha = \alpha'(x)\,dx$ is a density times Lebesgue measure; if $\alpha$ is the step function at the origin then $\mu_\alpha$ is the Dirac measure, which gives mass $1$ to any set containing $0$ ; and if $\alpha$ is the Cantor function then $\mu_\alpha$ is continuous (no atoms) yet lives entirely on the Cantor set, a set of Lebesgue measure zero. These three are exactly the absolutely continuous, pure point and singular continuous pieces into which the spectral measure of a self-adjoint operator decomposes, and they correspond physically to scattering states, bound states and the exotic intermediate behaviour found in quasiperiodic systems.

Positive measure without interior. The fat Cantor set of the example is not a curiosity. Sets that are nowhere dense yet of positive measure are exactly what appears in Kolmogorov–Arnold–Moser theory, where the invariant tori surviving a small perturbation form such a set: the surviving set has positive measure, so a randomly chosen initial condition is likely to be quasiperiodic, but it contains no open set, so no initial condition is known to be quasiperiodic from a finite-precision measurement. Keeping "measure-theoretically large" and "topologically large" apart is the whole content of that distinction.

Why some things cannot be assigned a probability. The non-measurable set is the reason a probability space carries a $\sigma$ -field as part of its data rather than the full power set. In applications this is not a technicality to be waved away but the statement that the model specifies, in advance, which questions have answers.

Definition

Measurable functions

Adapted

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Defines a measurable function as one whose preimages of Borel sets are measurable, shows that the class survives sums, products, suprema and pointwise limits, and builds every non-negative measurable function as an increasing pointwise limit of simple functions — the staircase on which the Lebesgue integral is erected.

Prerequisites: Sigma-algebras and measures, Lebesgue measure, Limits of sequences and completeness, Continuity and homeomorphism

Definition(Measurable mappings and measurable functions)AdaptedAdapted from Folland (1999), §2.1, pp. 43–46; Reed & Simon (1980), §I.3, p. 15; Evans (2010), Appendix E.2, p. 685

Any map $f : X \to Y$ between sets pulls subsets back: $f^{-1}(E) = \{x \in X : f(x) \in E\}$ . Preimage commutes with every set operation — $f^{-1}(\bigcup_j E_j) = \bigcup_j f^{-1}(E_j)$ , $f^{-1}(E^{c}) = \bigl(f^{-1}(E)\bigr)^{c}$ — and this single fact drives the whole unit.

Measurable mapping. Let $(X,\mathcal{M})$ and $(Y,\mathcal{N})$ be measurable spaces. A map $f : X \to Y$ is $(\mathcal{M},\mathcal{N})$ -measurable if $f^{-1}(E) \in \mathcal{M}$ for every $E \in \mathcal{N}$ . Because preimage composes, $(g \circ f)^{-1}(E) = f^{-1}\bigl(g^{-1}(E)\bigr)$ , a composition of measurable mappings is measurable.

Measurable function. When $Y = \mathbb{R}$ or $\mathbb{C}$ we always take $\mathcal{N}$ to be the Borel $\sigma$ -algebra, and call $f$ $\mathcal{M}$ -measurable, or just measurable. On the line two different $\sigma$ -algebras compete for the domain: $f : \mathbb{R}\to\mathbb{C}$ is Borel measurable if it is $(\mathcal{B}_{\mathbb{R}}, \mathcal{B}_{\mathbb{C}})$ -measurable and Lebesgue measurable if it is $(\mathcal{L}, \mathcal{B}_{\mathbb{C}})$ -measurable, with $\mathcal{L}$ the Lebesgue measurable sets of Lebesgue measure. Since $\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{L}$ , Borel measurable implies Lebesgue measurable but not conversely.

Extended-real-valued functions. It is convenient to allow the values $\pm\infty$ . Put $\overline{\mathbb{R}} = [-\infty,\infty]$ and

\mathcal{B}_{\overline{\mathbb{R}}} = \bigl\{\, E \subseteq \overline{\mathbb{R}} \;:\; E \cap \mathbb{R} \in \mathcal{B}_{\mathbb{R}} \,\bigr\} ,

which is generated by the rays $(a,\infty]$ , $a \in \mathbb{R}$ , and also by the rays $[-\infty,a)$ . A function $f : X \to \overline{\mathbb{R}}$ is measurable if it is $(\mathcal{M}, \mathcal{B}_{\overline{\mathbb{R}}})$ -measurable.

Characteristic and simple functions. For $E \subseteq X$ the characteristic function $\chi_E$ takes the value $1$ on $E$ and $0$ off $E$ ; it is measurable exactly when $E \in \mathcal{M}$ , since its only non-trivial preimages are $E$ , $E^{c}$ , $X$ and $\varnothing$ . A simple function is any $\varphi$ that can be written as $\varphi = \sum_{j=1}^{n} c_j \chi_{A_j}$ for some finite $n$ , some $c_j \in \mathbb{C}$ and some $A_j \in \mathcal{M}$ . Equivalently, $\varphi$ is simple when it is measurable and its range is a finite subset of $\mathbb{C}$ ; in that case, writing the distinct values as $z_1,\dots,z_n$ and $E_j = \varphi^{-1}(\{z_j\})$ gives the standard representation $\varphi = \sum_{j=1}^{n} z_j \chi_{E_j}$ with the $E_j$ disjoint and measurable. Simple functions are never allowed the values $\pm\infty$ .

Positive and negative parts. For $f : X \to \overline{\mathbb{R}}$ set $f^{+} = \max(f,0)$ and $f^{-} = \max(-f,0)$ , so that $f = f^{+} - f^{-}$ and $\lvert f \rvert = f^{+} + f^{-}$ , with both parts non-negative. This decomposition and the simple-function approximation below are what reduce the Lebesgue integral to the integral of a non-negative simple function.

Note(A different-looking definition)AdaptedAdapted from Stakgold & Holst (2011), §0.7, p. 43

Stakgold and Holst (Stakgold & Holst (2011), §0.7, p. 43) do not build a $\sigma$ -algebra at all. They call a set $E \subseteq [0,1]$ of measure zero if for every $\varepsilon > 0$ it can be covered by finitely or countably many intervals of total length below $\varepsilon$ , and they then call $f$ measurable if for every $\varepsilon > 0$ its values can be changed on a set of measure less than $\varepsilon$ so as to produce a continuous function. Their two examples are a piecewise continuous $f$ with finitely many jumps, repaired by replacing it near each jump with a straight segment, and the Dirichlet function (one at the rationals, zero at the irrationals), repaired by resetting it to zero on the rationals, a set of measure zero.

On $[0,1]$ the two definitions agree for finite-valued $f$ . One direction is elementary. Apply their condition with $\varepsilon = 2^{-n}$ : there are a continuous $g_n$ and a set $A_n$ with $m^{*}(A_n) < 2^{-n}$ such that $f = g_n$ off $A_n$ . Put $B_K = \bigcup_{n \ge K}A_n$ and $A = \bigcap_{K \ge 1}B_K$ . Countable subadditivity gives $m^{*}(B_K) \le \sum_{n \ge K}2^{-n} = 2^{\,1-K}$ , hence $m^{*}(A) = 0$ , so $A$ is a Lebesgue null set. If $x \notin A$ then $x \notin B_K$ for some $K$ , so $x$ lies in no $A_n$ with $n \ge K$ and therefore $g_n(x) = f(x)$ for every $n \ge K$ . Thus $g_n \to f$ pointwise off the null set $A$ ; the $g_n$ are continuous, hence measurable, so $f$ is Lebesgue measurable by the limit statement below together with the completeness of Lebesgue measure. The converse direction is Lusin's theorem, which is not proved in this handbook. We use the $\sigma$ -algebra definition throughout, because it is the one that makes sense on an abstract measure space, where there is no notion of continuity to fall back on.

Proposition(It is enough to test a generating family)AdaptedAdapted from Folland (1999), Prop. 2.1 and Cor. 2.2, pp. 43–44, and Prop. 2.3, p. 44; Reed & Simon (1980), §I.3, pp. 15–16

Let $(X,\mathcal{M})$ and $(Y,\mathcal{N})$ be measurable spaces and suppose $\mathcal{N} = \mathcal{M}(\mathcal{E})$ is generated by $\mathcal{E}$ . Then $f : X \to Y$ is $(\mathcal{M},\mathcal{N})$ -measurable if and only if $f^{-1}(E) \in \mathcal{M}$ for every $E \in \mathcal{E}$ .

Three consequences:

a. Continuity suffices. Let $X$ and $Y$ carry topologies — they may be metric spaces, for instance. Any map between them that is continuous is then measurable for the two Borel $\sigma$ -algebras, that is, $(\mathcal{B}_X,\mathcal{B}_Y)$ -measurable. b. One family of rays suffices. For $f : X \to \mathbb{R}$ (or $\overline{\mathbb{R}}$ ), measurability is equivalent to each one of

f^{-1}\bigl((a,\infty]\bigr) \in \mathcal{M}, \quad f^{-1}\bigl([a,\infty]\bigr) \in \mathcal{M}, \quad f^{-1}\bigl([-\infty,a)\bigr) \in \mathcal{M}, \quad f^{-1}\bigl([-\infty,a]\bigr) \in \mathcal{M}

holding for all $a \in \mathbb{R}$ — only one of the four lists need be checked. In the notation of Evans (Evans (2010), Appendix E.2, p. 685), it is likewise enough that $f^{-1}(U) \in \mathcal{M}$ for every open $U \subseteq \mathbb{R}$ . c. Composition with a Borel function is safe. If $f : X \to \mathbb{R}$ is $\mathcal{M}$ -measurable and $h : \mathbb{R}\to\mathbb{R}$ is Borel measurable, then $h \circ f$ is $\mathcal{M}$ -measurable. Reed and Simon (Reed & Simon (1980), §I.3, p. 16) record this as the statement that a composition of two Borel functions is Borel.

Definition(Closure under the usual operations)AdaptedAdapted from Folland (1999), Props. 2.6 and 2.7 and Cors. 2.8 and 2.9, pp. 45–46; Reed & Simon (1980), §I.3, p. 16; Evans (2010), Appendix E.2, p. 685

Let $(X,\mathcal{M})$ be a measurable space.

a. Algebra. A function $f : X \to \mathbb{C}$ is measurable if and only if $\operatorname{Re}f$ and $\operatorname{Im}f$ are measurable (Folland (1999), Cor. 2.5, p. 45). If $f, g : X \to \mathbb{C}$ are measurable, so are $f + g$ and $fg$ , and so are $\lambda f$ for $\lambda \in \mathbb{C}$ and $\lvert f \rvert$ . For $\overline{\mathbb{R}}$ -valued functions the same holds once the indeterminate expressions are fixed by convention: $0 \cdot (\pm\infty) = 0$ , and the value of $f+g$ is assigned by hand at the points where $f$ and $g$ are infinite of opposite sign (Folland (1999), Exercise 2, p. 48). b. Lattice. If $f,g : X \to \overline{\mathbb{R}}$ are measurable, so are $\max(f,g)$ and $\min(f,g)$ ; in particular $f^{+}$ , $f^{-}$ and $\lvert f\rvert$ are measurable. c. Countable suprema and limits. If $f_1, f_2, \dots : X \to \overline{\mathbb{R}}$ are measurable, then so are

\sup_j f_j, \qquad \inf_j f_j, \qquad \limsup_{j\to\infty} f_j, \qquad \liminf_{j\to\infty} f_j ,

and if $f(x) = \lim_{j\to\infty} f_j(x)$ exists at every $x$ , then $f$ is measurable. The same holds for complex-valued $f_j$ with a pointwise limit, by applying the real case to the real and imaginary parts and recombining them with the equivalence in (a) (Folland (1999), Cor. 2.9, p. 46). d. The set where the limit exists is measurable: $\{x : \lim_j f_j(x) \text{ exists in } \mathbb{R}\} \in \mathcal{M}$ (Folland (1999), Exercise 3, p. 48).

Definition(Approximation by simple functions)AdaptedAdapted from Folland (1999), Thm 2.10, p. 47

Let $(X,\mathcal{M})$ be a measurable space.

a. Non-negative case. Let $f : X \to [0,\infty]$ be measurable. Then $f$ is the limit of a staircase that only ever rises: simple functions $\varphi_n$ , $n \ge 0$ , may be chosen so that

0 \le \varphi_0 \le \varphi_1 \le \varphi_2 \le \cdots, \qquad \varphi_n \le f \ \text{ for each } n, \qquad \lim_{n \to \infty}\varphi_n(x) = f(x) \ \text{ for each } x \in X,

and so that on any $A \subseteq X$ with $\sup_A f < \infty$ the convergence is uniform. b. Complex case. Let $f : X \to \mathbb{C}$ be measurable. Then simple $\varphi_n$ may be chosen whose moduli rise towards that of $f$ , $0 \le \lvert\varphi_0\rvert \le \lvert \varphi_1 \rvert \le \cdots \le \lvert f \rvert$ , with $\varphi_n(x) \to f(x)$ at each $x$ , the convergence again being uniform wherever $f$ stays bounded.

Idea of the proofOriginal

Everything here rests on one observation: a preimage is a perfect homomorphism of set operations. So the collection $\{E \subseteq Y : f^{-1}(E) \in \mathcal{M}\}$ is automatically a $\sigma$ -algebra, whatever $f$ is. To prove a function measurable one therefore never has to look at all Borel sets; it is enough to catch a generating family in that collection, and the $\sigma$ -algebra generated takes care of the rest. Continuity, the rays $(a,\infty]$ , and the open sets are three different ways of supplying such a family.

Closure under limits comes from writing a limit operation as a countable set operation. The point $x$ satisfies $\sup_j f_j(x) > a$ exactly when some $f_j(x) > a$ , which is a countable union; upper and lower limits are an infimum of suprema, hence two such operations in a row. That is the whole proof, and it is why countable families are the natural ones: an uncountable supremum would need an uncountable union, which a $\sigma$ -algebra does not supply.

Simple approximation is Lebesgue's original idea turned into a formula: instead of chopping the domain, chop the range into dyadic steps of height $2^{-n}$ and read off which step each value falls on, capping the result at height $2^{n}$ so that only finitely many steps are used. Refining $n$ halves the step height and doubles the cap, so the staircase climbs and the error on the region where $f$ is already below the cap is at most one step.

Proof(The generating criterion and its consequences)AdaptedAdapted from Folland (1999), Prop. 2.1 and Cor. 2.2, pp. 43–44, and Prop. 2.3, p. 44

The criterion. "Only if" is immediate, as $\mathcal{E} \subseteq \mathcal{N}$ . For "if", consider

\mathcal{G} = \bigl\{\, E \subseteq Y \;:\; f^{-1}(E) \in \mathcal{M} \,\bigr\} .

$\mathcal{G}$ is a $\sigma$ -algebra: $f^{-1}(Y) = X \in \mathcal{M}$ ; $f^{-1}(E^{c}) = \bigl(f^{-1}(E)\bigr)^{c}$ lies in $\mathcal{M}$ whenever $f^{-1}(E)$ does, since $\mathcal{M}$ is closed under complements; and $f^{-1}(\bigcup_j E_j) = \bigcup_j f^{-1}(E_j) \in \mathcal{M}$ for a sequence, since $\mathcal{M}$ is closed under countable unions. By hypothesis $\mathcal{E} \subseteq \mathcal{G}$ , so $\mathcal{N} = \mathcal{M}(\mathcal{E}) \subseteq \mathcal{G}$ because $\mathcal{M}(\mathcal{E})$ is the smallest $\sigma$ -algebra containing $\mathcal{E}$ . That is exactly measurability.

(a) Continuity. $\mathcal{B}_Y$ is generated by the open subsets of $Y$ , and continuity of $f$ says precisely that $f^{-1}(U)$ is open in $X$ — hence Borel — for every open $U \subseteq Y$ ; see continuity in topological terms. Apply the criterion with $\mathcal{E} = \{\text{open sets of } Y\}$ .

(b) Rays. $\mathcal{B}_{\mathbb{R}}$ is generated by each of the four families of rays, by the generator list; and $\mathcal{B}_{\overline{\mathbb{R}}}$ is generated by the rays $(a,\infty]$ , $a \in \mathbb{R}$ , because the $\sigma$ -algebra these generate contains $\{\infty\} = \bigcap_{n}(n,\infty]$ , hence also $\{-\infty\} = \bigl(\bigcup_n (-n,\infty]\bigr)^{c}$ , hence every Borel subset of $\mathbb{R}$ as well. Apply the criterion to each family in turn. The open-set version is the same argument with $\mathcal{E}$ the open sets.

(c) Composition. $h$ Borel measurable means $h^{-1}(B) \in \mathcal{B}_{\mathbb{R}}$ for Borel $B$ , and $f$ measurable means $f^{-1}(\mathcal{B}_{\mathbb{R}}) \subseteq \mathcal{M}$ ; so $(h \circ f)^{-1}(B) = f^{-1}\bigl(h^{-1}(B)\bigr) \in \mathcal{M}$ . $\square$

Proof(Algebra and lattice operations)Original

Folland (Folland (1999), Prop. 2.6, p. 45) derives these from the product $\sigma$ -algebra on $\mathbb{C}\times\mathbb{C}$ . The following direct argument on the line uses only the ray criterion and the countability of $\mathbb{Q}$ . The complex case is then reduced to the real one by splitting into real and imaginary parts, and that reduction needs both directions of the equivalence in (a), so we prove it first.

Real and imaginary parts. If $f : X \to \mathbb{C}$ is measurable, so are $\operatorname{Re}f$ and $\operatorname{Im}f$ : the maps $z \mapsto \operatorname{Re}z$ and $z \mapsto \operatorname{Im}z$ are continuous, hence Borel, and part (c) of the previous statement applies. Conversely, let $u, v : X \to \mathbb{R}$ be measurable and put $f = u + iv$ . Identify $\mathbb{C}$ with $\mathbb{R}^{2}$ and call a set of the form $R = (p,q)\times(r,s)$ with $p<q$ and $r<s$ all rational a rational rectangle. There are only countably many of them, and every open $W \subseteq \mathbb{C}$ is a countable union of rational rectangles: around each $w \in W$ put an open disc contained in $W$ , and inside that disc a rational rectangle containing $w$ . Rational rectangles are themselves open, hence Borel, so the $\sigma$ -algebra they generate sits inside $\mathcal{B}_{\mathbb{C}}$ ; and since every open set is a countable union of them, that $\sigma$ -algebra contains all open sets and therefore equals $\mathcal{B}_{\mathbb{C}}$ . Finally

f^{-1}(R) = u^{-1}\bigl((p,q)\bigr) \cap v^{-1}\bigl((r,s)\bigr) \in \mathcal{M},

each factor being measurable by the ray criterion, since $u^{-1}((p,q)) = \{u > p\}\setminus\{u \ge q\}$ . By the generating criterion, $f$ is measurable. This is Folland (Folland (1999), Cor. 2.5, p. 45), which he obtains from $\mathcal{B}_{\mathbb{C}} = \mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}$ ; the route just taken replaces the product $\sigma$ -algebra by the countable family of rational rectangles.

Sums. Let $f,g : X \to \mathbb{R}$ be measurable and $a \in \mathbb{R}$ . Then

\{x : f(x) + g(x) > a\} \;=\; \bigcup_{q \in \mathbb{Q}} \Bigl( \{x: f(x) > q\} \cap \{x : g(x) > a - q\} \Bigr).

Right to left: if $f(x) > q$ and $g(x) > a-q$ then $f(x)+g(x) > a$ . Left to right: if $f(x)+g(x) > a$ then $f(x) > a - g(x)$ , and the density of $\mathbb{Q}$ in $\mathbb{R}$ supplies a rational $q$ with $a - g(x) < q < f(x)$ , whence $f(x) > q$ and $g(x) > a-q$ . The union is over a countable index set, and each term is an intersection of two members of $\mathcal{M}$ by the ray criterion; so $\{f+g>a\} \in \mathcal{M}$ for every $a$ , and the ray criterion applied in the other direction makes $f+g$ measurable.

Scalar multiples. Let $c \in \mathbb{R}$ and $f$ real measurable. If $c > 0$ then $\{cf > a\} = \{f > a/c\}$ ; if $c < 0$ then $\{cf > a\} = \{f < a/c\}$ , measurable because the rays $[-\infty,a)$ test measurability just as well; if $c = 0$ then $cf \equiv 0$ and $\{cf > a\}$ is $X$ or $\varnothing$ . So $cf$ is measurable for every real $c$ . In particular $-g$ is measurable, and so is $\tfrac14 h$ for any measurable $h$ .

Squares, hence products. For $a < 0$ , $\{f^{2} > a\} = X$ ; for $a \ge 0$ , $\{f^{2} > a\} = \{f > \sqrt{a}\} \cup \{f < -\sqrt{a}\}$ , a union of two measurable sets. So $f^{2}$ is measurable, and the polarisation identity

fg = \tfrac14\bigl[(f+g)^{2} - (f-g)^{2}\bigr]

makes $fg$ measurable: $f+g$ and $f - g = f + (-1)g$ are measurable by the sum and scalar cases, their squares by the previous sentence, the difference of the squares by the sum and scalar cases again, and the factor $\tfrac14$ by the scalar case.

Lattice. $\{\max(f,g) > a\} = \{f>a\}\cup\{g>a\}$ and $\{\min(f,g) > a\} = \{f>a\}\cap\{g>a\}$ , both measurable. Taking $g \equiv 0$ gives $f^{+}$ , and $f^{-} = \max(-f,0)$ ; then $\lvert f \rvert = f^{+}+f^{-}$ . This lattice paragraph holds verbatim for $\overline{\mathbb{R}}$ -valued functions, the rays being $(a,\infty]$ : no arithmetic is performed in it. The sum and product paragraphs do not transfer so cheaply, because $f+g$ and $fg$ must first be defined where $\infty - \infty$ or $0\cdot\infty$ occurs; with the conventions listed in statement (a) in force, measurability still holds (Folland (1999), Exercise 2, p. 48). $\square$

Proof(Suprema, limits, and the set where a limit exists)AdaptedAdapted from Folland (1999), Prop. 2.7, p. 45, and Exercise 3, p. 48

Suprema and infima. For measurable $f_j : X \to \overline{\mathbb{R}}$ and $a \in \mathbb{R}$ ,

\Bigl\{\sup_j f_j > a\Bigr\} = \bigcup_{j \ge 1}\{f_j > a\}, \qquad \Bigl\{\inf_j f_j < a\Bigr\} = \bigcup_{j \ge 1}\{f_j < a\} .

The first holds because a supremum exceeds $a$ exactly when some member does; the second is its mirror image. Each right-hand side is a countable union of measurable sets, so by the ray criterion — $(a,\infty]$ for the first, $[-\infty,a)$ for the second — both $\sup_j f_j$ and $\inf_j f_j$ are measurable. This is where countability is used, and it is used nowhere else in the proof.

Upper and lower limits. Set $h_k = \sup_{j > k} f_j$ , measurable by the previous paragraph applied to the sequence $(f_{k+1}, f_{k+2},\dots)$ . Then $\limsup_j f_j = \inf_k h_k$ is measurable, being an infimum of countably many measurable functions. Likewise $\liminf_j f_j = \sup_k \inf_{j>k} f_j$ .

Limits. If the pointwise limit $f(x) = \lim_j f_j(x)$ exists for every $x$ , then $f = \limsup_j f_j$ , which has just been shown measurable.

The set where the limit exists. A real sequence converges in $\mathbb{R}$ if and only if its upper and lower limits are equal and finite. Write $L = \limsup_j f_j$ and $\ell = \liminf_j f_j$ , both measurable by the previous paragraphs. Then

\Bigl\{x : \lim_j f_j(x) \text{ exists in } \mathbb{R}\Bigr\} = \{x : L(x) = \ell(x)\} \cap \{x : \lvert L(x)\rvert < \infty\} .

The second set is $L^{-1}(\mathbb{R})$ , a preimage of a Borel subset of $\overline{\mathbb{R}}$ . For the first, note that $\ell \le L$ everywhere, so the two differ at $x$ exactly when some rational lies strictly between them:

\{x : L(x) \ne \ell(x)\} = \bigcup_{q \in \mathbb{Q}} \bigl(\{\ell < q\} \cap \{q < L\}\bigr),

a countable union of measurable sets, so its complement $\{L = \ell\}$ is measurable. We do not write the first set as $\{L - \ell = 0\}$ : that difference is undefined at points where $L$ and $\ell$ are both $+\infty$ or both $-\infty$ . $\square$

Proof(Approximation by simple functions)AdaptedAdapted from Folland (1999), Thm 2.10, p. 47

(a) The construction. Following Folland (Folland (1999), Thm 2.10, p. 47), for $n = 0, 1, 2, \dots$ and $0 \le k \le 2^{2n}-1$ put

E^{k}_{n} = f^{-1}\Bigl(\bigl(k2^{-n},\,(k+1)2^{-n}\bigr]\Bigr), \qquad F_n = f^{-1}\bigl((2^{n},\infty]\bigr), \qquad \varphi_n = \sum_{k=0}^{2^{2n}-1} k2^{-n}\,\chi_{E^{k}_{n}} \;+\; 2^{n}\chi_{F_n}.

In words: cut the range into steps of height $2^{-n}$ up to the height $2^{n}$ , round each value down to the step below it, and flatten everything above $2^{n}$ to $2^{n}$ . All the sets are measurable, being preimages of Borel subsets of $[0,\infty]$ , and the intervals $(k2^{-n},(k+1)2^{-n}]$ for $0 \le k \le 2^{2n}-1$ together with $(2^{n},\infty]$ partition $(0,\infty]$ ; adding $f^{-1}(\{0\})$ , on which $\varphi_n = 0$ , partitions $X$ . So each $\varphi_n$ is a simple function with values among the $2^{2n}+1$ numbers $0, 2^{-n}, \dots, 2^{n}-2^{-n}, 2^{n}$ , and $\varphi_n \ge 0$ .

Step 1: $\varphi_n \le f$ . On $E^{k}_{n}$ we have $f > k2^{-n} = \varphi_n$ ; on $F_n$ we have $f > 2^{n} = \varphi_n$ ; on $f^{-1}(\{0\})$ both sides vanish.

Step 2: $\varphi_n \le \varphi_{n+1}$ . Fix $x$ and compare the two roundings.

If $f(x) = 0$ both are $0$ .
If $0 < f(x) \le 2^{n}$ , let $k$ be the index with $f(x) \in (k2^{-n},(k+1)2^{-n}]$ , so $\varphi_n(x) = k2^{-n}$ . That interval is the disjoint union of $\bigl(2k\,2^{-(n+1)}, (2k+1)2^{-(n+1)}\bigr]$ and $\bigl((2k+1)2^{-(n+1)}, (2k+2)2^{-(n+1)}\bigr]$ , so at level $n+1$ the index of $f(x)$ is $2k$ or $2k+1$ , and both lie below $2^{2(n+1)}-1$ because $k \le 2^{2n}-1$ . Hence $\varphi_{n+1}(x)$ equals $2k\,2^{-(n+1)} = k2^{-n} = \varphi_n(x)$ or $(2k+1)2^{-(n+1)} > \varphi_n(x)$ .
If $2^{n} < f(x) \le 2^{n+1}$ , then $\varphi_n(x) = 2^{n}$ , while at level $n+1$ the index $k'$ of $f(x)$ satisfies $k'2^{-(n+1)} \ge 2^{n}$ — because $k'$ is the largest index with $k'2^{-(n+1)} < f(x)$ , and $2^{2n+1}2^{-(n+1)} = 2^{n} < f(x)$ forces $k' \ge 2^{2n+1}$ . So $\varphi_{n+1}(x) \ge 2^{n} = \varphi_n(x)$ .
If $f(x) > 2^{n+1}$ , then $\varphi_n(x) = 2^{n} < 2^{n+1} = \varphi_{n+1}(x)$ .

Step 3: the error bound. Suppose $f(x) \le 2^{n}$ . If $f(x) = 0$ then $f(x)-\varphi_n(x) = 0$ ; otherwise $f(x) \in (k2^{-n},(k+1)2^{-n}]$ for some $k \le 2^{2n}-1$ and

0 \;<\; f(x) - \varphi_n(x) \;=\; f(x) - k2^{-n} \;\le\; 2^{-n} .

So $0 \le f - \varphi_n \le 2^{-n}$ on the set $\{f \le 2^{n}\}$ .

Step 4: pointwise and uniform convergence. If $f(x) < \infty$ , then $f(x) \le 2^{n}$ for all large $n$ , and step 3 gives $0 \le f(x)-\varphi_n(x) \le 2^{-n} \to 0$ . If $f(x) = \infty$ , then $x \in F_n$ for every $n$ and $\varphi_n(x) = 2^{n} \to \infty = f(x)$ . If $f \le M$ on a set $A$ , then for every $n$ with $2^{n}\ge M$ step 3 gives $\sup_{A}\lvert f - \varphi_n\rvert \le 2^{-n}$ , which is uniform convergence on $A$ .

(b) The complex case. Write $f = g + ih$ with $g = \operatorname{Re}f$ and $h = \operatorname{Im}f$ measurable, and split each into positive and negative parts, obtaining four non-negative measurable functions $g^{+},g^{-},h^{+},h^{-}$ . Apply part (a) to each, getting increasing sequences $\psi^{\pm}_n \uparrow g^{\pm}$ and $\zeta^{\pm}_n \uparrow h^{\pm}$ of simple functions, and set $\varphi_n = (\psi^{+}_n - \psi^{-}_n) + i(\zeta^{+}_n - \zeta^{-}_n)$ , a simple function. Then $\varphi_n \to f$ pointwise, and uniformly wherever $f$ is bounded, because each of the four parts converges that way. For the monotonicity of $\lvert \varphi_n\rvert$ , note that at each point at most one of $g^{+},g^{-}$ is non-zero and at most one of $h^{+},h^{-}$ is, so $\lvert\varphi_n\rvert^{2} = (\psi^{+}_n+\psi^{-}_n)^{2} + (\zeta^{+}_n+\zeta^{-}_n)^{2}$ increases in $n$ and is bounded by $g^{2}+h^{2} = \lvert f\rvert^{2}$ . $\square$

Remark(Where the hypotheses are used)Original

Countability of the index set is used exactly once, in the identity $\{\sup_j f_j > a\} = \bigcup_j \{f_j > a\}$ , and it cannot be removed: an uncountably indexed supremum of measurable functions need not be measurable at all (Folland (1999), Exercise 6, p. 48). A one-line example: let $N \subseteq \mathbb{R}$ be the non-measurable set of Lebesgue measure and let $f_t = \chi_{\{t\}}$ for $t \in N$ . Each $f_t$ is Lebesgue measurable, since a singleton is Borel, but $\sup_{t \in N} f_t = \chi_N$ is not.
Measurability of the domain $\sigma$ -algebra, not of the range, is what is being asserted. The definition does not require $f(E)$ to be measurable for measurable $E$ ; images behave badly, preimages do not.
The generating criterion needs $\mathcal{E}$ to generate, not merely to be contained in, $\mathcal{N}$ . Checking $f^{-1}((a,\infty])\in\mathcal{M}$ for $a$ in a dense set $D$ of reals is still enough, because for any real $a$ one may pick $a_n \in D$ with $a_n > a$ and $a_n \to a$ , and then $(a,\infty] = \bigcup_{n}(a_n,\infty]$ — but checking it for finitely many $a$ is not. Folland (Folland (1999), Exercise 4, p. 48) poses this for $D = \mathbb{Q}$ ; the one-line argument just given needs only density, so we state it for a general dense $D$ .
Composition is safe on the right, not on the left. If $g$ is measurable and $h$ is Borel, $h\circ g$ is measurable. If $h$ is merely Lebesgue measurable the conclusion fails, even for continuous $g$ : $h^{-1}(B)$ is then only in $\mathcal{L}$ , and $g^{-1}$ of a set in $\mathcal{L}$ need not be in $\mathcal{L}$ . Folland (Folland (1999), §2.1, p. 44 and Exercise 9, p. 48) gives the counterexample built from the Cantor function, and Reed and Simon (Reed & Simon (1980), §I.3, p. 16) cite exactly this as their reason for working with Borel sets and Borel functions only. The asymmetry traces back to the fact that $\mathcal{L}$ is a completion and so is tied to one particular measure.
Completeness is what makes "almost everywhere" harmless. If $f$ is measurable and $g = f$ off a null set, or if $f_n \to f$ only almost everywhere, then $g$ and $f$ are measurable provided the measure is complete — and these implications hold for all such $g$ and $f_n$ only when it is (Folland (1999), Prop. 2.11, p. 47). Without completeness, the exceptional set can hide a non-measurable subset. This is one reason the Lebesgue $\sigma$ -algebra $\mathcal{L}$ , rather than $\mathcal{B}_{\mathbb{R}}$ , is the default on the line, and it is why the modes of convergence compared in modes of convergence are all stated almost everywhere.
In the approximation theorem the two hypotheses do different jobs. Non-negativity is what lets the staircase increase; for a signed or complex $f$ one must split first, and then only $\lvert\varphi_n\rvert$ increases, not $\varphi_n$ . Boundedness of $f$ on a set is what upgrades pointwise to uniform convergence; where $f$ is unbounded the cap $2^{n}$ is binding and the error is $f - 2^{n}$ , which is as large as one likes. No measure appears anywhere in the construction: the theorem is about $(X,\mathcal{M})$ alone.

Method(Showing that a given function is measurable)Original

Identify the $\sigma$ -algebra on the domain. Borel or Lebesgue on the line; in probability, the $\sigma$ -field of the model. The answer can differ between the two.
Try continuity first. A continuous function is Borel measurable, hence Lebesgue measurable. A function that is continuous at every point outside a countable set $D$ is measurable too: if $X = A \cup B$ with $A, B \in \mathcal{M}$ , a function on $X$ is measurable as soon as it is measurable on $A$ and on $B$ (Folland (1999), Exercise 5, p. 48), so take $A = D$ and $B = X\setminus D$ . On $D$ every subset is Borel, so $f$ is measurable there; on $X \setminus D$ the restriction of $f$ is continuous.
Otherwise test one family of rays. Show $\{f > a\} \in \mathcal{M}$ for every real $a$ — or for every rational $a$ , which is enough. This is usually the shortest route for a function defined by a formula or by cases.
Build it from measurable pieces. Sums, products, scalar multiples, maxima, minima, absolute values, positive and negative parts, and compositions $h\circ f$ with $h$ Borel all stay inside the class.
Or exhibit it as a countable limit. $\sup$ , $\inf$ , $\limsup$ , $\liminf$ and pointwise limits of sequences of measurable functions are measurable. For an almost-everywhere limit, check that the measure is complete, or modify the function on the exceptional null set first.
For an integral, pass to the staircase. Apply the approximation theorem to $f^{+}$ and $f^{-}$ separately; everything in the Lebesgue integral and monotone convergence is then proved for simple functions first and transported by the increasing limit.
Do not try to prove measurability by looking at images, and do not compose two Lebesgue measurable functions without checking that the inner one is Borel.

Example(Building the staircase for f(x) = x^2 on [0,3])Original

Take $X = [0,3]$ with the Lebesgue measurable sets and $f(x) = x^{2}$ . It is continuous, hence Borel measurable, so the construction of the approximation theorem applies.

The level sets at $n = 2$ . Here $2^{-n} = \tfrac14$ and the cap is $2^{n} = 4$ , so $k$ runs from $0$ to $2^{4}-1 = 15$ and

E^{k}_{2} = \Bigl\{x \in [0,3] : \tfrac{k}{4} < x^{2} \le \tfrac{k+1}{4}\Bigr\} = \Bigl(\tfrac{\sqrt{k}}{2},\ \tfrac{\sqrt{k+1}}{2}\Bigr], \qquad F_2 = \{x : x^{2} > 4\} = (2,3] .

For instance $E^{0}_{2} = (0,\tfrac12]$ with $m = 0.5$ , $E^{1}_{2} = (\tfrac12, \tfrac{\sqrt2}{2}]$ with $m = 0.2071068\ldots$ , and $E^{15}_{2} = \bigl(\tfrac{\sqrt{15}}{2}, 2\bigr] = (1.9364917\ldots,\, 2]$ with $m = 0.0635083\ldots$ .

Check that the pieces account for all of $[0,3]$ . The measures telescope:

\sum_{k=0}^{15} m\bigl(E^{k}_{2}\bigr) = \sum_{k=0}^{15}\frac{\sqrt{k+1}-\sqrt{k}}{2} = \frac{\sqrt{16}-\sqrt{0}}{2} = 2,

and $m(F_2) = 1$ , while the remaining point $f^{-1}(\{0\}) = \{0\}$ has measure $0$ . The total is $2 + 1 + 0 = 3 = m([0,3])$ , as it must be.

Values at a point. Take $x = 1.3$ , so $f(x) = 1.69$ .

$n$	cap $2^{n}$	step $2^{-n}$	index $k$ with $k2^{-n} < 1.69 \le (k+1)2^{-n}$	$\varphi_n(1.3)$	$f-\varphi_n$
$0$	$1$	$1$	— ( $1.69 > 1$ , so $x \in F_0$ )	$1$	$0.69$
$1$	$2$	$0.5$	$3$	$1.5$	$0.19$
$2$	$4$	$0.25$	$6$	$1.5$	$0.19$
$3$	$8$	$0.125$	$13$	$1.625$	$0.065$
$4$	$16$	$0.0625$	$27$	$1.6875$	$0.0025$

The column of values increases, as step 2 of the proof requires, and from $n = 1$ onwards — the first $n$ with $2^{n} \ge f(1.3)$ — the error obeys the bound $f - \varphi_n \le 2^{-n}$ of step 3: $0.19 \le 0.5$ , $0.19 \le 0.25$ , $0.065 \le 0.125$ , $0.0025 \le 0.0625$ .

Where the cap bites. At $x = 2.5$ we have $f(x) = 6.25 > 4 = 2^{2}$ , so $x \in F_2$ and $\varphi_2(2.5) = 4$ , an error of $2.25$ — far larger than the step height $0.25$ . On $[0,2]$ , where $f \le 4 = 2^{2}$ , the error is at most $0.25$ everywhere, so $\varphi_2$ approximates $f$ uniformly there; on all of $[0,3]$ it does not, and uniformity returns only for $n \ge 4$ , when $2^{n} \ge 9 = \max_{[0,3]} f$ . This is exactly the restriction "on any set on which $f$ is bounded" in the statement: here $f$ is bounded on $[0,3]$ , so uniform convergence does hold on the whole interval — but only once $n$ is large enough that the cap has cleared the maximum.

ApplicationOriginal

Random variables are measurable functions. A real random variable on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ is by definition an $\mathcal{F}$ -measurable $X : \Omega \to \mathbb{R}$ , and the ray criterion is why the distribution function $F(a) = \mathbb{P}(X \le a)$ determines the whole law: the events $\{X \le a\}$ generate everything one can ask about $X$ . Closure under pointwise limits is what makes limit theorems well posed — $\limsup_n X_n$ is an honest random variable, so statements like "the sample average converges" are about measurable sets and have probabilities. See random variables and distributions.

Quantisation and detector binning. A simple function is a measurement reported to finite precision: finitely many values, each on a measurable set of the sample space. The approximation theorem says every non-negative measurable observable is the increasing limit of such reports, with the error bounded by one quantisation step wherever the signal is below the full-scale range, and uncontrolled above it. That is precisely the behaviour of an analogue-to-digital converter: resolution $2^{-n}$ , full scale $2^{n}$ , and clipping beyond it. The requirement that the staircase increase is what makes monotone convergence applicable to the resulting integrals, so that finer and finer quantisation gives an increasing and convergent sequence of estimates.

Why a measurability hypothesis appears in every theorem of this chapter. The convergence theorems — Fatou and monotone convergence, dominated convergence — each assume the functions involved are measurable, and part (c) above is what makes the assumption cheap: a limit of measurable functions is measurable automatically, so one never has to re-verify it along a limiting argument. In the theory of partial differential equations this is used constantly; Evans (Evans (2010), Appendix E.2, pp. 685–686) lists it alongside the fact that the Lebesgue integral of a non-negative measurable function is defined by approximation with simple functions.

Lebesgue's idea restated. Stakgold and Holst (Stakgold & Holst (2011), §0.7, p. 42) describe the contrast with Riemann's construction as a shopkeeper counting a day's takings either transaction by transaction or by sorting the money into denominations first. Partitioning the range rather than the domain is exactly what the sets $E^{k}_{n}$ do, and it is why the construction never had to assume anything about the regularity of $f$ as a function of $x$ — only that the sorting itself is legitimate, that is, that each $E^{k}_{n}$ is a measurable set.

Definition

The Lebesgue integral

Adapted

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Builds the integral against a measure in three stages — finite sums for simple functions, a supremum for non-negative measurable functions, a difference of two such for the rest — records the properties that make it a linear order-preserving functional on $L^1$ , and shows that on a compact interval it reproduces every proper Riemann integral while reaching functions the Riemann theory cannot.

Prerequisites: Measurable functions, Lebesgue measure

Definition(The integral, in three stages)AdaptedAdapted from Folland (1999), §2.2, pp. 49–50 and §2.3, pp. 52–53; Grimmett & Stirzaker (2020), §5.6(B), items (7)–(11), pp. 200–201

Fix a measure space $(X, \Sigma, \mu)$ : a set $X$ , a $\sigma$ -algebra $\Sigma$ of subsets of $X$ and a measure $\mu$ on it, as in Sigma-algebras and measures. Write $\mathcal{L}^{+}$ for the set of measurable functions $f : X \to [0,\infty]$ ; values $+\infty$ are allowed, and throughout we use the convention

0 \cdot \infty = 0 ,

so that a set of infinite measure on which the integrand vanishes contributes nothing.

Stage 1: simple functions. A simple function is a measurable $\varphi : X \to \mathbb{R}$ taking only finitely many values. Listing its distinct non-zero values as $a_1, \dots, a_n$ and putting $E_j = \varphi^{-1}(\{a_j\}) \in \Sigma$ gives the standard representation

\varphi = \sum_{j=1}^{n} a_j \, \mathbf{1}_{E_j},

with the $E_j$ disjoint. For $\varphi \ge 0$ simple we define

\int \varphi \, d\mu \;=\; \sum_{j=1}^{n} a_j \, \mu(E_j) \;\in\; [0,\infty] .

This is a finite sum of areas of rectangles: each value $a_j$ is weighted by the measure of the set where it is taken.

Stage 2: non-negative measurable functions. For $f \in \mathcal{L}^{+}$ ,

\int f \, d\mu \;=\; \sup\Bigl\{ \int \varphi \, d\mu \;:\; \varphi \ \text{simple},\ 0 \le \varphi \le f \Bigr\} \;\in\; [0,\infty] .

No finiteness is assumed and none is concluded. Because a simple $\varphi \ge 0$ is one of its own competitors, the two definitions agree where both apply.

Stage 3: the general case. For measurable $f : X \to [-\infty, \infty]$ split off the positive and negative parts

f^{+} = \max(f, 0), \qquad f^{-} = \max(-f, 0),

both in $\mathcal{L}^{+}$ , with $f = f^{+} - f^{-}$ and $\lvert f \rvert = f^{+} + f^{-}$ . When at least one of $\int f^{+}$ , $\int f^{-}$ is finite, set

\int f \, d\mu \;=\; \int f^{+} d\mu \;-\; \int f^{-} d\mu .

If both are finite — equivalently $\int \lvert f \rvert \, d\mu < \infty$ — we call $f$ integrable and write $f \in L^{1}(\mu)$ . A complex-valued measurable $f$ is integrable when $\int \lvert f \rvert \, d\mu < \infty$ , which happens exactly when $\operatorname{Re} f$ and $\operatorname{Im} f$ are both integrable, and then $\int f = \int \operatorname{Re} f + i \int \operatorname{Im} f$ .

Integrating over a set. For $A \in \Sigma$ , $\displaystyle \int_{A} f \, d\mu = \int f \, \mathbf{1}_{A} \, d\mu$ . When $X = \mathbb{R}^{n}$ and $\mu$ is Lebesgue measure $m$ , the resulting integral is the Lebesgue integral, written $\int_{A} f(x)\, dx$ .

Note(Two words that are not synonyms, one notation that is a quotient, and two forward references)AdaptedAdapted from Folland (1999), §2.3, pp. 53–54; Evans (2010), App. E.2, p. 686; Reed & Simon (1980), §1.3, p. 17

Integrable versus summable. Our stage 3 uses the vocabulary of Folland (Folland (1999), §2.3, p. 53), for whom integrable means $\int \lvert f \rvert < \infty$ . Evans (Evans (2010), App. E.2, p. 686) splits the word in two: for him $f$ is integrable as soon as one of $\int f^{\pm}$ is finite, so that $\int f$ has a value in $[-\infty,\infty]$ , and summable when that value is finite. The two books therefore use "integrable" for different classes. We follow Folland; where Evans would say summable, we say integrable. A function whose integral exists but equals $+\infty$ or $-\infty$ — stage 3 asks only that one of $\int f^{\pm}$ be finite, so such an $f$ need not be of one sign — is never called integrable here.

$L^1$ is a space of equivalence classes. Two integrable functions that agree outside a null set have the same integral over every measurable set, so $\int \lvert f - g \rvert \, d\mu = 0$ does not force $f = g$ pointwise. Both Folland (Folland (1999), §2.3, p. 54) and Reed and Simon (Reed & Simon (1980), §1.3, p. 17) therefore redefine $L^{1}(\mu)$ as the set of equivalence classes under equality almost everywhere, which is what makes $\lVert f \rVert_{1} = \int \lvert f \rvert \, d\mu$ a norm rather than a seminorm. The cost is that for $f \in L^{1}$ the symbol $f(x)$ at one point $x$ carries no information; the gain is the completeness used in L^p spaces. We keep the usual abuse of writing $f \in L^{1}(\mu)$ for a function.

Two forward references. Three places below — step 4 of the proof of the elementary properties, step 3 of the Riemann comparison, and step 4 of the method — use the monotone or the dominated convergence theorem, which belong to Monotone convergence and Fatou's lemma and Dominated convergence. Those units are not prerequisites of this one and are not listed as such: both are proved from the three stages defined above and from the monotonicity of the integral on $\mathcal{L}^{+}$ (claim 1 below), and neither uses the statements they are invoked to prove. The dependence therefore runs one way only, and nothing here is circular.

Definition(Elementary properties)AdaptedAdapted from Folland (1999), §2.2, Prop. 2.13 and the display after it, pp. 49–50, Prop. 2.16, p. 51, and §2.3, Props. 2.21, 2.22 and 2.23b, pp. 53–54; Reed & Simon (1980), §1.3, Thm 1.9, p. 16

Let $(X,\Sigma,\mu)$ be a measure space.

Monotonicity and scaling on $\mathcal{L}^{+}$ . If $f, g \in \mathcal{L}^{+}$ and $f \le g$ then $\int f \le \int g$ ; and $\int c f = c \int f$ for every $c \in [0,\infty)$ .
Additivity on simple functions. If $\varphi, \psi \ge 0$ are simple then $\int(\varphi + \psi) = \int \varphi + \int \psi$ .
Each non-negative simple function induces a measure. For $\varphi \ge 0$ simple, $A \mapsto \int_{A} \varphi \, d\mu$ is a measure on $\Sigma$ .
Linearity on $L^{1}$ . The integrable real-valued functions form a real vector space and $f \mapsto \int f \, d\mu$ is linear on it; the complex-valued ones form a complex vector space on which the integral is complex-linear.
The triangle inequality. If $f \in L^{1}(\mu)$ , possibly complex-valued, then $\bigl\lvert \int f \, d\mu \bigr\rvert \le \int \lvert f \rvert \, d\mu$ .
Null sets are invisible. For $f \in \mathcal{L}^{+}$ : $\int f \, d\mu = 0$ if and only if $f = 0$ almost everywhere. For $f, g \in L^{1}(\mu)$ the three statements " $\int_{E} f = \int_{E} g$ for every $E \in \Sigma$ ", " $\int \lvert f - g \rvert = 0$ " and " $f = g$ a.e." are equivalent.
A crude bound. If $f$ is measurable and bounded on a finite interval $(a,b) \subset \mathbb{R}$ , then $f$ is Lebesgue integrable there and $\bigl\lvert \int_{a}^{b} f \, dx \bigr\rvert \le (b-a) \sup_{a < x < b} \lvert f(x) \rvert$ .

Definition(Comparison with the Riemann integral)AdaptedAdapted from Folland (1999), §2.3, Thm 2.28 and the discussion on pp. 57–58

Write $m$ for Lebesgue measure, fix a compact interval $[a,b]$ , and let $f : [a,b] \to \mathbb{R}$ be bounded.

Suppose $f$ possesses a Riemann integral. Then $f$ belongs to the Lebesgue measurable functions; being bounded on a set of finite measure it is also Lebesgue integrable there; and the two numbers coincide, $\int_{a}^{b} f(x) \, dx \;=\; \int_{[a,b]} f \, dm .$
Conversely, what characterises the Riemann integrable $f$ among the bounded ones is a smallness condition on their discontinuities: $f$ has a Riemann integral exactly when $\{x \in [a,b] : f \text{ is not continuous at } x\}$ is a set of measure zero.

So the proper Riemann integral is a special case of the Lebesgue integral. Improper Riemann integrals are not all covered. Suppose $f$ has a Riemann integral over $[0,b]$ for each $b > 0$ ; if $f$ is also Lebesgue integrable over $[0,\infty)$ then $\int_{[0,\infty)} f \, dm = \lim_{b \to \infty} \int_{0}^{b} f(x)\, dx$ , yet that limit may converge for an $f$ possessing no Lebesgue integral at all. Folland's example (Folland (1999), §2.3, p. 57) is $f = \sum_{n \ge 1} (-1)^{n} n^{-1} \mathbf{1}_{(n, n+1]}$ , whose positive and negative parts both have infinite integral.

Idea of the proofOriginal

The Riemann integral chops the domain and asks how high the function is over each piece; the Lebesgue integral chops the range and asks how wide the function is at each height. Stakgold and Holst (Stakgold & Holst (2011), §0.7, p. 42) put it as a shopkeeper counting a day's takings: transaction by transaction, or by sorting the coins into denominations first and multiplying each denomination by how many there are. Sorting by denomination is what the three stages do. A simple function has finitely many denominations and its total is a finite sum, so stage 1 is forced on us once we agree that a rectangle of height $a$ over a set of measure $\mu(E)$ should contribute $a\,\mu(E)$ . Stage 2 says that the only honest value for a general non-negative $f$ is the best a simple function underneath it can do; monotonicity then comes for free, because enlarging $f$ enlarges the family of competitors. Stage 3 is bookkeeping: integrate the positive and negative parts separately and subtract, which is legitimate exactly when the subtraction is not $\infty - \infty$ .

The price of sorting by denomination is that the level sets $f^{-1}([c,d))$ may be complicated, so a theory of measure has to come first; the reward is that a limit of functions perturbs each level set a little rather than perturbing the whole domain, which is why the convergence theorems of Monotone convergence and Fatou's lemma and Dominated convergence are available here and not for the Riemann integral.

Proof(The elementary properties)AdaptedAdapted from Folland (1999), §2.2, proof of Prop. 2.13, p. 50, proof of Prop. 2.16, p. 51, and §2.3, proofs of Props. 2.21 and 2.22, p. 53

Step 1: monotonicity and scaling on $\mathcal{L}^{+}$ (claim 1). If $f \le g$ then every simple $\varphi$ with $0 \le \varphi \le f$ also satisfies $0 \le \varphi \le g$ , so the supremum defining $\int g$ is taken over a larger family and cannot be smaller. For $c > 0$ the map $\varphi \mapsto c\varphi$ is a bijection of the simple minorants of $f$ onto those of $cf$ , and it multiplies the stage-1 sum by $c$ because $(c\varphi)^{-1}(\{ca_j\}) = E_j$ ; for $c = 0$ both sides vanish by the convention $0\cdot\infty = 0$ .

Step 2: additivity on simple functions (claim 2). Let $\varphi = \sum_{j} a_j \mathbf{1}_{E_j}$ and $\psi = \sum_{k} b_k \mathbf{1}_{F_k}$ be standard representations, and adjoin to each the set where it vanishes, so that $\{E_j\}$ and $\{F_k\}$ are now finite measurable partitions of $X$ . The sets $E_j \cap F_k$ form a common refinement, on each of which $\varphi + \psi$ is the constant $a_j + b_k$ . Finite additivity of $\mu$ gives $\mu(E_j) = \sum_{k} \mu(E_j \cap F_k)$ and $\mu(F_k) = \sum_{j} \mu(E_j \cap F_k)$ , so

\int \varphi + \int \psi = \sum_{j,k} a_j \,\mu(E_j \cap F_k) + \sum_{j,k} b_k\, \mu(E_j \cap F_k) = \sum_{j,k} (a_j + b_k)\, \mu(E_j \cap F_k),

and the last sum is the stage-1 integral of $\varphi + \psi$ , collecting together the cells on which $a_j + b_k$ takes the same value — legitimate because $\mu$ is finitely additive and the cells are disjoint.

Step 3: each simple function induces a measure (claim 3). $\int_{A} \varphi \, d\mu = \sum_{j} a_j\, \mu(A \cap E_j)$ is $0$ at $A = \emptyset$ , and for a disjoint sequence $A_1, A_2, \dots$ countable additivity of $\mu$ gives $\mu\bigl(\bigcup_i A_i \cap E_j\bigr) = \sum_{i} \mu(A_i \cap E_j)$ for each of the finitely many $j$ ; interchanging the finite sum over $j$ with the sum over $i$ — allowed, all terms being non-negative — gives countable additivity of $A \mapsto \int_A \varphi$ .

Step 4: linearity on $L^{1}$ (claim 4). Scalar multiplication is step 1 applied to $f^{\pm}$ , with the roles of $f^{+}$ and $f^{-}$ exchanged when $c < 0$ . For additivity the obstacle is that $(f+g)^{\pm}$ is not $f^{\pm} + g^{\pm}$ . Write $h = f + g$ ; then $h^{+} - h^{-} = f^{+} - f^{-} + g^{+} - g^{-}$ , and rearranging so that only non-negative functions appear,

h^{+} + f^{-} + g^{-} \;=\; h^{-} + f^{+} + g^{+} .

Both sides are in $\mathcal{L}^{+}$ , so additivity of the integral on $\mathcal{L}^{+}$ — which is where the monotone convergence theorem enters, applied to the increasing simple approximations supplied by Measurable functions — turns this into $\int h^{+} + \int f^{-} + \int g^{-} = \int h^{-} + \int f^{+} + \int g^{+}$ . Every term is finite because $\lvert h \rvert \le \lvert f \rvert + \lvert g \rvert$ is integrable, so the terms may be moved across the equality sign, and regrouping gives $\int h = \int f + \int g$ . The complex case follows by taking real and imaginary parts.

Step 5: the triangle inequality (claim 5). For real $f$ , $\lvert \int f \rvert = \lvert \int f^{+} - \int f^{-} \rvert \le \int f^{+} + \int f^{-} = \int \lvert f \rvert$ . For complex $f$ with $\int f \ne 0$ , rotate: let $\alpha$ be the unit complex number with $\alpha \int f = \lvert \int f \rvert$ . Then $\lvert \int f \rvert = \int \alpha f$ is real, so it equals its own real part, and by linearity

\Bigl\lvert \int f \Bigr\rvert = \int \operatorname{Re}(\alpha f) \le \int \lvert \operatorname{Re} (\alpha f) \rvert \le \int \lvert \alpha f \rvert = \int \lvert f \rvert ,

using the real case in the middle and $\lvert \alpha \rvert = 1$ at the end.

Step 6: null sets (claim 6). Let $f \in \mathcal{L}^{+}$ . If $f = 0$ a.e. and $\varphi = \sum_j a_j \mathbf{1}_{E_j} \le f$ is simple, then each $E_j$ with $a_j > 0$ lies in the null set $\{f > 0\}$ , so $\mu(E_j) = 0$ and $\int \varphi = 0$ ; the supremum over such $\varphi$ is $0$ . Conversely suppose $f$ is not $0$ a.e. Since $\{f > 0\} = \bigcup_{n \ge 1} A_n$ with $A_n = \{f > 1/n\}$ , continuity of $\mu$ from below forces $\mu(A_n) > 0$ for some $n$ ; then $\mathbf{1}_{A_n}/n$ is a simple minorant of $f$ and $\int f \ge \mu(A_n)/n > 0$ . For the second part of claim 6: if $\int \lvert f - g \rvert = 0$ then for every $E$ , $\lvert \int_E f - \int_E g\rvert = \lvert \int (f-g)\mathbf{1}_E \rvert \le \int \lvert f - g\rvert = 0$ by claim 5. If instead $f \ne g$ on a set of positive measure, then one of the four non-negative functions $(\operatorname{Re}(f-g))^{\pm}$ , $(\operatorname{Im}(f-g))^{\pm}$ is positive on a set $E$ of positive measure; on that $E$ the opposite part vanishes, so by the first part $\int_E (f-g)$ has non-zero real or imaginary part, and $\int_E f \ne \int_E g$ .

Step 7: the crude bound (claim 7). With $M = \sup_{(a,b)} \lvert f \rvert < \infty$ , the constant $M$ is a simple function on $(a,b)$ with integral $M(b-a) < \infty$ , so $\int_a^b \lvert f \rvert \le M(b-a)$ by claim 1 and $f$ is integrable there; claim 5 gives the bound. $\square$

Proof(Comparison with the Riemann integral)AdaptedAdapted from Folland (1999), §2.3, proof of Thm 2.28, p. 57

We prove part 1 and indicate where part 2 comes from.

Step 1: turn Darboux sums into integrals of simple functions. A partition $P : a = t_0 < t_1 < \dots < t_n = b$ carries the two step functions

G_P = \sum_{j=1}^{n} M_j \,\mathbf{1}_{(t_{j-1}, t_j]} , \qquad g_P = \sum_{j=1}^{n} m_j \,\mathbf{1}_{(t_{j-1}, t_j]} , \qquad M_j = \sup_{[t_{j-1},t_j]} f, \quad m_j = \inf_{[t_{j-1},t_j]} f ,

which are simple because the intervals are Borel sets and $f$ is bounded, so the suprema and infima are finite. Since $m(\,(t_{j-1},t_j]\,) = t_j - t_{j-1}$ , stage 1 of the definition gives $\int G_P \, dm = S_P f$ and $\int g_P \, dm = s_P f$ , the upper and lower Darboux sums. Note that $g_P \le f \le G_P$ at every point of $(a,b]$ but not at the left endpoint, where both step functions vanish while $f(a)$ may not; since $\{a\}$ is null this costs nothing, and all the squeezes below are read almost everywhere.

Step 2: choose a nested sequence of partitions. Riemann integrability means $\inf_P S_P f = \sup_P s_P f = \int_a^b f(x)\,dx =: I$ . For each $k \ge 1$ pick a partition $Q_k$ realising the infimum to within $1/k$ , so $S_{Q_k} f \le I + 1/k$ , and a partition $R_k$ realising the supremum to within $1/k$ , so $s_{R_k} f \ge I - 1/k$ . Write $U_k$ for the uniform partition of $[a,b]$ into $k$ equal pieces, and define the $P_k$ recursively as sets of partition points,

P_1 = Q_1 \cup R_1 \cup U_1, \qquad P_k = P_{k-1} \cup Q_k \cup R_k \cup U_k \quad (k \ge 2).

The recursion is what makes the sequence nested: $P_k \supseteq P_{k-1}$ by construction, whereas the uniform partitions by themselves are not nested (the point $\tfrac12(a+b)$ of $U_2$ is absent from $U_3$ ). Refining a partition can only raise an infimum and lower a supremum over each piece, so $g_{P_k}$ increases pointwise with $k$ and $G_{P_k}$ decreases. Refining also only improves Darboux sums, so $I \le S_{P_k} f \le S_{Q_k} f \le I + 1/k$ and $I \ge s_{P_k} f \ge s_{R_k} f \ge I - 1/k$ ; hence both $S_{P_k} f$ and $s_{P_k} f$ converge to $I$ . The sets $U_k$ are carried along only to force the mesh of $P_k$ to zero, which step 6 uses.

Step 3: pass to the limit. Set $G = \lim_k G_{P_k}$ and $g = \lim_k g_{P_k}$ , which exist pointwise by monotonicity and are measurable as pointwise limits of simple functions. Each $\lvert G_{P_k} \rvert$ and $\lvert g_{P_k} \rvert$ is bounded by the constant $\sup_{[a,b]} \lvert f\rvert$ , which is integrable on the bounded interval $[a,b]$ , so dominated convergence applies to both sequences and yields

\int_{[a,b]} G \, dm = \lim_k S_{P_k} f = \int_a^b f(x)\,dx = \lim_k s_{P_k} f = \int_{[a,b]} g \, dm .

(No circularity, as the note at the head of the unit records: dominated convergence is proved from the definition above and from Fatou's lemma, neither of which mentions the Riemann integral.)

Step 4: identify $f$ with $G$ . By step 1, $g_{P_k} \le f \le G_{P_k}$ on $(a,b]$ for each $k$ , hence $g \le f \le G$ there and $G - g \ge 0$ on $(a,b]$ — and $G \ge g$ everywhere, the two being limits of $G_{P_k} \ge g_{P_k}$ . Step 3 gives $\int (G-g) \, dm = 0$ , so $G = g$ almost everywhere by claim 6 of the elementary properties, and therefore $G = f$ almost everywhere, the squeeze forcing equality at every point of $(a,b]$ where $G = g$ , and $\{a\}$ being null.

Step 5: conclude. $G$ is measurable, and $f$ differs from $G$ only on a null set. Lebesgue measure is complete — every subset of a null set is measurable, which is the point at which the completeness established in Lebesgue measure is used — so $\{f > c\}$ differs from $\{G > c\}$ by a subset of a null set and is measurable for each $c$ ; thus $f$ is Lebesgue measurable. Being bounded on a set of finite measure it is integrable, and $\int_{[a,b]} f \, dm = \int_{[a,b]} G \, dm = \int_a^b f(x)\,dx$ .

Step 6: part 2. Folland obtains the criterion by comparing $G$ and $g$ with the upper and lower envelopes $H(x) = \lim_{\delta \downarrow 0} \sup_{\lvert y - x\rvert < \delta} f(y)$ and $h(x) = \lim_{\delta \downarrow 0} \inf_{\lvert y - x \rvert < \delta} f(y)$ : one has $H = h$ exactly at the points of continuity of $f$ , while $H = G$ and $h = g$ off the (countable, hence null) set of partition points, so $\int (H - h)\,dm = \int (G-g)\,dm$ . Riemann integrability is $\int(G-g)\,dm = 0$ , which by claim 6 is $H = h$ a.e., that is, continuity a.e. We follow Folland (Folland (1999), §2.3, Thm 2.28b and Exercise 23, pp. 57, 59), who states the criterion as part of the theorem and leaves this route as an exercise. $\square$

Remark(Where the hypotheses are used)Original

$0 \cdot \infty = 0$ is a convention, not a theorem, and it is used in stage 1 whenever a simple function vanishes on a set of infinite measure. Without it, $\int 0 \, dm$ over $\mathbb{R}$ would be undefined. It is also what makes $\int_A f$ vanish for every $f$ when $\mu(A) = 0$ , however large $f$ is on $A$ — even $+\infty$ .
Non-negativity is what makes stage 2 unambiguous. The supremum of a set of non-negative numbers always exists in $[0,\infty]$ , so no hypothesis is needed and no limit has to be shown to exist. Grimmett and Stirzaker (Grimmett & Stirzaker (2020), §5.6(B), items (8)–(9), pp. 200–201) instead define the second stage as $\lim_n \mathbb{E}(X_n)$ along one increasing sequence of simple variables and then have to argue that the limit does not depend on the sequence chosen; the supremum form avoids that step, and the monotone convergence theorem of Monotone convergence and Fatou's lemma shows the two agree.
Stage 3 needs one of $\int f^{+}$ , $\int f^{-}$ to be finite, or the difference is $\infty - \infty$ . The standard witness is $f(x) = 1/x$ on $\mathbb{R} \setminus \{0\}$ , where $\int f^{+} = \int f^{-} = \infty$ : the symmetric limit $\lim_{\varepsilon \downarrow 0}\bigl(\int_{-1}^{-\varepsilon} + \int_{\varepsilon}^{1}\bigr) f = 0$ exists as a principal value, but $f$ has no Lebesgue integral, because that limit depends on the symmetry of the exhaustion and the Lebesgue integral depends on no exhaustion at all.
Boundedness in the Riemann comparison is used twice: in step 1 to make the $M_j$ finite, so that $G_P$ is a genuine simple function, and in step 3 to supply the constant dominating function. Both fail for unbounded $f$ , which is why the theorem is about proper Riemann integrals only. The example $f = \sum_{n}(-1)^n n^{-1}\mathbf{1}_{(n,n+1]}$ above shows the other half of the gap: an improper Riemann integral may converge by cancellation, while the Lebesgue integral, built from $f^{+}$ and $f^{-}$ separately, never cancels.
Completeness of $m$ is used only in step 5. Over an incomplete measure space the same argument shows that $f$ agrees a.e. with a measurable function, which is all one ever needs, but $f$ itself need not be measurable.
Lebesgue theory does not turn divergent integrals into finite numbers. Stakgold and Holst (Stakgold & Holst (2011), §0.7, Example 3 and the discussion following it, pp. 43–44) make the point with $f(x) = x^{-a}$ on $(0,1]$ : for $0 < a < 1$ the improper Riemann integral and the Lebesgue integral agree, while for $a \ge 1$ their verdict is that no integral of either sort is available — neither in the improper Riemann sense nor in the Lebesgue sense. That verdict belongs to their construction, which defines the integral of a non-negative $f$ as a greatest lower bound over sequences of continuous approximants whose Riemann integrals have a common bound, so that no such bound means no integral. Under the construction this unit follows, the two differ: $x^{-a}$ is non-negative and measurable, so stage 2 assigns it an integral in $[0,\infty]$ unconditionally, and that integral exists and equals $+\infty$ for $a \ge 1$ . What fails is integrability, not existence. The arithmetic is the same either way, and nothing elsewhere in the handbook depends on the phrasing; we keep the Folland convention of stage 2 throughout.
Enlarging the class of integrands is not, by itself, the main gain. Folland (Folland (1999), §2.3, p. 58) names exactly two genuine gains of the Lebesgue theory, and lists them in this order. The first is the strength of its convergence theorems, monotone and dominated convergence among them. The second is that more functions can be integrated — about which he observes at once that classical computations almost never need the extra generality, so that what makes it matter is a different thing, which he calls its crucial effect: function spaces whose metric is written as an integral turn out to be complete when their members are Lebesgue integrable and incomplete when only Riemann integrable functions are admitted. That crucial effect is exactly the defect Kreyszig (Kreyszig (1978), §2.2, Example 2.2-7, p. 62) exhibits: the continuous functions on $[a,b]$ are not complete under $\lVert x \rVert = (\int_a^b x(t)^2\,dt)^{1/2}$ , and their completion is $L^2[a,b]$ . Reed and Simon (Reed & Simon (1980), §1.3, pp. 12–13) open their treatment with the same observation for the $L^1$ norm.

Method(Computing a Lebesgue integral)Original

Check measurability first. Nothing below is meaningful otherwise. Continuity, monotonicity, or being a pointwise limit of measurable functions all suffice (Measurable functions).
If the integrand is Riemann integrable on a compact interval, stop. By the comparison theorem the answer is the Riemann integral, so use calculus. This covers almost every explicit integral.
Otherwise look for a null set to discard. If $f = g$ outside a set of measure zero and $g$ is easy, then $\int f = \int g$ by claim 6. This disposes of integrands defined by a different rule on the rationals, on a Cantor set, or at the endpoints.
For a genuinely non-negative integrand, sort by height. Use the dyadic truncation $\varphi_n = \sum_{k=0}^{n2^{n}-1} \frac{k}{2^{n}} \,\mathbf{1}_{\{ k2^{-n} \,\le\, f \,<\, (k+1)2^{-n}\}} \;+\; n\,\mathbf{1}_{\{f \ge n\}} ,$ which is simple, satisfies $0 \le \varphi_n \le f$ , increases with $n$ and converges pointwise to $f$ . Each $\int \varphi_n$ is a finite sum of measures of level sets, and $\int \varphi_n \uparrow \int f$ by the monotone convergence theorem of Monotone convergence and Fatou's lemma — the forward reference flagged in the note at the head of the unit. On $\{f < n\}$ the error obeys $0 \le f - \varphi_n < 2^{-n}$ , so the truncation level $n$ and the mesh $2^{-n}$ control the accuracy.
For a signed or complex integrand, split and check finiteness. Compute $\int \lvert f \rvert$ first. If it is infinite, there is no integral to compute unless one of $\int f^{\pm}$ is finite, in which case the answer is $\pm\infty$ . If it is finite, integrate $f^{+}$ and $f^{-}$ , or real and imaginary parts, separately and combine by linearity.
Sanity-check with the crude bound. On a set of finite measure, $\lvert \int_A f \rvert \le \mu(A)\, \sup_A \lvert f \rvert$ catches most arithmetic slips.

Example(A function with no Riemann integral, integrated from the definition)Original

On $X = [0,1]$ with Lebesgue measure, let $\mathbb{Q}_1 = \mathbb{Q} \cap [0,1]$ and define

f(x) = \begin{cases} 2, & x \in \mathbb{Q}_1, \\ x^{2}, & x \notin \mathbb{Q}_1 .\end{cases}

It is measurable and bounded. $f = 2\,\mathbf{1}_{\mathbb{Q}_1} + x^{2}\mathbf{1}_{[0,1]\setminus\mathbb{Q}_1}$ is a sum of products of a continuous function with indicators of Borel sets, and $0 \le f \le 2$ .

It is not Riemann integrable. Every subinterval contains both rationals and irrationals, so on each subinterval the supremum of $f$ is $2$ while the infimum is at most $1$ . Hence $S_P f \ge 2$ and $s_P f \le 1$ for every partition $P$ , and the Darboux sums cannot meet. Equivalently, $f$ is discontinuous at every point of $[0,1]$ , a set of measure $1 \ne 0$ , so part 2 of the comparison theorem refuses it. Stakgold and Holst (Stakgold & Holst (2011), §0.7, Example 2, pp. 43–44) make the same point with the Dirichlet function, the case $f = \mathbf{1}_{\mathbb{Q}_1}$ .

Step 1: discard the null set. $\mathbb{Q}_1$ is countable; enclosing its $k$ th member in an interval of length $\varepsilon 2^{-k}$ covers it by a total length $\varepsilon \sum_{k \ge 1} 2^{-k} = \varepsilon$ , for every $\varepsilon > 0$ , so $m(\mathbb{Q}_1) = 0$ . Thus $f(x) = x^{2}$ a.e. and claim 6 gives $\int_{[0,1]} f \, dm = \int_{[0,1]} x^{2} \, dm = \tfrac13$ , the last step by the comparison theorem applied to the continuous function $x^2$ .

Step 2: confirm it from the definition, by sorting by height. Apply step 4 of the method to $g(x) = x^{2}$ on $[0,1)$ with $n \ge 1$ , so that the truncation at height $n$ never bites. For $0 \le k < 2^{n}$ ,

\bigl\{\, k2^{-n} \le x^{2} < (k+1)2^{-n} \,\bigr\} = \bigl[\, 2^{-n/2}\sqrt{k},\ 2^{-n/2}\sqrt{k+1} \,\bigr), \qquad \text{of measure } 2^{-n/2}\bigl(\sqrt{k+1} - \sqrt{k}\bigr),

because $x \mapsto x^{2}$ is increasing on $[0,1)$ . Stage 1 of the definition therefore gives the finite sum

\int \varphi_n \, dm \;=\; 2^{-3n/2} \sum_{k=0}^{2^{n}-1} k\bigl(\sqrt{k+1} - \sqrt{k}\bigr) .

At $n = 2$ the four terms are $0$ , $\sqrt2 - 1$ , $2(\sqrt3-\sqrt2)$ and $3(2-\sqrt3)$ , which sum to $5 - \sqrt2 - \sqrt3$ , so

\int \varphi_2 \, dm = \frac{5 - \sqrt2 - \sqrt3}{8} = 0.2317169\ldots

Step 3: check the error bound and the limit. On $\{g < n\}$ we have $0 \le g - \varphi_n < 2^{-n}$ , so $0 \le \tfrac13 - \int \varphi_n \le 2^{-n}$ . At $n = 2$ : $\tfrac13 - 0.2317169 = 0.1016164 \le 0.25$ , as required. Continuing,

$n$	$\int \varphi_n \, dm$	$\tfrac13 - \int \varphi_n \, dm$	$2^{-n}$
$2$	$0.2317170$	$0.1016164$	$0.25$
$4$	$0.3051688$	$0.0281645$	$0.0625$
$8$	$0.3314303$	$0.0019030$	$0.00390625$
$12$	$0.3332121$	$0.0001213$	$0.000244$

The supremum over all simple minorants is therefore $\tfrac13$ , confirming

\int_{[0,1]} f \, dm = \frac13 ,

a number the Riemann theory cannot produce for this $f$ at all.

ApplicationOriginal

Expectation in probability is one Lebesgue integral. On a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ the expectation of a random variable is $\mathbb{E}(X) = \int_{\Omega} X \, d\mathbb{P}$ , built by exactly the three stages above: finitely many outcomes weighted by their probabilities, then a supremum, then the split into $X^{+}$ and $X^{-}$ . As Grimmett and Stirzaker (Grimmett & Stirzaker (2020), §5.6(B), items (7)–(11) and (15), pp. 200–202) point out, this single definition replaces the separate sum-over-values and integral-against-a-density formulas of the discrete and continuous cases, and specialises to the Lebesgue–Stieltjes integral $\int g(x)\,dF(x)$ when the integration is pushed forward onto $\mathbb{R}$ . The condition $\mathbb{E}\lvert X \rvert < \infty$ in every theorem about expectations is stage 3's finiteness requirement, and random variables with no mean (a Cauchy variable, for instance) are precisely those for which stage 3 is $\infty - \infty$ . See Expectation, variance and covariance.

Quantum states live in a completion. A wavefunction is an element of $L^{2}(\mathbb{R}^{3})$ , and what makes that space usable — the existence of limits of Cauchy sequences, hence of spectral decompositions and of convergent variational schemes — is its completeness, which holds for the Lebesgue integral and fails for the Riemann integral on the same formula, as Kreyszig (Kreyszig (1978), §2.2, Example 2.2-7, p. 62) shows for continuous functions on an interval. Two wavefunctions differing on a set of measure zero are the same physical state, which is the equivalence-class convention of the note above, not a physical postulate.

Signal energy and the "almost everywhere" clause in engineering statements. For a finite-energy signal $s$ one writes $E = \int_{\mathbb{R}} \lvert s(t)\rvert^{2}\,dt$ . Because null sets contribute nothing, an idealised signal may be left undefined at the switching instants of a sampler or at the jumps of a square wave without any effect on its energy, its Fourier coefficients, or the reconstruction theorem of The sampling theorem. This is why statements such as "the sampled signal is recovered almost everywhere" are the right ones to prove; the pointwise statement is false at a measure-zero set of instants and would be useless even if true.

Variational problems need the larger class. A minimisation method such as Rayleigh–Ritz produces a minimiser as a limit of a minimising sequence, and the limit of a sequence of smooth functions is in general only measurable. Formulating energies as Lebesgue integrals, over the Sobolev spaces built from them, is what makes the limit admissible; over the Riemann-integrable functions the minimising sequence may simply leave the space.

Convergence theorems

Theorem

Monotone convergence and Fatou's lemma

Adapted

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For non-negative measurable integrands, an increasing pointwise limit may always be taken outside the integral, a non-negative series may always be integrated term by term, and for an arbitrary sequence the integral of the lower limit never exceeds the lower limit of the integrals.

Prerequisites: Sigma-algebras and measures, Measurable functions, The Lebesgue integral

Definition(The set-up)AdaptedAdapted from Folland (1999), §2.2, p. 49 and §2.3, p. 54; Reed & Simon (1980), §1.3, p. 17

Throughout, $(X, \Sigma, \mu)$ is a measure space and $\mathcal{L}^{+}$ is the set of measurable functions $f : X \to [0,\infty]$ , with $\int f \, d\mu \in [0,\infty]$ as defined in The Lebesgue integral. No integrand below is assumed to have a finite integral.

Write $f_n \uparrow f$ for $f_n$ increases pointwise to $f$ : $f_n(x) \le f_{n+1}(x)$ for every $n$ and every $x$ , and $f_n(x) \to f(x)$ . Since an increasing sequence in $[0,\infty]$ always converges in $[0,\infty]$ , the limit function is $f = \sup_n f_n$ , and it is measurable whenever the $f_n$ are (Measurable functions).

A property holds almost everywhere (a.e.) when the set where it fails is contained in a set of measure zero. For a sequence $(g_n)$ of measurable functions,

\liminf_{n \to \infty} g_n \;=\; \lim_{k \to \infty} \ \inf_{n \ge k} g_n \;=\; \sup_{k \ge 1} \ \inf_{n \ge k} g_n ,

the inner infima increasing with $k$ ; it is measurable, and is defined pointwise even when $(g_n(x))$ has no limit.

Theorem(Monotone convergence theorem)AdaptedAdapted from Folland (1999), §2.2, Thm 2.14 and Cor. 2.17, pp. 50–51; Evans (2010), App. E.3, Thm 4, p. 686

Let $(f_n)_{n \ge 1}$ be a sequence in $\mathcal{L}^{+}$ with $f_n \uparrow f$ . Then

\int f \, d\mu \;=\; \lim_{n \to \infty} \int f_n \, d\mu ,

both sides being allowed the value $+\infty$ . The conclusion is unchanged if the hypothesis holds only almost everywhere, provided the limit is still required to be an integrand: if $(f_n) \subset \mathcal{L}^{+}$ , $f \in \mathcal{L}^{+}$ , and $f_n(x) \uparrow f(x)$ for a.e. $x$ , then still $\int f = \lim_n \int f_n$ . The measurability of $f$ has to be assumed here and not deduced: off the exceptional null set $f$ is unconstrained, so on a measure space that is not complete $f$ may fail to be measurable and $\int f$ may have no meaning. Folland (Folland (1999), §2.2, Cor. 2.17, p. 51) carries $f \in \mathcal{L}^{+}$ in the hypotheses for exactly this reason.

Three things are not assumed: that any $\int f_n$ is finite, that $f$ is finite anywhere, or that the convergence is uniform or even dominated.

Theorem(Termwise integration of a non-negative series)AdaptedAdapted from Folland (1999), §2.2, Thm 2.15, p. 51; Grimmett & Stirzaker (2020), §5.6(B), item (13), p. 201

Let $(f_n)_{n \ge 1}$ be a finite or infinite sequence in $\mathcal{L}^{+}$ and $f = \sum_n f_n$ pointwise. Then

\int \sum_{n} f_n \, d\mu \;=\; \sum_{n} \int f_n \, d\mu .

Again both sides may be $+\infty$ , and no convergence hypothesis on the series is needed: the terms are non-negative, so the partial sums increase and the series converges in $[0,\infty]$ at every point. Taking $X = \mathbb{N}$ with counting measure turns the theorem into the statement that a non-negative double series may be summed in either order — but only once the integral against counting measure has been identified with the sum $\sum_{k} f(k)$ , which Folland (Folland (1999), §2.2, Exercise 22, p. 59) leaves as an exercise and we do not prove here; see Fubini and Tonelli theorems, where the interchange of two sums is obtained as a case of Tonelli's theorem.

Theorem(Fatou's lemma)AdaptedAdapted from Folland (1999), §2.2, Thm 2.18 and Cor. 2.19, p. 52; Evans (2010), App. E.3, Thm 3, p. 686

Let $(f_n)_{n \ge 1}$ be any sequence in $\mathcal{L}^{+}$ . Then

\int \Bigl( \liminf_{n \to \infty} f_n \Bigr) d\mu \;\le\; \liminf_{n \to \infty} \int f_n \, d\mu .

No monotonicity, no convergence and no domination is required. Consequently, if $(f_n) \subset \mathcal{L}^{+}$ , $f \in \mathcal{L}^{+}$ and $f_n \to f$ a.e., then $\int f \, d\mu \le \liminf_n \int f_n \, d\mu$ : mass can be lost in the limit, never gained. As in the monotone convergence theorem, $f \in \mathcal{L}^{+}$ is a hypothesis of this corollary and not a consequence of the others, which is how Folland (Folland (1999), §2.2, Cor. 2.19, p. 52) states it.

Idea of the proofOriginal

In the monotone convergence theorem the inequality $\int f \ge \lim_n \int f_n$ is free: the limit function dominates each term, and the integral is monotone. (Fatou's lemma has no such free half — for an arbitrary sequence the $\liminf$ dominates nothing — which is why it is an inequality and not an identity.) The content of monotone convergence is the other direction, and one device supplies it. To beat the supremum defining $\int f$ one must beat every simple $\varphi$ with $0 \le \varphi \le f$ . Since $f_n \uparrow f$ , the set where $f_n$ has caught up with $\varphi$ grows to all of $X$ — but only if we give ourselves a margin, because $f_n$ may approach $\varphi$ from below without ever reaching it. Shrinking $\varphi$ to $a\varphi$ with $a < 1$ buys that margin, and then the sets $\{f_n \ge a\varphi\}$ increase to $X$ , so continuity of a measure from below delivers $\lim_n \int f_n \ge a \int \varphi$ . Letting $a \uparrow 1$ and taking the supremum over $\varphi$ finishes it.

The other two results are consequences rather than new arguments. A series of non-negative terms is an increasing sequence of partial sums, so termwise integration is monotone convergence plus additivity on two summands. And Fatou's lemma is monotone convergence applied to the increasing sequence $g_k = \inf_{n \ge k} f_n$ , each $g_k$ being a lower bound for all later $f_n$ and hence for their integrals.

Proof(Monotone convergence)AdaptedAdapted from Folland (1999), §2.2, proof of Thm 2.14, p. 50 and proof of Cor. 2.17, p. 51

Step 1: the easy inequality. $f_n \le f_{n+1} \le f$ for every $n$ , so by monotonicity of the integral the numbers $\int f_n$ increase and are bounded above by $\int f$ . An increasing sequence in $[0,\infty]$ converges in $[0,\infty]$ , so $L := \lim_n \int f_n$ exists and $L \le \int f$ .

Step 2: fix a competitor and a margin. Let $\varphi = \sum_{j=1}^{N} a_j \mathbf{1}_{E_j}$ be simple with $0 \le \varphi \le f$ , and fix $a \in (0,1)$ . Put

A_n \;=\; \{\, x \in X : f_n(x) \ge a\,\varphi(x) \,\} \in \Sigma .

Step 3: the sets $A_n$ increase to $X$ . They increase because the $f_n$ do. They exhaust $X$ : given $x$ , either $\varphi(x) = 0$ , and then $x \in A_1$ since $f_1 \ge 0$ ; or $\varphi(x) > 0$ , and then $a\varphi(x) < \varphi(x) \le f(x)$ strictly, so some $f_n(x)$ exceeds $a\varphi(x)$ , because $f_n(x) \uparrow f(x)$ . This is the one place the margin $a < 1$ is used, and it is needed: at a point where $f_n(x) < \varphi(x) = f(x)$ for every $n$ , the sets $\{f_n \ge \varphi\}$ would miss $x$ altogether.

Step 4: continuity from below. By monotonicity of the integral and then by restriction to $A_n$ ,

\int f_n \, d\mu \;\ge\; \int_{A_n} f_n \, d\mu \;\ge\; a \int_{A_n} \varphi \, d\mu \;=\; a \sum_{j=1}^{N} a_j\, \mu(E_j \cap A_n) .

For each fixed $j$ the sets $E_j \cap A_n$ increase with $n$ and their union is $E_j$ , so $\mu(E_j \cap A_n) \uparrow \mu(E_j)$ by continuity of a measure from below (Sigma-algebras and measures). The sum over $j$ is finite, so letting $n \to \infty$ in the display gives $L \ge a \int \varphi \, d\mu$ .

Step 5: remove the margin and take the supremum. The inequality $L \ge a\int\varphi$ holds for every $a \in (0,1)$ , so letting $a \uparrow 1$ gives $L \ge \int \varphi \, d\mu$ . This holds for every simple $\varphi$ between $0$ and $f$ , and $\int f$ is by definition the supremum of such $\int \varphi$ , so $L \ge \int f$ . With step 1, $L = \int f$ .

Step 6: the almost-everywhere version. Suppose $f, f_n \in \mathcal{L}^{+}$ and $f_n \uparrow f$ on $X \setminus Z$ with $\mu(Z) = 0$ , and set $E = X \setminus Z$ . (Measurability of $f$ is where the extra hypothesis is spent: without it $\int f$ is not defined, and the identity to be proved has no left-hand side.) Then $f_n \mathbf{1}_E \uparrow f \mathbf{1}_E$ everywhere, so the version just proved gives $\int f\mathbf{1}_E = \lim_n \int f_n \mathbf{1}_E$ . Finally $f\mathbf{1}_Z$ and $f_n\mathbf{1}_Z$ vanish off $Z$ , hence are $0$ a.e., hence have zero integral by claim 6 of The Lebesgue integral. Splitting $f = f\mathbf{1}_E + f\mathbf{1}_Z$ then needs additivity of the integral on $\mathcal{L}^{+}$ , which is step 1 of the next proof; that step uses only the everywhere version established in steps 1–5, so invoking it here is not circular. It gives $\int f = \int f\mathbf{1}_E + \int f\mathbf{1}_Z = \int f \mathbf{1}_E$ and likewise $\int f_n = \int f_n \mathbf{1}_E$ for each $n$ , so the two limits coincide. $\square$

Proof(Termwise integration of a non-negative series)AdaptedAdapted from Folland (1999), §2.2, proof of Thm 2.15, p. 51

Step 1: two summands. Let $f, g \in \mathcal{L}^{+}$ . By the approximation theorem of Measurable functions choose simple $\varphi_k \uparrow f$ and $\psi_k \uparrow g$ with all terms non-negative. Then $\varphi_k + \psi_k$ is simple, non-negative, and increases to $f + g$ . Monotone convergence applies to all three sequences, and additivity of the integral on simple functions (The Lebesgue integral) applies at each fixed $k$ :

\int (f + g) \,d\mu = \lim_{k} \int (\varphi_k + \psi_k)\, d\mu = \lim_{k} \Bigl( \int \varphi_k + \int \psi_k \Bigr) = \int f \, d\mu + \int g \, d\mu ,

the last step because both limits exist in $[0,\infty]$ and are approached monotonically, so no $\infty - \infty$ can occur.

Step 2: finitely many summands. Induction on step 1 gives $\int \sum_{n=1}^{N} f_n = \sum_{n=1}^{N} \int f_n$ for every $N$ .

Step 3: pass to the infinite series. Put $S_N = \sum_{n=1}^{N} f_n$ . Each $f_n \ge 0$ , so $S_N \uparrow \sum_{n \ge 1} f_n$ pointwise in $[0,\infty]$ . Monotone convergence, applied now to $(S_N)_N$ , together with step 2 gives

\int \sum_{n \ge 1} f_n \, d\mu = \lim_{N \to \infty} \int S_N \, d\mu = \lim_{N \to \infty} \sum_{n=1}^{N} \int f_n \, d\mu = \sum_{n \ge 1} \int f_n \, d\mu ,

the final equality being the definition of the sum of a series of terms in $[0,\infty]$ . $\square$

Proof(Fatou's lemma)AdaptedAdapted from Folland (1999), §2.2, proof of Thm 2.18 and of Cor. 2.19, p. 52

Step 1: build an increasing sequence. For $k \ge 1$ set $g_k = \inf_{n \ge k} f_n$ , measurable and non-negative. Taking the infimum over a smaller index set can only increase it, so $g_k \le g_{k+1}$ , and by definition $g_k \uparrow \liminf_n f_n$ .

Step 2: compare each $g_k$ with the tail. For every $j \ge k$ we have $g_k \le f_j$ pointwise, hence $\int g_k \le \int f_j$ by monotonicity. The right-hand side is free to vary over $j \ge k$ , so

\int g_k \, d\mu \;\le\; \inf_{j \ge k} \int f_j \, d\mu .

Step 3: let $k \to \infty$ . The left-hand side tends to $\int \liminf_n f_n$ by monotone convergence applied to $g_k \uparrow \liminf_n f_n$ ; the right-hand side tends, by definition, to $\liminf_n \int f_n$ . Both sides are monotone in $k$ , so the inequality survives the limit:

\int \Bigl(\liminf_{n} f_n\Bigr) d\mu \;=\; \lim_{k} \int g_k \, d\mu \;\le\; \lim_{k}\ \inf_{j \ge k} \int f_j \, d\mu \;=\; \liminf_{n} \int f_n \, d\mu .

Step 4: the almost-everywhere corollary. Assume $f \in \mathcal{L}^{+}$ , so that $\int f$ is defined, and $f_n \to f$ off a null set $Z$ . Then $h := \liminf_n f_n$ is in $\mathcal{L}^{+}$ by step 1, and $h = f$ on $E := X \setminus Z$ . Both $f\mathbf{1}_Z$ and $h\mathbf{1}_Z$ vanish a.e., so both have zero integral by claim 6 of The Lebesgue integral; additivity on $\mathcal{L}^{+}$ (step 1 of the previous proof) then gives $\int f = \int f\mathbf{1}_E + \int f \mathbf{1}_Z = \int h \mathbf{1}_E + \int h\mathbf{1}_Z = \int h$ . Steps 1–3 give $\int h \le \liminf_n \int f_n$ , so $\int f \le \liminf_n \int f_n$ . $\square$

Remark(Where the hypotheses are used, and how the books differ)Original

Monotonicity is used three times in the first proof: twice in step 1, to make the numbers $\int f_n$ increase (so that $L = \lim_n \int f_n$ exists) and to give $L \le \int f$ ; and once in step 3, to make the sets $A_n$ increase. The third use is the essential one — the first two only organise the easy inequality — and it cannot be dropped. Folland (Folland (1999), §2.2, p. 51) gives the two ways it fails on $\mathbb{R}$ with Lebesgue measure: $\mathbf{1}_{(n,n+1)} \to 0$ pointwise with the area walking off to the right, and $n\,\mathbf{1}_{(0,1/n)} \to 0$ pointwise with the area growing taller as it narrows. Both sequences have every integral equal to $1$ while the limit has integral $0$ , so monotone convergence would be false and Fatou's inequality is strict: $0 < 1$ .
The margin $a < 1$ is not a technicality. Take $X = \{p\}$ a single point of measure $1$ and $f_n \equiv 1 - 1/n$ , so $f \equiv 1$ . With $\varphi = \mathbf{1}_{X}$ the sets $\{f_n \ge \varphi\}$ are all empty, and the proof would give nothing; the sets $\{f_n \ge a\varphi\}$ are eventually all of $X$ for each $a < 1$ .
Non-negativity is used in Fatou's lemma to guarantee that every integral appearing is defined. Grimmett and Stirzaker (Grimmett & Stirzaker (2020), §5.6(B), item (14), p. 202) state the lemma under the weaker hypothesis that $X_n \ge Y$ almost surely for a single $Y$ with $\mathbb{E}\lvert Y \rvert < \infty$ , noting that the case $Y = 0$ is the one normally used; our statement is that case. Stakgold and Holst (Stakgold & Holst (2011), §0.7, Thm 0.7.3, p. 47) state it for a sequence of measurable functions on a measurable $\Omega \subseteq \mathbb{R}^{n}$ , their Thm 0.7.2 immediately above carrying the condition $0 \le f_i \le f_{i+1}$ . We follow Folland (Folland (1999), §2.2, Thm 2.18, p. 52) and Evans (Evans (2010), App. E.3, Thm 3, p. 686), who require the $f_n$ to be non-negative, and in our proof the hypothesis enters twice: at step 1, so that $g_k \in \mathcal{L}^{+}$ and $\int g_k$ is defined without a finiteness assumption, and at step 3, so that monotone convergence may be applied to $(g_k)$ . Some lower bound is genuinely needed, since the sequence $-n\,\mathbf{1}_{(0,1/n)}$ on $\mathbb{R}$ has $\liminf = 0$ a.e. but every integral equal to $-1$ .
Two forms of the monotone convergence theorem are in circulation. Ours, from Folland and Evans, assumes only $f_n \uparrow f$ and concludes $\int f_n \to \int f$ in $[0,\infty]$ . Reed and Simon (Reed & Simon (1980), §1.3, Thm 1.10, p. 17) add the hypothesis $\int f_n \le C$ for all $n$ and strengthen the conclusion to $f \in \mathcal{L}^{1}$ with $\int \lvert f - f_n \rvert \to 0$ . These are not in conflict: under their extra hypothesis our version gives $\int f = \lim \int f_n \le C < \infty$ , so $f$ is integrable, and then $\lvert f - f_n \rvert = f - f_n$ has integral $\int f - \int f_n \to 0$ by linearity on $L^{1}$ — a subtraction that is legitimate precisely because $C$ is finite. Without a finite bound their conclusion is empty, since $f$ need not be integrable, while ours still says $\int f_n \to \infty$ . We state the Folland–Evans form and use the Reed–Simon form where $L^{1}$ convergence is wanted.
Equality in Fatou's lemma is not the typical case, and the lemma is normally used in exactly one direction: to bound the integral of a limit from above by what the sequence does. That is why it is the engine behind lower semicontinuity of energy functionals, a use Stakgold and Holst (Stakgold & Holst (2011), §0.7, p. 47) single out.
The three results are one result in disguise. Monotone convergence proves termwise integration and Fatou's lemma, as above, and Fatou's lemma in turn proves dominated convergence. Monotone convergence also closes a gap left in The Lebesgue integral: additivity $\int (f+g) = \int f + \int g$ on $\mathcal{L}^{+}$ is step 1 of the second proof and is not available directly from the supremum definition, because the simple minorants of $f + g$ are not all of the form $\varphi + \psi$ .

Method(Interchanging a limit or a sum with an integral, for non-negative integrands)Original

Check the sign first. If the integrand is non-negative, no integrability hypothesis of any kind is needed for an increasing limit or for a series: go straight to step 2 or 3. If it changes sign, these theorems do not apply; use dominated convergence.
Increasing limit. Verify $f_n \le f_{n+1}$ pointwise and identify $f = \lim_n f_n$ . Holding a.e. is enough, provided $f$ is separately known to be measurable: off the exceptional null set it is unconstrained, and on an incomplete space its measurability cannot be deduced. Then $\int f = \lim_n \int f_n$ . Both sides may be infinite, and this is often the cheapest way to prove an integral infinite.
Series. Write the integrand as $\sum_n f_n$ with every $f_n \ge 0$ — typically by expanding a geometric series $\frac{1}{1 - u} = \sum_{n \ge 0} u^{n}$ with $0 \le u < 1$ , or $\frac{1}{e^{x} - 1} = \sum_{n \ge 1} e^{-nx}$ for $x > 0$ . Integrate each term and add. No uniform convergence is required, and the interchange is valid even where the series diverges.
Only a bound wanted? Use Fatou's lemma. To show $\int f \le C$ when $f = \lim_n f_n$ a.e. — with $f$ again assumed measurable, for the same reason — and each $\int f_n \le C$ , no monotonicity is needed.
Check for escaping mass. If the answer comes out smaller than expected, look for a sequence whose mass escapes to infinity or concentrates at a point: those are the two failure modes, and they are exactly the cases in which Fatou's inequality is strict.

Example(A logarithmic integral summed term by term)Original

Evaluate

I = \int_{0}^{1} \frac{\ln(1/x)}{1 - x} \, dx .

The integrand is unbounded at the left end, behaving like $\ln(1/x)$ as $x \downarrow 0$ , while at the right end it is harmless, tending to $1$ as $x \uparrow 1$ . The singularity at $0$ is enough to make an interchange theorem that needs a bounded integrand awkward here. Non-negativity makes it routine.

Step 1: the integrand is non-negative and measurable. For $0 < x < 1$ both $\ln(1/x) > 0$ and $1 - x > 0$ , and the integrand is continuous on $(0,1)$ , hence measurable. The endpoints form a null set and may be ignored.

Step 2: expand in a non-negative series. For $0 < x < 1$ the geometric series gives $\frac{1}{1-x} = \sum_{n \ge 0} x^{n}$ , so

\frac{\ln(1/x)}{1-x} = \sum_{n \ge 0} f_n(x), \qquad f_n(x) = x^{n}\ln(1/x) \ \ge\ 0 \quad (0 < x < 1).

Every term is non-negative, which is the only hypothesis the termwise theorem asks for.

Step 3: integrate one term. Substituting $x = e^{-t}$ , so $dx = -e^{-t}dt$ and $\ln(1/x) = t$ , turns $(0,1)$ into $(0,\infty)$ :

\int_{0}^{1} x^{n} \ln(1/x)\,dx = \int_{0}^{\infty} t\, e^{-(n+1)t} \, dt = \frac{1}{(n+1)^{2}} \int_{0}^{\infty} u \,e^{-u}\,du = \frac{1}{(n+1)^{2}} ,

using $u = (n+1)t$ and $\int_{0}^{\infty} u e^{-u} du = \Gamma(2) = 1$ .

Step 4: add. By the termwise theorem,

I = \sum_{n \ge 0} \frac{1}{(n+1)^{2}} = \sum_{m \ge 1} \frac{1}{m^{2}} = \zeta(2) = \frac{\pi^{2}}{6} = 1.6449340668\ldots

Step 5: check. The partial sums $\sum_{m \le M} m^{-2}$ are $1.4636111$ at $M = 5$ , $1.5961632$ at $M = 20$ and $1.6349839$ at $M = 100$ , climbing towards $1.6449341$ from below as they must, the terms being positive. For the Basel value $\zeta(2) = \pi^{2}/6$ see Riemann zeta function and Bernoulli numbers. Numerical quadrature of the original integral gives $1.644934067$ , agreeing to nine decimals.

Step 6: what monotone convergence adds. The same computation read backwards shows the partial sums $S_N(x) = \sum_{n < N} x^{n}\ln(1/x)$ increase pointwise to the integrand, so $\int_{0}^{1} S_N \uparrow I$ : the sequence of partial sums of the series is itself an increasing sequence of integrands, and that is all the termwise theorem is.

ApplicationOriginal

The Stefan–Boltzmann law. Planck's law makes the energy density of black-body radiation per unit frequency proportional to $\nu^{3}/(e^{h\nu/kT} - 1)$ . Integrating over all frequencies and substituting $x = h\nu/kT$ leaves the pure number

J = \int_{0}^{\infty} \frac{x^{3}}{e^{x} - 1}\,dx .

The integrand is non-negative on $(0,\infty)$ , so no domination has to be checked. For $x > 0$ we have $e^{-x} < 1$ and

\frac{1}{e^{x}-1} = \frac{e^{-x}}{1 - e^{-x}} = \sum_{n \ge 1} e^{-nx} ,

all terms positive. Termwise integration and the substitution $u = nx$ give $\int_{0}^{\infty} x^{3} e^{-nx}\,dx = n^{-4}\int_{0}^{\infty} u^{3}e^{-u}\,du = 3!/n^{4} = 6/n^{4}$ , whence

J = 6 \sum_{n \ge 1} \frac{1}{n^{4}} = 6\,\zeta(4) = 6 \cdot \frac{\pi^{4}}{90} = \frac{\pi^{4}}{15} = 6.4939394023\ldots

This is the constant that converts Planck's law into $u = aT^{4}$ and fixes $\sigma = 2\pi^{5}k^{4}/(15 h^{3}c^{2})$ . The only analytic input beyond the substitution is the right to integrate a positive series term by term. See Riemann zeta function and Bernoulli numbers for $\zeta(4) = \pi^4/90$ .

Expectations of sums of non-negative random variables. Grimmett and Stirzaker (Grimmett & Stirzaker (2020), §5.6(B), item (13), p. 201) derive from monotone convergence that for non-negative $Z_1, Z_2, \dots$ with finite expectations, $\mathbb{E}\bigl(\sum_i Z_i\bigr) = \sum_i \mathbb{E}(Z_i)$ . The termwise theorem above drops even that hypothesis: the identity holds for any non-negative $Z_i$ , both sides being allowed the value $+\infty$ . This is what licenses counting arguments such as $\mathbb{E}(\text{number of records in } n \text{ draws}) = \sum_{i=1}^{n} 1/i$ , in which the summands are indicators and the interchange is never questioned. It is also the step behind Tonelli's theorem for counting measure in Fubini and Tonelli theorems.

Lower semicontinuity in the calculus of variations. The direct method minimises an energy $E(u) = \int_{\Omega} F(x, u, \nabla u)\,dx$ over an admissible class by taking a minimising sequence $(u_k)$ , extracting a limit $u$ , and then trying to prove

E(u) \;\le\; \liminf_{k \to \infty} E(u_k) \;=\; \inf E ,

which makes $u$ a minimiser. Fatou's lemma is the tool for that inequality, and it is worth being exact about what it does and does not supply. Write $F_k(x) = F(x, u_k(x), \nabla u_k(x))$ and suppose $F \ge 0$ . Fatou gives $\int_{\Omega} \liminf_k F_k \le \liminf_k \int_{\Omega} F_k = \liminf_k E(u_k)$ with no hypothesis beyond non-negativity. The left-hand side is $E(u)$ only when $\liminf_k F_k = F(x,u,\nabla u)$ almost everywhere — that is, when the integrands converge, which by the a.e. corollary above needs $\nabla u_k \to \nabla u$ a.e., not merely $u_k \to u$ . Supplying that convergence is the hard part of the direct method, since compactness normally yields only weak convergence of the gradients.

The gap is real, not pedantic. Take $\Omega = (0,1)$ and the Bolza energy

E(u) \;=\; \int_{0}^{1} \Bigl[ \bigl(u'(x)^{2} - 1\bigr)^{2} + u(x)^{2} \Bigr] dx

over the $u$ that are continuous and piecewise smooth with $u(0) = u(1) = 0$ ; the integrand is non-negative, so $F \ge 0$ as required. Let $u_k$ be the sawtooth that is linear on each of the $2k$ intervals $\bigl(\tfrac{j-1}{2k}, \tfrac{j}{2k}\bigr)$ , $j = 1, \dots, 2k$ , with slope $+1$ for $j$ odd and $-1$ for $j$ even, starting from $u_k(0) = 0$ ; it rises to $1/(2k)$ at the odd endpoints $j/(2k)$ and returns to $0$ at the even ones, so in particular $u_k(0) = u_k(1) = 0$ , and $\lVert u_k \rVert_{\infty} = 1/(2k) \to 0$ , while $u_k'(x) = \pm 1$ off the finite set of corners. The first term of the energy therefore vanishes identically, and the second is the same integral on each of the $2k$ pieces:

E(u_k) = \int_{0}^{1} u_k^{2}\,dx = 2k \int_{0}^{1/(2k)} t^{2}\,dt = \frac{1}{12k^{2}} \ \longrightarrow\ 0 ,

so $(u_k)$ is a minimising sequence and $\inf E = 0$ . That infimum is not attained: $E(u) = 0$ would force $u = 0$ almost everywhere and $u'(x)^{2} = 1$ at all but finitely many $x$ , whereas a piecewise-smooth $u$ vanishing almost everywhere vanishes identically on each interval of smoothness, where its derivative is $0$ and the first term contributes $1$ . So here the direct method fails at the first hurdle — no minimiser exists — and it fails for the reason Fatou's lemma cannot repair. The uniform limit of the minimising sequence is $u \equiv 0$ , with $E(0) = \int_{0}^{1} 1 \, dx = 1 > \inf E$ : the functional is not lower semicontinuous along $(u_k)$ . Fatou's lemma is not violated, merely silent. Here $F_k(x) = \bigl(u_k'(x)^2 - 1\bigr)^2 + u_k(x)^2 = u_k(x)^{2} \le 1/(4k^{2})$ for a.e. $x$ , so $\liminf_k F_k = 0$ a.e. and the lemma reads $0 \le 0$ — the left-hand side is not $E(0) = 1$ , precisely because the integrands do not converge to $F(x,0,0) = 1$ . The derivatives $u_k'$ oscillate between $+1$ and $-1$ and converge at no point.

So Fatou's lemma delivers lower semicontinuity exactly when the integrands converge a.e.; it does not by itself turn the limit of a minimising sequence into a minimiser, and when the gradients only converge weakly a convexity hypothesis on $F$ in its last argument has to be supplied instead. That Fatou's lemma is the engine behind such lower semicontinuity arguments is the use Stakgold and Holst (Stakgold & Holst (2011), §0.7, p. 47) highlight, and it is the reason variational energies in continuum mechanics are written as non-negative integrands wherever possible.

Partition functions and infinite level sums. In statistical mechanics $Z = \sum_{j} e^{-\beta E_j}$ is interchanged with phase-space integrals as a matter of course. Because every term $e^{-\beta E_j}$ is positive, the interchange is the termwise theorem, valid even when $Z$ diverges — as it does for the hydrogen atom's bound states, where the divergence of $\sum_n n^{2} e^{-\beta E_n}$ with $E_n \to 0$ is a statement the theorem makes rigorous rather than one it forbids.

Theorem

Dominated convergence

Adapted

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If a sequence of measurable functions converges almost everywhere and is held under one fixed integrable envelope, the limit is integrable and the integrals converge — indeed in the $L^1$ norm — and the same envelope condition licenses taking a limit, or a derivative in a parameter, inside an integral.

Prerequisites: Monotone convergence and Fatou's lemma

Theorem(Dominated convergence theorem)AdaptedAdapted from Folland (1999), §2.3, Thm 2.24, p. 54; Evans (2010), App. E.3, Thm 5, p. 687; Reed & Simon (1980), §1.3, Thm 1.11, p. 17

Let $(X,\Sigma,\mu)$ be a measure space and let $(f_n)_{n \ge 1}$ be a sequence in $L^{1}(\mu)$ , real- or complex-valued, such that

(a) $f_n \to f$ almost everywhere, and
(b) there is a single non-negative $g \in L^{1}(\mu)$ , independent of $n$ , with $\lvert f_n \rvert \le g$ almost everywhere, for every $n$ .

Then $f \in L^{1}(\mu)$ and

\int f \, d\mu \;=\; \lim_{n \to \infty} \int f_n \, d\mu .

Moreover the convergence holds in the $L^{1}$ norm:

\lVert f_n - f \rVert_{1} = \int \lvert f_n - f \rvert \, d\mu \;\longrightarrow\; 0 ,

which is the form in which Reed and Simon (Reed & Simon (1980), §1.3, Thm 1.11, p. 17) state the conclusion. Integrability of the limit is part of the conclusion, not a hypothesis, a point Stakgold and Holst (Stakgold & Holst (2011), §0.7, p. 45) emphasise. Read literally, " $f \in L^{1}(\mu)$ " asserts in particular that $f$ is measurable, and that does not follow from (a) alone when $\mu$ is not complete, since (a) leaves $f$ free on a null set; as in Folland (Folland (1999), §2.3, Thm 2.24 and its proof, pp. 54–55), the conclusion is to be read in the $L^{1}$ of a.e.-defined equivalence classes set up in The Lebesgue integral, where $f$ is identified with the measurable function it equals almost everywhere. Step 0 of the proof says this precisely.

Bounded convergence. If $\mu(X) < \infty$ and $\lvert f_n \rvert \le c$ a.e. for a constant $c$ , condition (b) holds with $g \equiv c$ , whose integral is $c\,\mu(X) < \infty$ . On a probability space this is Grimmett and Stirzaker's (Grimmett & Stirzaker (2020), §5.6(B), item (12)(c), p. 201) bounded convergence theorem: $X_n \to X$ a.s. and $\lvert X_n \rvert \le c$ a.s. give $\mathbb{E}(X_n) \to \mathbb{E}(X)$ .

Theorem(Termwise integration of an absolutely summable series)AdaptedAdapted from Folland (1999), §2.3, Thm 2.25, p. 55

Let $(f_j)_{j \ge 1} \subset L^{1}(\mu)$ satisfy $\sum_{j \ge 1} \int \lvert f_j \rvert \, d\mu < \infty$ . Then $\sum_{j \ge 1} f_j$ converges almost everywhere, its sum is in $L^{1}(\mu)$ , and

\int \sum_{j \ge 1} f_j \, d\mu \;=\; \sum_{j \ge 1} \int f_j \, d\mu .

This is the signed counterpart of the non-negative termwise theorem of Monotone convergence and Fatou's lemma: there, no hypothesis at all; here, absolute summability of the integrals.

Theorem(Limits and derivatives in a parameter)AdaptedAdapted from Folland (1999), §2.3, Thm 2.27 and the paragraph following it, p. 56

Let $-\infty < a < b < \infty$ , let $f : X \times [a,b] \to \mathbb{C}$ be such that $f(\cdot, t) \in L^{1}(\mu)$ for each $t$ , and set

F(t) = \int_{X} f(x,t) \, d\mu(x) .

Continuity. Suppose there is $g \in L^{1}(\mu)$ with $\lvert f(x,t) \rvert \le g(x)$ for all $x$ and all $t$ . If $\lim_{t \to t_0} f(x,t) = f(x,t_0)$ for every $x$ , then $\lim_{t\to t_0} F(t) = F(t_0)$ . In particular, should every map $t \mapsto f(x,t)$ be continuous, $F$ inherits continuity throughout $[a,b]$ .
Differentiation. Suppose $\partial f/\partial t$ exists at every point of $X \times [a,b]$ and there is $g \in L^{1}(\mu)$ with $\lvert (\partial f/\partial t)(x,t)\rvert \le g(x)$ for all $x$ and all $t$ . Then $F$ is differentiable at every $t \in [a,b]$ — one-sidedly at $t = a$ and $t = b$ , where the difference quotient is only defined from inside the interval, and $\partial f/\partial t$ is likewise a one-sided derivative there — and $F'(t) \;=\; \int_{X} \frac{\partial f}{\partial t}(x,t) \, d\mu(x) .$

The interval $[a,b]$ may be a compact subinterval of a larger interval $I$ on which $f(x,\cdot)$ is defined. If the hypotheses hold on every compact $[a,b] \subset I$ , with a dominating function allowed to depend on $a$ and $b$ , then $F$ is continuous, respectively differentiable, on all of $I$ , continuity and differentiability being local properties. Compare Differentiation under the integral sign, which asks for joint continuity of $f$ and $\partial f/\partial t$ on a product with a compact interval and needs no dominating function.

Idea of the proofOriginal

Only one inequality is available for an arbitrary sequence of non-negative integrands, Fatou's lemma, and it points one way: the integral of the limit is at most the lower limit of the integrals. To get an equality out of a one-sided tool, apply it twice, to two different non-negative sequences built from the same data. The envelope $g$ is exactly what makes this possible: $g + f_n \ge 0$ and $g - f_n \ge 0$ , so Fatou applies to both. Applied to $g + f_n$ it bounds $\int f$ above by $\liminf_n \int f_n$ ; applied to $g - f_n$ , where the minus sign turns the $\liminf$ on the right into $-\limsup$ , it bounds $\int f$ below by $\limsup_n \int f_n$ . Since always $\liminf \le \limsup$ , the chain $\int f \le \liminf_n \int f_n \le \limsup_n \int f_n \le \int f$ closes on itself, every inequality in it is an equality, and the limit exists.

Folland (Folland (1999), §2.3, p. 54) describes the hypothesis geometrically: the graphs of all the $\lvert f_n \rvert$ are confined to one region of finite area, so the area cannot leak away to infinity as it does for $\mathbf{1}_{(n,n+1)}$ or climb out of sight as it does for $n\mathbf{1}_{(0,1/n)}$ . The $L^{1}$ -norm form of the conclusion comes free: $\lvert f_n - f \rvert \to 0$ a.e. and is dominated by $2g$ , so the theorem applies to that sequence too.

Proof(Dominated convergence)AdaptedAdapted from Folland (1999), §2.3, proof of Thm 2.24, p. 55

Step 0: dispose of the null sets and of the complex case. The hypotheses hold off a countable union of null sets, which is null; call its union $Z$ and let $\tilde f_n = f_n \mathbf{1}_{X \setminus Z}$ , $\tilde g = g\mathbf{1}_{X\setminus Z}$ and $\tilde f = \lim_n \tilde f_n$ , the limit now existing at every point. Redefining a function on a null set changes no integral (The Lebesgue integral), so $\int \tilde f_n = \int f_n$ and $\int \tilde g = \int g$ , while $\tilde f$ is measurable as a pointwise limit of measurable functions (Measurable functions) and $\tilde f = f$ off $Z$ .

Here is the one place where a care is needed that the statement hides. On the exceptional set $Z$ the hypotheses say nothing about $f$ , so $f$ itself need not be measurable unless $\mu$ is complete: what the argument produces is a measurable $\tilde f$ agreeing with $f$ almost everywhere. That is exactly enough, because $L^{1}(\mu)$ was redefined in The Lebesgue integral as a space of equivalence classes of a.e.-defined functions, in which $f$ and $\tilde f$ are the same element; and it is how Folland (Folland (1999), §2.3, proof of Thm 2.24, p. 55) reads his own conclusion: he allows $f$ to be altered on a null set before calling it measurable, and appeals to the completion of $\mu$ to do so. We work with $\tilde f$ and drop the tildes. Since $\lvert \tilde f \rvert = \lim_n \lvert \tilde f_n \rvert \le \tilde g$ and $\tilde g$ is integrable, $\tilde f \in L^{1}(\mu)$ — this is the first of the two uses of $g$ , and it is what makes every integral below finite. Finally, $\lvert \operatorname{Re} h \rvert \le \lvert h \rvert$ and $\lvert \operatorname{Im} h\rvert \le \lvert h \rvert$ , so the complex case follows from the real one applied to real and imaginary parts; assume from here that all functions are real-valued.

Step 1: two non-negative sequences. By (b), $g + f_n \ge 0$ and $g - f_n \ge 0$ pointwise, and both are measurable. This is the second and essential use of $g$ : Fatou's lemma (Monotone convergence and Fatou's lemma) needs non-negative integrands, and nothing in the data is non-negative until $g$ is added.

Step 2: Fatou on $g + f_n$ . Since $g + f_n \to g + f$ pointwise, $\liminf_n (g+f_n) = g+f$ , and Fatou gives $\int (g+f) \le \liminf_n \int (g + f_n)$ . All the integrals are finite, so linearity on $L^{1}$ lets us cancel the common $\int g$ :

\int g + \int f \;\le\; \int g + \liminf_{n} \int f_n \qquad\Longrightarrow\qquad \int f \;\le\; \liminf_{n} \int f_n ,

using $\liminf_n (c + a_n) = c + \liminf_n a_n$ for a finite constant $c$ .

Step 3: Fatou on $g - f_n$ . Likewise $\int (g - f) \le \liminf_n \int (g - f_n)$ . Now $\liminf_n(-a_n) = -\limsup_n a_n$ , so the right-hand side is $\int g - \limsup_n \int f_n$ , and cancelling $\int g$ again gives

-\int f \;\le\; -\limsup_{n} \int f_n \qquad\Longrightarrow\qquad \limsup_{n} \int f_n \;\le\; \int f .

Step 4: squeeze. Combining steps 2 and 3 with the general inequality $\liminf \le \limsup$ ,

\int f \;\le\; \liminf_{n} \int f_n \;\le\; \limsup_{n} \int f_n \;\le\; \int f ,

so all four quantities are equal; in particular $\lim_n \int f_n$ exists and equals $\int f$ .

Step 5: convergence in $L^{1}$ . Apply steps 0–4 to the sequence $h_n = \lvert f_n - f \rvert$ . It is measurable and non-negative, $h_n \to 0$ a.e. by (a), and $h_n \le \lvert f_n \rvert + \lvert f \rvert \le 2g$ a.e., with $2g$ integrable. The theorem just proved gives $\int h_n \to \int 0 = 0$ , which is $\lVert f_n - f\rVert_1 \to 0$ . The first conclusion follows from this one as well, since $\lvert \int f_n - \int f \rvert \le \int \lvert f_n - f\rvert$ . $\square$

Proof(The series corollary and the parameter theorems)AdaptedAdapted from Folland (1999), §2.3, proofs of Thm 2.25, p. 55 and Thm 2.27, p. 56

Series. Put $G = \sum_{j \ge 1} \lvert f_j \rvert$ , a pointwise sum of non-negative measurable functions. The termwise theorem for non-negative integrands (Monotone convergence and Fatou's lemma) gives $\int G = \sum_j \int \lvert f_j \rvert < \infty$ by hypothesis, so $G \in L^{1}(\mu)$ and therefore $G < \infty$ almost everywhere — if $\mu(\{G = \infty\}) > 0$ then $\int G = \infty$ . At each point where $G(x) < \infty$ the series $\sum_j f_j(x)$ is absolutely convergent, hence convergent. Let $S_N = \sum_{j \le N} f_j$ be the partial sums. Then $S_N \to \sum_j f_j$ a.e., and $\lvert S_N \rvert \le \sum_{j \le N}\lvert f_j\rvert \le G$ for every $N$ , so $G$ is a dominating function. Dominated convergence applied to $(S_N)_N$ gives both that the sum is in $L^{1}$ and that $\int \sum_j f_j = \lim_N \int S_N = \lim_N \sum_{j\le N} \int f_j = \sum_j \int f_j$ , the middle equality by linearity on $L^{1}$ .

Continuity in the parameter. Dominated convergence speaks about sequences, so convert: $\lim_{t \to t_0}F(t) = F(t_0)$ holds if and only if $F(t_n) \to F(t_0)$ for every sequence $t_n \to t_0$ in $[a,b]$ . Fix such a sequence and set $f_n(x) = f(x,t_n)$ . Then $f_n \to f(\cdot, t_0)$ pointwise by the hypothesis on $f(x,\cdot)$ , and $\lvert f_n \rvert \le g$ for every $n$ by hypothesis. Dominated convergence gives $F(t_n) = \int f_n \to \int f(\cdot,t_0) = F(t_0)$ . As the sequence was arbitrary, $F$ is continuous at $t_0$ .

Differentiation in the parameter. Fix $t_0 \in [a,b]$ and a sequence $t_n \to t_0$ with $t_n \ne t_0$ , and set

h_n(x) = \frac{f(x,t_n) - f(x,t_0)}{t_n - t_0} .

Pointwise limit. By hypothesis $\partial f/\partial t$ exists everywhere, so $h_n(x) \to (\partial f/\partial t)(x,t_0)$ for every $x$ . In particular $\partial f/\partial t\,(\cdot, t_0)$ is measurable, being a pointwise limit of measurable functions.
Domination. Each $h_n$ is a difference quotient of $t \mapsto f(x,t)$ , which is differentiable on $[a,b]$ . For real-valued $f$ the mean value theorem on the interval between $t_0$ and $t_n$ gives $\lvert h_n(x)\rvert \le \sup_{t\in[a,b]} \lvert(\partial f/\partial t)(x,t)\rvert \le g(x)$ directly, which is the bound Folland (Folland (1999), §2.3, proof of Thm 2.27b, p. 56) writes. For complex-valued $f$ the mean value theorem has to be applied to the real and the imaginary part separately, and the two intermediate points differ, so what it yields is $\lvert h_n(x)\rvert \le \lvert \operatorname{Re} h_n(x)\rvert + \lvert \operatorname{Im} h_n(x)\rvert \le 2 g(x)$ . Nothing is lost: $2g \in L^{1}(\mu)$ as well, and $2g$ is all the next bullet needs. Either way the hypothesis is used as stated: the bound on the derivative must be uniform in $t$ , and it is the derivative, not $f$ itself, that has to be dominated.
Conclusion. Dominated convergence applied to $(h_n)$ , with envelope $2g$ , gives $\frac{F(t_n) - F(t_0)}{t_n - t_0} = \int h_n \, d\mu \;\longrightarrow\; \int \frac{\partial f}{\partial t}(x,t_0)\, d\mu(x) .$ The sequence $t_n \to t_0$ was arbitrary, so the difference quotient of $F$ has that limit, which is the claim. $\square$

Remark(Where the hypotheses are used)Original

The envelope is used three times. In step 0 it puts $f$ in $L^{1}$ ; in step 1 it makes the two sequences non-negative, so that Fatou's lemma applies at all; and in steps 2 and 3 the quantity $\int g$ appears on both sides of each inequality and is cancelled, which is legitimate only because $\int g < \infty$ . That last use is easy to overlook and is where integrability of $g$ , as opposed to mere domination, earns its keep: with $\int g = \infty$ the cancellation would be an $\infty - \infty$ . All three uses need $g$ to be the same function for every $n$ .
An envelope that is not integrable is no envelope. The pointwise supremum $\sup_n \lvert f_n \rvert$ always dominates; the content of (b) is that some dominating function is integrable. Reed and Simon (Reed & Simon (1980), §1.3, p. 17) give $f_n = n^{-1}\mathbf{1}_{(-n,n)}$ on $\mathbb{R}$ : here $f_n \to 0$ at every point, uniformly in fact, yet $\int \lvert f_n \rvert = 2$ for every $n$ , so $\int f_n \to 2 \ne 0$ . The smallest envelope is $\sup_n \lvert f_n\rvert(x) = 1/(\lfloor \lvert x\rvert\rfloor + 1)$ — the supremum is over those $n$ with $n > \lvert x \rvert$ , and the smallest such $n$ is $\lfloor \lvert x\rvert\rfloor + 1$ — which decays like $1/\lvert x\rvert$ and so is not integrable. The complementary failure is $n\,\mathbf{1}_{(0,1/n)}$ , whose smallest envelope behaves like $1/x$ near $0$ .
Uniform convergence is neither necessary nor sufficient. The theorem it is measured against is Stakgold and Holst's (Stakgold & Holst (2011), §0.6, Thm 0.6.2, p. 38) Thm 0.6.2: if $s_n \to s$ uniformly on a bounded interval $E$ and all the functions are integrable there, then $\int_E s_n \to \int_E s$ . Not sufficient, because the boundedness of $E$ cannot be dropped: the example just given converges uniformly on $\mathbb{R}$ and the conclusion fails, and Stakgold and Holst (Stakgold & Holst (2011), §0.6, p. 39) make the same point with $s_n(x) = n^{-1}f(x/n)$ on $[0,\infty)$ , which tends to $0$ uniformly while $\int_0^\infty s_n = \int_0^\infty f = 1$ for every $n$ . Not necessary, because on a bounded interval domination is weaker than uniformity: their own $s_n(x) = nxe^{-nx}$ on $(0,1]$ (Eq. (0.6.1), p. 37) has a bump of the fixed height $1/e$ at $x = 1/n$ , so it converges to $0$ pointwise but not uniformly, yet $\lvert s_n\rvert \le 1/e$ there and the constant $1/e$ is integrable on a set of measure $1$ — dominated convergence applies and gives $\int_0^1 s_n \, dx \to 0$ , as the exact value $n^{-1}\bigl(1 - (1+n)e^{-n}\bigr)$ confirms ( $0.2642$ , $0.1919$ , $0.0500$ , $0.0100$ at $n = 1, 5, 20, 100$ ). Stakgold and Holst (Stakgold & Holst (2011), §0.7, p. 45) sum up the comparison: against the uniform-convergence theorem it replaces, dominated convergence needs only pointwise convergence, only almost everywhere, on a possibly unbounded interval, and it delivers integrability of the limit instead of assuming it.
"Almost everywhere" is enough in both hypotheses, since a countable union of null sets is null; this is the weakening Grimmett and Stirzaker (Grimmett & Stirzaker (2020), §5.6(B), item (12), p. 201) record for the probabilistic form. Stakgold and Holst (Stakgold & Holst (2011), §0.7, p. 45) relax (b) further to "for all sufficiently large $n$ ", which costs nothing: changing finitely many terms of a sequence changes neither its limit nor the limit of its integrals.
The envelope may be allowed to move. Folland (Folland (1999), §2.3, Exercise 20, p. 59) records a generalised form: if $f_n \to f$ and $g_n \to g$ a.e. with $\lvert f_n \rvert \le g_n$ , all in $L^{1}$ , and if $\int g_n \to \int g$ , then $\int f_n \to \int f$ . The fixed- $g$ statement is the case $g_n = g$ . Folland's Exercise 21 (Folland (1999), §2.3, p. 59) adds the sharp criterion for $L^{1}$ convergence: for $f_n, f \in L^{1}$ with $f_n \to f$ a.e., $\lVert f_n - f\rVert_1 \to 0$ holds iff $\int \lvert f_n \rvert \to \int \lvert f \rvert$ — so no envelope is needed at all if the norms are known to converge.
The parameter theorems need the bound uniformly in $t$ , and on the right function. In part 1 it is $f$ that is dominated; in part 2 it is $\partial f/\partial t$ , and no bound on $f$ itself is required beyond $f(\cdot,t) \in L^{1}$ . Dropping uniformity in $t$ breaks the mean value theorem step. The restriction to a compact $[a,b]$ is what allows the supremum over $t$ to be finite in practice; as the statement notes, a dominating function depending on $[a,b]$ suffices because differentiability is local.
The sequential device is not optional. Dominated convergence is a statement about sequences, so both parameter theorems are proved along arbitrary sequences $t_n \to t_0$ and then invoke the sequential characterisation of a limit of a function of a real variable. Folland (Folland (1999), §2.3, p. 56) notes that one afterwards writes "let $t \to t_0$ " with this understood.
It is dominated convergence that makes $L^{1}$ complete. In Reed and Simon's (Reed & Simon (1980), §1.3, Thm 1.12, p. 18) proof of the Riesz–Fischer theorem, monotone convergence puts the telescoping series $\sum_k \lvert f_k - f_{k+1}\rvert$ of a rapidly Cauchy subsequence into $L^{1}$ , and dominated convergence then converts the resulting pointwise a.e. limit into an $L^{1}$ limit. See L^p spaces.

Method(Taking a limit or a derivative inside an integral)Original

Identify the pointwise limit of the integrand for each fixed $x$ , allowing a null set of exceptions. If it does not exist a.e., stop; no version of the theorem applies.
Hunt for an envelope. Find one function $g(x)$ , free of $n$ (or of the parameter $t$ ), with $\lvert f_n(x)\rvert \le g(x)$ a.e. Standard supplies: $\lvert \sin u \rvert \le \min(1, \lvert u\rvert)$ ; $(1+u)^{-1} \le 1$ ; $\lvert e^{iu}\rvert = 1$ ; $u^{k}e^{-u} \le C_k e^{-u/2}$ on $[0,\infty)$ ; and, when the sequence is monotone, $g = \lim_n \lvert f_n \rvert$ itself.
Verify $\int g < \infty$ . This is the step that actually does work, and the step most often skipped. A bound such as $g(x) = 1$ on $\mathbb{R}$ , or $g(x) = 1/x$ near $0$ , is useless.
Conclude, and if the $L^{1}$ statement is wanted, note that $\lvert f_n - f\rvert \le 2g$ gives it at once.
For a parameter integral $F(t) = \int f(x,t)\,d\mu(x)$ : to prove $F$ continuous, dominate $f$ ; to differentiate under the integral, dominate $\partial f/\partial t$ , uniformly over $t$ in a compact interval around the point of interest. Then $F' (t) = \int \partial_t f\, d\mu$ , and the resulting ordinary differential equation in $t$ is often easier than the integral.
If no envelope exists, fall back: non-negative integrands need none (Monotone convergence and Fatou's lemma); an inequality rather than an equality may suffice, and then Fatou's lemma needs none either; or check whether $\int \lvert f_n\rvert \to \int \lvert f \rvert$ , which by Folland's Exercise 21 is equivalent to $L^{1}$ convergence.

Example(Two uses of one envelope)Original

Part A: a limit under the integral sign. Evaluate

L = \lim_{n \to \infty} \int_{0}^{\infty} \frac{n \sin(x/n)}{x\,(1 + x^{2})} \, dx .

Step 1: the pointwise limit. Write $f_n(x) = \dfrac{n\sin(x/n)}{x\,(1+x^{2})}$ for the integrand. For fixed $x > 0$ , $n\sin(x/n)/x = \sin(x/n)/(x/n) \to 1$ as $n \to \infty$ , since $\sin u / u \to 1$ as $u \to 0$ . So $f_n(x) \to f(x) := 1/(1+x^{2})$ at every $x > 0$ .

Step 2: the envelope. Hypothesis (b) bounds a modulus, so the inequality to use is the two-sided $\lvert \sin u \rvert \le u$ for $u \ge 0$ , not $\sin u \le u$ , which says nothing where $\sin u < 0$ (at $u = 4$ the quotient $\sin u / u$ equals $-0.1892$ , so the one-sided bound would leave the claim " $0 < n\sin(x/n)/x$ " false). Writing $u = x/n > 0$ ,

\Bigl\lvert \frac{n\sin(x/n)}{x}\Bigr\rvert = \Bigl\lvert \frac{\sin u}{u}\Bigr\rvert \le 1 \qquad (n \ge 1,\ x > 0),

so the integrand is dominated in modulus by

g(x) = \frac{1}{1+x^{2}} , \qquad \int_{0}^{\infty} g(x) \, dx = \bigl[\arctan x\bigr]_{0}^{\infty} = \frac{\pi}{2} < \infty .

The envelope does not depend on $n$ and is integrable, so hypothesis (b) holds.

Step 3: conclude.

L = \int_{0}^{\infty} \frac{dx}{1+x^{2}} = \frac{\pi}{2} = 1.5707963268\ldots

Step 4: check. This integral has a closed form. Writing $a = 1/n$ and using $\int_{0}^{\infty} \frac{\sin(ax)}{x(1+x^{2})}\,dx = \frac{\pi}{2}\bigl(1 - e^{-a}\bigr)$ for $a > 0$ , the $n$ th integral equals $\frac{n\pi}{2}\bigl(1 - e^{-1/n}\bigr)$ , which indeed increases to $\pi/2$ as $n \to \infty$ because $n(1-e^{-1/n}) \uparrow 1$ . Numerically, and confirmed to twelve digits by oscillatory quadrature of the original integral:

$n$	$\frac{n\pi}{2}(1-e^{-1/n})$	$\pi/2 - {}$ that
$1$	$0.9929327$	$0.5778637$
$5$	$1.4236853$	$0.1471110$
$50$	$1.5551926$	$0.0156038$
$500$	$1.5692266$	$0.0015697$

Step 5: why a uniform-convergence theorem would still not have served. Here the convergence happens to be uniform. From $0 \le 1 - \frac{\sin u}{u} \le \min\bigl(2, \frac{u^{2}}{6}\bigr)$ for $u > 0$ , with $u = x/n$ ,

\lvert f_n(x) - f(x)\rvert \;=\; \frac{1 - \sin(x/n)\big/(x/n)}{1+x^{2}} \;\le\; \frac{\min\bigl(2,\ x^{2}/6n^{2}\bigr)}{1+x^{2}} \;\le\; \frac{1}{6n^{2}} ,

the last step because $x^{2}/(1+x^{2}) \le 1$ handles the first branch, while the second branch is in force only for $x > n\sqrt{12}$ , where $2/(1+x^{2}) < 1/(6n^{2})$ . The bound is tight: the exact suprema of $\lvert f_n - f\rvert$ are $0.10907$ , $0.0061021$ , $6.6074 \times 10^{-5}$ and $6.6607\times 10^{-7}$ at $n = 1, 5, 50, 500$ , against $1/(6n^{2}) = 0.16667$ , $0.0066667$ , $6.6667\times 10^{-5}$ and $6.6667 \times 10^{-7}$ .

Uniform convergence nevertheless delivers nothing here, because the interval is unbounded: the termwise-integration theorem built on uniform convergence, Stakgold and Holst's (Stakgold & Holst (2011), §0.6, Thm 0.6.2 and the discussion on p. 39) Thm 0.6.2, requires a bounded interval, and they show by example that the hypothesis cannot be dropped. Dominated convergence is what actually licenses the interchange above, and the envelope $1/(1+x^{2})$ is what it needed.

Part B: differentiating under the integral sign. Let

F(t) = \int_{0}^{\infty} e^{-x^{2}} \cos(tx) \, dx , \qquad t \in \mathbb{R} .

Step 1: hypotheses for part 2 of the parameter theorem. Here $\mu$ is Lebesgue measure on $(0,\infty)$ and $f(x,t) = e^{-x^{2}}\cos(tx)$ . For each $t$ , $\lvert f(\cdot,t)\rvert \le e^{-x^{2}} \in L^{1}(0,\infty)$ , so $f(\cdot,t)$ is integrable. The derivative $\partial f/\partial t = -x e^{-x^{2}}\sin(tx)$ exists everywhere and satisfies, for all $t \in \mathbb{R}$ ,

\Bigl\lvert \frac{\partial f}{\partial t}(x,t) \Bigr\rvert \le x\,e^{-x^{2}} =: g(x), \qquad \int_{0}^{\infty} x\,e^{-x^{2}}\,dx = \tfrac12 < \infty .

The bound is uniform in $t$ on every compact interval — indeed on all of $\mathbb{R}$ — so the theorem applies on each $[a,b]$ and hence on $\mathbb{R}$ .

Step 2: differentiate and integrate by parts.

F'(t) = -\int_{0}^{\infty} x\,e^{-x^{2}} \sin(tx)\,dx = \Bigl[\tfrac12 e^{-x^{2}}\sin(tx)\Bigr]_{0}^{\infty} - \frac{t}{2}\int_{0}^{\infty} e^{-x^{2}}\cos(tx)\,dx = -\frac{t}{2}\,F(t) ,

the bracket vanishing at both ends because $e^{-x^{2}}\sin(tx) \to 0$ as $x \to \infty$ and $\sin 0 = 0$ .

Step 3: solve. The relation $F' + \tfrac{t}{2}F = 0$ is linear, so multiply by the integrating factor $e^{t^{2}/4}$ rather than dividing by $F$ , whose vanishing is not yet excluded: $\bigl(e^{t^{2}/4}F(t)\bigr)' = e^{t^{2}/4}\bigl(F'(t) + \tfrac{t}{2}F(t)\bigr) = 0$ , so $e^{t^{2}/4}F(t)$ is constant and equals its value $F(0)$ at $t = 0$ . With $F(0) = \int_0^\infty e^{-x^{2}}dx = \sqrt{\pi}/2$ (Gaussian integrals),

F(t) = \frac{\sqrt{\pi}}{2}\, e^{-t^{2}/4} .

Step 4: a checkable number. At $t = 2$ ,

F(2) = \frac{\sqrt{\pi}}{2e} = 0.3260246661\ldots ,

and direct numerical quadrature of $\int_{0}^{\infty} e^{-x^{2}}\cos 2x \, dx$ returns $0.3260246661$ , agreeing to ten decimals.

ApplicationOriginal

The Fourier transform of an $L^{1}$ function is continuous and bounded. For $f \in L^{1}(\mathbb{R})$ the integrand of $\hat f(\xi) = \int f(x) e^{-i\xi x}dx$ obeys $\lvert f(x)e^{-i\xi x}\rvert = \lvert f(x)\rvert$ , one envelope serving every $\xi$ , so part 1 of the parameter theorem makes $\hat f$ continuous, with $\lVert \hat f\rVert_\infty \le \lVert f\rVert_1$ . If in addition $x f(x) \in L^{1}$ , then $x f(x)$ dominates $\partial_\xi (f(x)e^{-i\xi x})$ and part 2 gives $\hat f' (\xi) = \int (-ix) f(x) e^{-i\xi x}dx$ : decay of $f$ buys smoothness of $\hat f$ . This is the elementary half of the smoothness–decay duality of Smoothness and decay of Fourier coefficients, and the same argument with $\mathbb{P}$ in place of Lebesgue measure makes a characteristic function $\varphi_X(t) = \mathbb{E}(e^{itX})$ continuous always, and differentiable at $0$ with $\varphi_X'(0) = i\,\mathbb{E}(X)$ whenever $\mathbb{E}\lvert X\rvert < \infty$ (Characteristic functions).

Thermodynamic averages from a partition function. With $Z(\beta) = \int e^{-\beta E(x)}\,d\mu(x)$ over phase space, the internal energy is $U = -\partial_\beta \ln Z$ . Justifying the differentiation means producing an envelope for $\partial_\beta e^{-\beta E} = -E e^{-\beta E}$ valid on a closed interval $[\beta_1,\beta_2]$ of inverse temperatures with $\beta_1 > 0$ : if $E \ge 0$ , then $\lvert E e^{-\beta E}\rvert \le E e^{-\beta_1 E}$ , which is integrable exactly when $Z(\beta_1)$ converges with one derivative to spare. The physical content of the hypothesis is that the system have a finite mean energy throughout the temperature range; where it fails — near a phase transition, or for a system whose density of states grows too fast — differentiating under the integral is not a formality, and the thermodynamic quantity it defines may not exist.

Passing to the limit in a numerical scheme. A quadrature or a Galerkin method produces approximants $u_n$ converging pointwise a.e. to $u$ . Convergence of the energies $\int \lvert u_n \rvert^{2}$ to $\int \lvert u \rvert^{2}$ needs an envelope, and a uniform-in- $n$ bound such as a maximum principle or an a priori estimate $\lvert u_n \rvert \le g$ with $g \in L^{2}$ is precisely what supplies it. Without such an estimate the limit may lose energy, exactly as in $n\mathbf{1}_{(0,1/n)}$ , and the scheme converges pointwise to the wrong thing in the mean.

Expectations of truncated random variables. For $X$ with $\mathbb{E}\lvert X\rvert < \infty$ , the truncations $X_n = X\,\mathbf{1}_{\{\lvert X\rvert \le n\}}$ satisfy $X_n \to X$ a.s. and $\lvert X_n\rvert \le \lvert X \rvert \in L^{1}$ , so $\mathbb{E}(X_n) \to \mathbb{E}(X)$ . Truncation is the first move in most proofs of the law of large numbers and in the construction of conditional expectation, and dominated convergence is what lets the truncation be removed at the end.

Theorem

Fubini and Tonelli theorems

Adapted

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Gives the conditions under which a double integral may be computed as an iterated integral in either order — non-negativity for Tonelli, absolute integrability for Fubini, sigma-finiteness for both — and shows by counterexample what goes wrong when each is dropped.

Prerequisites: Dominated convergence, Monotone convergence and Fatou's lemma, The Lebesgue integral, Measurable functions, Lebesgue measure, Sigma-algebras and measures

Definition(Product sigma-algebra, sections, product measure)AdaptedAdapted from Folland (1999), §2.5, pp. 64–65; Reed & Simon (1980), Thm I.22, p. 26

Let $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ be measure spaces. A rectangle is a set $A \times B$ with $A \in \mathcal{M}$ and $B \in \mathcal{N}$ , and $\mathcal{M} \otimes \mathcal{N}$ denotes the smallest $\sigma$ -algebra on $X \times Y$ containing every rectangle (see sigma-algebras and measures).

For $E \subseteq X \times Y$ and points $x \in X$ , $y \in Y$ , the sections of $E$ are

E_x = \{\, y \in Y : (x,y) \in E \,\} \subseteq Y , \qquad E^y = \{\, x \in X : (x,y) \in E \,\} \subseteq X ,

and for a function $f$ on $X \times Y$ the sections $f_x$ and $f^y$ are the one-variable functions $f_x(y) = f^y(x) = f(x,y)$ . Sections of measurable objects are measurable: if $E \in \mathcal{M}\otimes\mathcal{N}$ then every $E_x \in \mathcal{N}$ and every $E^y \in \mathcal{M}$ , and if $f$ is $\mathcal{M}\otimes\mathcal{N}$ -measurable then every $f_x$ is $\mathcal{N}$ -measurable and every $f^y$ is $\mathcal{M}$ -measurable.

The rectangles are not closed under unions or complements, but they do form an elementary family: the empty set is a rectangle; an intersection of two rectangles is one, because $(A\times B)\cap(A'\times B') = (A\cap A')\times(B\cap B')$ ; and the complement of a rectangle is a finite disjoint union of rectangles, namely

(A\times B)^{c} \;=\; (A^{c}\times B) \ \sqcup\ (A\times B^{c}) \ \sqcup\ (A^{c}\times B^{c}) .

The third condition is the one that does the work, and by Folland's (Folland (1999), Prop. 1.7, p. 23) criterion an elementary family has the property that its finite disjoint unions form an algebra. So the finite disjoint unions of rectangles form an algebra $\mathcal{A}$ , and the $\sigma$ -algebra it generates is $\mathcal{M}\otimes\mathcal{N}$ . On $\mathcal{A}$ set

\pi\Bigl(\ \bigsqcup_{j=1}^{n} A_j \times B_j \ \Bigr) \;=\; \sum_{j=1}^{n} \mu(A_j)\,\nu(B_j) , \qquad 0\cdot\infty = 0 .

This is well defined and countably additive on $\mathcal{A}$ — it is a premeasure — so the Carathéodory machinery of Lebesgue measure extends it to a measure on $\mathcal{M}\otimes\mathcal{N}$ , the product measure $\mu \times \nu$ . Two consequences are used below, and they carry different hypotheses.

The covering formula. The Carathéodory construction defines the outer measure of an arbitrary $E \subseteq X\times Y$ as the infimum of $\sum_j \pi(C_j)$ over countable coverings of $E$ by members of $\mathcal{A}$ ; since each member of $\mathcal{A}$ is a finite union of rectangles, this is the same as $(\mu\times\nu)(E) = \inf \ \sum_{j=1}^{\infty} \mu(A_j)\,\nu(B_j) ,$ the infimum running over all countable coverings $E \subseteq \bigcup_j A_j \times B_j$ by rectangles. This is a definition, not a theorem, so it holds whether or not the factors are $\sigma$ -finite — a point the counterexample in the remark below depends on.
Uniqueness. If $\mu$ and $\nu$ are $\sigma$ -finite, then so is the premeasure $\pi$ , and $\mu\times\nu$ is the unique measure on $\mathcal{M}\otimes\mathcal{N}$ with $(\mu\times\nu)(A\times B) = \mu(A)\nu(B)$ for every rectangle. Reed and Simon (Reed & Simon (1980), Thm I.22) state the uniqueness and the covering formula together, under their standing $\sigma$ -finiteness hypothesis.

Recall that $\mu$ is $\sigma$ -finite if $X$ is a countable union of sets of finite $\mu$ -measure. If $\mu$ and $\nu$ are $\sigma$ -finite, so is $\mu\times\nu$ , since the products of the two exhausting families exhaust $X\times Y$ . Write $L^{+}$ for the measurable functions with values in $[0,\infty]$ .

Theorem(Measure of a set through its sections)AdaptedAdapted from Folland (1999), Thm 2.36, pp. 66–67

Let $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ be $\sigma$ -finite. If $E \in \mathcal{M}\otimes\mathcal{N}$ , then $x \mapsto \nu(E_x)$ is $\mathcal{M}$ -measurable, $y \mapsto \mu(E^y)$ is $\mathcal{N}$ -measurable, and

(\mu\times\nu)(E) \;=\; \int_X \nu(E_x)\,d\mu(x) \;=\; \int_Y \mu(E^y)\,d\nu(y) .

In words: the measure of $E$ is obtained by sweeping a slice across $X$ and adding up the slice measures, and the answer does not depend on which variable is swept.

Theorem(The Fubini-Tonelli theorem)AdaptedAdapted from Folland (1999), Thm 2.37 and (2.38), p. 67

Let $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ be $\sigma$ -finite measure spaces.

Tonelli. If $f \in L^{+}(X\times Y)$ , then $g(x) = \int_Y f_x \,d\nu$ lies in $L^{+}(X)$ , $h(y) = \int_X f^y \,d\mu$ lies in $L^{+}(Y)$ , and $\int_{X\times Y} f \,d(\mu\times\nu) = \int_X\!\Bigl[\int_Y f(x,y)\,d\nu(y)\Bigr] d\mu(x) = \int_Y\!\Bigl[\int_X f(x,y)\,d\mu(x)\Bigr] d\nu(y) .$ No integrability is assumed: all three quantities may equal $+\infty$ , and the assertion is that they are equal.
Fubini. If $f \in L^{1}(\mu\times\nu)$ , then $f_x \in L^{1}(\nu)$ for almost every $x$ , $f^y \in L^{1}(\mu)$ for almost every $y$ , the almost-everywhere defined functions $g$ and $h$ above lie in $L^{1}(\mu)$ and $L^{1}(\nu)$ respectively, and the same chain of equalities holds.

No continuity, and no topology on $X$ or $Y$ , enters either part.

Note(Three other formulations in the library)AdaptedAdapted from Reed & Simon (1980), Thm I.21, p. 23; Stakgold & Holst (2011), Thm 0.7.4, p. 47; Lee (2013), Thm C.27, p. 654

The books in the library state the theorem in forms that differ in which hypothesis is taken as given. None contradicts the version above; they differ in convenience.

Reed and Simon (Reed & Simon (1980), Thm I.21) state it as an equivalence for a measurable $f$ on $M\times N$ : the iterated integral $\int_M\bigl(\int_N |f(m,n)|\,d\nu(n)\bigr)d\mu(m)$ is finite if and only if $\int_N\bigl(\int_M |f(m,n)|\,d\mu(m)\bigr)d\nu(n)$ is finite, and when one of them (hence both) is finite, the two iterated integrals of $f$ itself agree. Their $\sigma$ -finiteness hypothesis is not repeated in the statement because they have already declared that every measure they use is $\sigma$ -finite. This "if and only if" is just Tonelli applied to $|f|$ : it is the practical form, because it says that either iterated integral of $|f|$ may be used as the test.
Stakgold and Holst (Stakgold & Holst (2011), Thm 0.7.4) state the concrete case $X\times Y = \mathbb{R}^{m}\times\mathbb{R}^{n}$ for measurable $f$ : if one or more of the double integral $I_1 = \int_{\mathbb{R}^{m+n}}|f|$ and the two iterated integrals $I_2, I_3$ of $|f|$ exists and is finite, then $f(\cdot,y) \in L^1(\mathbb{R}^m)$ for almost every $y$ , $f(x,\cdot) \in L^1(\mathbb{R}^n)$ for almost every $x$ , the two partial integrals are themselves integrable, and $I_1 = I_2 = I_3$ . Lebesgue measure on $\mathbb{R}^k$ is $\sigma$ -finite, so no extra hypothesis is needed.
Lee (Lee (2013), Thm C.27) gives the version that needs no measure theory at all: for $f$ continuous on a closed rectangle $A = [a_1,b_1]\times\cdots\times[a_n,b_n] \subset \mathbb{R}^n$ , the integral over $A$ equals the $n$ -fold iterated integral. A continuous function on a compact rectangle is bounded, hence absolutely integrable, so this is a special case of Fubini — but it is proved directly from the Riemann theory and is all that is needed in differential geometry.

Idea of the proofOriginal

Everything reduces to indicator functions. For $f = \chi_E$ the statement is that the measure of $E$ can be computed by slicing, and that statement is true for rectangles by the very definition of $\mu\times\nu$ . The family of sets for which it is true is closed under increasing unions and decreasing intersections — the first by monotone convergence, the second by dominated convergence — so it is a monotone class containing the algebra of finite disjoint unions of rectangles, and the monotone class lemma makes it all of $\mathcal{M}\otimes\mathcal{N}$ . From indicators one passes to non-negative simple functions by linearity, and from simple functions to all of $L^{+}$ by monotone convergence, which is what makes Tonelli a theorem about non-negative functions and nothing else. Fubini is then bookkeeping: split a complex integrable $f$ into four non-negative pieces, apply Tonelli to each, and note that the pieces have finite integrals so the subtractions are legitimate. The two places where a hypothesis is genuinely consumed are the decreasing-intersection step, which needs finiteness and hence $\sigma$ -finiteness, and the subtraction, which needs $f$ to be absolutely integrable.

Proof(Sections: the monotone class argument)AdaptedAdapted from Folland (1999), Thm 2.36 with Lemma 2.35, pp. 66–67

Step 1: reduce to finite measures. Suppose first that $\mu(X) < \infty$ and $\nu(Y) < \infty$ . Let $\mathcal{C}$ be the collection of all $E \in \mathcal{M}\otimes\mathcal{N}$ for which the two section functions are measurable and the displayed identity holds.

Step 2: rectangles and the algebra they generate. For $E = A\times B$ the sections are $E_x = B$ if $x \in A$ and $E_x = \varnothing$ otherwise, so $\nu(E_x) = \nu(B)\chi_A(x)$ , a measurable function whose integral is $\mu(A)\nu(B)$ ; symmetrically for $E^y$ . Both integrals therefore equal $(\mu\times\nu)(A\times B)$ , so every rectangle lies in $\mathcal{C}$ . Sections and integrals are additive over disjoint unions, so every finite disjoint union of rectangles lies in $\mathcal{C}$ ; that is, $\mathcal{A} \subseteq \mathcal{C}$ .

Step 3: $\mathcal{C}$ is closed under increasing unions. Let $E_1 \subseteq E_2 \subseteq \cdots$ lie in $\mathcal{C}$ with union $E$ . Taking sections commutes with unions, so $\mu\bigl((E_n)^y\bigr)$ increases to $\mu(E^y)$ by continuity of a measure from below. A pointwise increasing limit of measurable functions is measurable, and the monotone convergence theorem gives

\int_Y \mu(E^y)\,d\nu(y) = \lim_n \int_Y \mu\bigl((E_n)^y\bigr)d\nu(y) = \lim_n (\mu\times\nu)(E_n) = (\mu\times\nu)(E) ,

the last equality again by continuity from below, now for $\mu\times\nu$ . The same argument works in the other variable, so $E \in \mathcal{C}$ .

Step 4: $\mathcal{C}$ is closed under decreasing intersections. Let $E_1 \supseteq E_2 \supseteq \cdots$ lie in $\mathcal{C}$ with intersection $E$ . Here the sections decrease, so we need continuity from above, which requires a finite dominating quantity: the function $y \mapsto \mu\bigl((E_1)^y\bigr)$ is bounded above by the constant $\mu(X) < \infty$ , and a bounded function on a space of finite measure is integrable because $\nu(Y) < \infty$ , so it belongs to $L^1(\nu)$ and dominates every $\mu\bigl((E_n)^y\bigr)$ . Dominated convergence then passes the limit through the integral, and $E \in \mathcal{C}$ . This is the step that uses finiteness of the measures.

Step 5: the monotone class lemma. By steps 3 and 4, $\mathcal{C}$ is a monotone class; by step 2 it contains the algebra $\mathcal{A}$ . Folland's (Folland (1999), Lemma 2.35, p. 66) monotone class lemma says that the monotone class generated by an algebra is already the $\sigma$ -algebra generated by it, so $\mathcal{C} \supseteq \mathcal{M}\otimes\mathcal{N}$ , which is the theorem for finite measures.

Step 6: back to $\sigma$ -finite measures. Choose rectangles $X_j \times Y_j$ of finite product measure increasing to $X\times Y$ ; this is possible precisely because both factors are $\sigma$ -finite. Step 5 applies to each restricted space and gives

(\mu\times\nu)\bigl(E \cap (X_j\times Y_j)\bigr) = \int_X \chi_{X_j}(x)\,\nu(E_x \cap Y_j)\,d\mu(x) = \int_Y \chi_{Y_j}(y)\,\mu(E^y \cap X_j)\,d\nu(y) .

Let $j \to \infty$ . On the left, the sets $E \cap (X_j\times Y_j)$ increase to $E$ , so $(\mu\times\nu)\bigl(E\cap(X_j\times Y_j)\bigr) \to (\mu\times\nu)(E)$ by continuity of $\mu\times\nu$ from below. On the right, the sections of the truncations increase to the sections of $E$ , so both integrands increase pointwise to $\nu(E_x)$ and $\mu(E^y)$ , and monotone convergence carries the limit inside the two integrals. Equating the three limits gives the identity. $\square$

Proof(Tonelli, then Fubini)AdaptedAdapted from Folland (1999), Thm 2.37, p. 67

Step 1: indicators. For $f = \chi_E$ with $E \in \mathcal{M}\otimes\mathcal{N}$ we have $f_x = \chi_{E_x}$ and $f^y = \chi_{E^y}$ , so $g(x) = \nu(E_x)$ and $h(y) = \mu(E^y)$ and the previous theorem is exactly Tonelli for $f$ .

Step 2: simple functions. A non-negative simple function is a finite non-negative combination $\sum_k c_k \chi_{E_k}$ . Sectioning, the integral over $X\times Y$ , and both iterated integrals are all linear in $f$ with non-negative coefficients, so step 1 extends to this case. Measurability of $g$ and $h$ is preserved because finite sums of measurable functions are measurable.

Step 3: general $f \in L^{+}$ . By the simple-function approximation of measurable functions, choose simple $0 \le f_1 \le f_2 \le \cdots$ with $f_n \to f$ pointwise on $X\times Y$ . For each fixed $x$ the sections $(f_n)_x$ increase to $f_x$ , so monotone convergence in the $Y$ variable gives $g_n(x) \uparrow g(x)$ ; in particular $g$ is a pointwise limit of measurable functions and so is measurable, and likewise $h_n \uparrow h$ . Applying monotone convergence twice more,

\int_X g\,d\mu = \lim_n \int_X g_n \,d\mu = \lim_n \int_{X\times Y} f_n\,d(\mu\times\nu) = \int_{X\times Y} f\,d(\mu\times\nu) ,

where the middle equality is step 2. The same three lines in the other order give $\int_Y h\,d\nu = \int f\,d(\mu\times\nu)$ . This is Tonelli. Beyond the $\sigma$ -finiteness already consumed in the previous proof, no hypothesis on $f$ other than $f \ge 0$ and measurability is used, which is why Tonelli is the test one applies first.

Step 4: integrable sections. Suppose $f \in L^{+}$ with $\int f \,d(\mu\times\nu) < \infty$ . By step 3, $\int_X g\,d\mu < \infty$ , and a non-negative function with finite integral is finite almost everywhere. Hence $g(x) = \int_Y f_x\,d\nu < \infty$ for almost every $x$ , that is $f_x \in L^1(\nu)$ for almost every $x$ ; symmetrically $f^y \in L^1(\mu)$ for almost every $y$ .

Step 5: Fubini. Let $f \in L^1(\mu\times\nu)$ be complex valued and decompose it as

f = (\operatorname{Re} f)^{+} - (\operatorname{Re} f)^{-} + i (\operatorname{Im} f)^{+} - i(\operatorname{Im} f)^{-} ,

four functions in $L^{+}$ each dominated by $|f|$ and hence each of finite integral. Steps 3 and 4 apply to each of the four: their sections are integrable almost everywhere, their partial integrals are integrable functions, and for each of them the double integral equals both iterated integrals. Adding the four identities with the coefficients $1,-1,i,-i$ gives the result for $f$ . The subtractions are legitimate because each of the four integrals is finite — this is where $f \in L^1$ , rather than merely $f$ measurable, is used; and the exceptional null sets are discarded by taking the union of the four, which is again null. $\square$

Remark(Where the hypotheses are used, and what fails without them)AdaptedAdapted from Folland (1999), the remarks following Thm 2.37 and Exercises 46–48, pp. 68–69

$\sigma$ -finiteness is consumed at step 4 of the first proof, where dominated convergence needs a finite majorant, and again at step 6 where the exhausting rectangles are built. It cannot be dropped. Folland (Folland (1999), Exercise 46, pp. 68–69) gives the standard witness: take $X = Y = [0,1]$ with the Borel sets, $\mu$ Lebesgue measure and $\nu$ counting measure, and let $D = \{(x,x)\}$ be the diagonal. Counting measure on an uncountable set is not $\sigma$ -finite. Each section $D^y = \{y\}$ has $\mu(D^y)=0$ , so $\int_Y \mu(D^y)\,d\nu(y) = 0$ ; each section $D_x = \{x\}$ has $\nu(D_x)=1$ , so $\int_X \nu(D_x)\,d\mu(x) = 1$ . For the product measure itself we use the covering formula of the definition, which is legitimate here precisely because it is the Carathéodory outer measure attached to the premeasure $\pi$ and so does not presuppose the $\sigma$ -finiteness that has just failed. Suppose $D \subseteq \bigcup_j A_j\times B_j$ with $\sum_j \mu(A_j)\nu(B_j) < \infty$ . Every $j$ with $B_j$ infinite has $\nu(B_j)=\infty$ and hence $\mu(A_j)=0$ ; the union $S$ of the finite $B_j$ is countable, hence $\mu(S)=0$ . Any $x \notin S$ lies in some $A_j$ with $B_j$ infinite, so $[0,1]\setminus S$ is covered by countably many $\mu$ -null sets, which is impossible. Hence $(\mu\times\nu)(D) = \infty$ , and the three numbers are $0$ , $1$ and $\infty$ : all different, as Folland asserts.
Non-negativity or integrability is consumed at step 5 of the second proof. Folland notes two separate ways the conclusion can fail without it: the sections can all be measurable and both iterated integrals exist while $f$ is not even product-measurable, and $f$ can be product-measurable with both iterated integrals existing while $\int|f|\,d(\mu\times\nu) = \infty$ . In both cases the two iterated integrals need not agree. The second failure already happens with counting measure on $\mathbb{N}\times\mathbb{N}$ , a perfectly $\sigma$ -finite space; see the example below.
Measurability of $f$ on the product is not implied by measurability of all its sections, and the theorem says nothing about functions that are not product-measurable.
Completeness is not inherited. Folland points out that $\mu\times\nu$ is almost never complete even when $\mu$ and $\nu$ are: if $\mu(A) = 0$ for some non-empty $A$ and $\mathcal{N}$ is not all of $\mathcal{P}(Y)$ , then for $E \notin \mathcal{N}$ the set $A \times E$ is a subset of the null set $A\times Y$ but is not in $\mathcal{M}\otimes\mathcal{N}$ . Lebesgue measure on $\mathbb{R}^2$ is the completion of the product of the two one-dimensional Lebesgue measures, not the product itself; for that setting Folland's Thm 2.39 restates Fubini–Tonelli for functions measurable with respect to the completion, concluding only that $f_x$ is $\mathcal{N}$ -measurable for almost every $x$ and $f^y$ is $\mathcal{M}$ -measurable for almost every $y$ . That is the version implicitly used whenever one integrates over $\mathbb{R}^2$ .
The conclusion of Tonelli is an identity between possibly infinite numbers, so it never asserts convergence. It is used to decide convergence, not to assume it.

Example(A

\sigma

-finite counterexample: the two orders differ)AdaptedAdapted from Folland (1999), Exercise 48, p. 69

Let $X = Y = \mathbb{N}$ with counting measure, which is $\sigma$ -finite, so the only hypothesis available to fail is absolute integrability. The array used here is the one in Folland's (Folland (1999), Exercise 48, p. 69) exercise, with the roles of the two indices interchanged; the computation and the discussion are ours. Integration is summation, and $\mathcal{M}\otimes\mathcal{N}$ is the whole power set, so every $f$ is measurable. Put

a_{m,n} = \begin{cases} \ \ 1, & n = m, \\ -1, & n = m+1, \\ \ \ 0, & \text{otherwise.}\end{cases}

Summing over $n$ first. For each fixed $m$ the row has exactly the two non-zero entries $a_{m,m}=1$ and $a_{m,m+1}=-1$ , so $\sum_n a_{m,n} = 0$ and therefore $\sum_m \sum_n a_{m,n} = 0$ .

Summing over $m$ first. The column $n=1$ contains only $a_{1,1}=1$ , so its sum is $1$ . For $n \ge 2$ the column contains $a_{n,n}=1$ and $a_{n-1,n}=-1$ and nothing else, so its sum is $0$ . Therefore $\sum_n\sum_m a_{m,n} = 1$ .

The two iterated integrals are $0$ and $1$ . Both exist, every section is integrable, and the space is $\sigma$ -finite; what fails is that $\sum_{m,n}|a_{m,n}| = \sum_m 2 = \infty$ , so $f \notin L^1(\mu\times\nu)$ and Fubini does not apply. Tonelli applied to $|a_{m,n}|$ detects this immediately: both iterated sums of $|a_{m,n}|$ equal $\infty$ , in agreement with Tonelli and in warning against Fubini.

Method(Interchanging the order of integration)Original

Check product-measurability of $f$ . For an explicit formula built from continuous functions and countably many measurable sets this is automatic; it is a genuine hypothesis only for exotic constructions.
Check that both factors are $\sigma$ -finite. Lebesgue measure on $\mathbb{R}^k$ , any probability measure and counting measure on a countable set all are. Counting measure on an uncountable set is the usual offender.
Apply Tonelli to $|f|$ . Compute $\int\!\int |f|$ in whichever order is easier; Tonelli guarantees the answer is the same in either order and equals $\int |f| \,d(\mu\times\nu)$ , finite or not.
If that number is finite, $f \in L^1(\mu\times\nu)$ and Fubini licenses the interchange for $f$ itself. If it is infinite, stop: the interchange is not justified, and the example above shows it can genuinely be false.
If $f \ge 0$ already, skip steps 3 and 4: Tonelli applies directly, and an infinite answer is a legitimate conclusion rather than an obstruction.
Term-by-term integration of a series is the case $\mu$ = counting measure on $\mathbb{N}$ : $\int \sum_n f_n = \sum_n \int f_n$ holds for non-negative $f_n$ with no further hypothesis, and for general $f_n$ as soon as $\sum_n \int |f_n| < \infty$ .

Example(Evaluating a double integral by Tonelli)Original

Compute

I = \int_0^1\!\!\int_0^1 \frac{dx\,dy}{1 - xy} .

Step 1: the integrand is non-negative. On the open square $(0,1)^2$ we have $0 < xy < 1$ , so $1/(1-xy) > 0$ , and it is continuous there, hence Borel measurable. Lebesgue measure on $(0,1)$ is finite, so both factors are $\sigma$ -finite. Tonelli therefore applies with no integrability check.

Step 2: expand in a geometric series. For $0 < xy < 1$ ,

\frac{1}{1-xy} = \sum_{k=0}^{\infty} (xy)^k = \sum_{k=0}^{\infty} x^k y^k ,

a series of non-negative terms. Counting measure on $\{0,1,2,\dots\}$ is $\sigma$ -finite, so Tonelli applies again, now to the product of counting measure with the two copies of Lebesgue measure: the sum and the two integrals may be done in any order.

Step 3: integrate term by term.

I = \sum_{k=0}^{\infty} \int_0^1 x^k \,dx \int_0^1 y^k \,dy = \sum_{k=0}^{\infty} \frac{1}{k+1}\cdot\frac{1}{k+1} = \sum_{k=0}^{\infty}\frac{1}{(k+1)^2} = \sum_{n=1}^{\infty}\frac{1}{n^2} .

Step 4: the number. The last series is $\zeta(2) = \pi^2/6$ , so

I = \frac{\pi^2}{6} = 1.6449340668\ldots

Numerical quadrature of the double integral gives $1.6449340668$ , agreeing to ten decimals.

What would go wrong without Tonelli. Nothing here is absolutely convergent by inspection: the integrand is unbounded near the corner $(1,1)$ , and the naive justification "the series converges uniformly" is false on $(0,1)^2$ . Non-negativity is the whole licence. Had the series been $\sum_k (-1)^k (xy)^k = 1/(1+xy)$ one would first apply Tonelli to the absolute values, find $\sum_k 1/(k+1)^2 = \pi^2/6 < \infty$ , and only then interchange, obtaining $\int_0^1\!\int_0^1 (1+xy)^{-1} dx\,dy = \sum_{k\ge 0}(-1)^k/(k+1)^2 = \pi^2/12 = 0.8224670334\ldots$

ApplicationOriginal

Convolution. That $f * g (x) = \int f(x-y) g(y)\,dy$ is defined almost everywhere and satisfies $\|f*g\|_1 \le \|f\|_1\|g\|_1$ for $f,g \in L^1(\mathbb{R}^n)$ is proved by applying Tonelli to $|f(x-y)g(y)|$ on $\mathbb{R}^n\times\mathbb{R}^n$ : the iterated integral in $x$ first factorises into $\|f\|_1\|g\|_1$ , which is finite, and Fubini then gives the rest. Every manipulation of the convolution theorem and of Green's-function solutions written as convolutions rests on this.

Fourier analysis in several variables. Interchanging $\int_{\mathbb{R}^n}$ with $\int_{\mathbb{R}^n}$ is what turns the Fourier transform of a convolution into a product, and what lets the $n$ -dimensional transform of the transform on $\mathbb{R}^n$ be computed as $n$ one-dimensional transforms when the integrand factorises. For a Gaussian the factorisation plus Fubini reduces an $n$ -dimensional integral to the $n$ th power of a one-dimensional one.

Statistical mechanics. A classical partition function $Z = \int e^{-\beta H(q,p)} dq\,dp$ with $H = \sum_i H_i(q_i,p_i)$ factorises into $\prod_i \int e^{-\beta H_i}$ exactly by Tonelli, the integrand being positive; this is the mathematical content of "non-interacting degrees of freedom contribute multiplicatively to $Z$ and additively to $\log Z$ ".

Probability. Independence means exactly that the joint law factorises as a product measure, so $\mathbb{E}[XY] = \mathbb{E}[X]\mathbb{E}[Y]$ for independent integrable $X, Y$ is Fubini on the product space, and the density of a sum is the convolution of the densities. The same theorem justifies the ubiquitous identity $\mathbb{E}[X] = \int_0^\infty \mathbb{P}(X > t)\,dt$ for $X \ge 0$ , obtained by applying Tonelli to $\chi_{\{X(\omega) > t\}}$ on $\Omega \times (0,\infty)$ . See joint distributions.

Changing the order in physics calculations. Swapping a time integral with a space integral, a sum over modes with an integral over the domain, or a radial with an angular integral are all instances; the practical discipline step 3 of the method imposes — test the absolute value first — is what separates a safe interchange from a divergent one in, for instance, retarded-potential and perturbation-series calculations.

Definition

Modes of convergence

Adapted

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Defines almost-everywhere, almost-uniform, in-measure and $L^p$ convergence, settles exactly which of them implies which, proves Egorov's theorem and the almost-everywhere subsequence theorem, and gives the four standard counterexamples that block every remaining implication.

Prerequisites: Dominated convergence, Monotone convergence and Fatou's lemma, The Lebesgue integral, Measurable functions, Lebesgue measure, Sigma-algebras and measures, Uniform convergence

Definition(The four modes)AdaptedAdapted from Folland (1999), §2.4, p. 61; Grimmett & Stirzaker (2020), Definition (1), p. 352

Let $(X,\mathcal{M},\mu)$ be a measure space and let $f, f_1, f_2, \dots$ be measurable complex-valued functions on $X$ (see measurable functions). We say that $f_n \to f$

almost everywhere (a.e.) if $\mu\bigl(\{x : f_n(x) \not\to f(x)\}\bigr) = 0$ ;
almost uniformly if for every $\varepsilon > 0$ there is a measurable $E$ with $\mu(E) < \varepsilon$ such that $f_n \to f$ uniformly on $X \setminus E$ ;
in measure if for every $\varepsilon > 0$ , $\mu\bigl(\{x : |f_n(x) - f(x)| \ge \varepsilon\}\bigr) \longrightarrow 0 \quad\text{as } n \to \infty ;$
in $L^p$ ( $1 \le p < \infty$ ) if every $f_n$ and $f$ have $\int |f_n|^p\,d\mu < \infty$ and $\|f_n - f\|_p = \bigl(\int |f_n - f|^p d\mu\bigr)^{1/p} \to 0$ . For $p = 1$ this is called convergence in mean, for $p = 2$ convergence in mean square; see $L^p$ spaces.

The sequence is Cauchy in measure if $\mu(\{x : |f_n(x) - f_m(x)| \ge \varepsilon\}) \to 0$ as $m, n \to \infty$ , for every $\varepsilon > 0$ . A note on the inequality: Folland (Folland (1999), §2.4, p. 61) and Grimmett and Stirzaker (Grimmett & Stirzaker (2020), Definition (1c), p. 352) write these two definitions with the strict inequality $|f_n - f| > \varepsilon$ . The two forms define the same notion, because both are quantified over every $\varepsilon > 0$ and $\{|f_n - f| > \varepsilon\} \subseteq \{|f_n - f| \ge \varepsilon\} \subseteq \{|f_n - f| > \varepsilon/2\}$ ; we use $\ge$ throughout the definitions, the statement and the examples, and keep the strict form only inside the proof of part 4, where the sets $E_n(k)$ are written as Folland writes them; either inequality serves there. Convergence in measure and almost-everywhere convergence determine the limit only up to a null set, so all four modes are statements about equivalence classes of functions equal a.e.

Two warnings about the shape of the definitions. Almost-uniform convergence is not "uniform convergence off a null set": the exceptional set is small but need not be null, and it is allowed to grow as $\varepsilon$ shrinks. And convergence in measure is a statement about the size of the set where the error is large, with no control at all of how large the error is on that set — which is the opposite emphasis to uniform convergence, where the error is controlled everywhere.

Note(The same four modes in probability)AdaptedAdapted from Grimmett & Stirzaker (2020), Definition (1) and the notes following it, pp. 352–353, and Thms (3) and (4), p. 354

On a probability space $(\Omega,\mathcal{F},\mathbb{P})$ the measure-theoretic modes acquire probabilists' names: a.e. convergence becomes almost sure convergence $X_n \xrightarrow{\text{a.s.}} X$ (also "with probability 1"), convergence in measure becomes convergence in probability $X_n \xrightarrow{P} X$ , and $L^r$ convergence becomes convergence in $r$ th mean, written $X_n \xrightarrow{r} X$ and defined by $\mathbb{E}|X_n - X|^r \to 0$ with $\mathbb{E}|X_n^r| < \infty$ for every $n$ .

Grimmett and Stirzaker (Grimmett & Stirzaker (2020), §7.2, Definition (1d) and Thm (3), pp. 352, 354) add a fourth mode with no analogue in a general measure space: convergence in distribution, $\mathbb{P}(X_n \le x) \to \mathbb{P}(X \le x)$ at every continuity point of the limit distribution function. It is a statement about distribution functions alone, so it does not even require the $X_n$ to live on one space. Their chart of implications is

\bigl(X_n \xrightarrow{\text{a.s.}} X\bigr) \Rightarrow \bigl(X_n \xrightarrow{P} X\bigr) \Rightarrow \bigl(X_n \xrightarrow{D} X\bigr), \qquad \bigl(X_n \xrightarrow{r} X\bigr) \Rightarrow \bigl(X_n \xrightarrow{P} X\bigr),

together with $\bigl(X_n \xrightarrow{r} X\bigr) \Rightarrow \bigl(X_n \xrightarrow{s} X\bigr)$ for $r > s \ge 1$ , and they state that no other implications hold in general. Their partial converses are: convergence in distribution to a constant implies convergence in probability; convergence in probability with a uniform bound $\mathbb{P}(|X_n| \le k) = 1$ implies $r$ th-mean convergence for every $r \ge 1$ ; and $\sum_n \mathbb{P}(|X_n - X| \ge \varepsilon) < \infty$ for every $\varepsilon > 0$ implies almost sure convergence.

The first arrow of their chart is stronger than what holds on a general measure space, and the reason is that a probability measure is finite: see the last sentence of part 5 of the statement below, which is that arrow, and part 4 (Egorov), which is what supplies it. The $r > s$ implication likewise uses $\mathbb{P}(\Omega) = 1$ and fails on $\mathbb{R}$ . These are not disagreements with Folland but consequences of a standing hypothesis that probability carries for free. See convergence of random variables.

Definition(Which mode implies which)AdaptedAdapted from Folland (1999), Prop. 2.29, Thm 2.30, Cor. 2.32 and Thm 2.33, pp. 61–62; Evans (2010), Thm 2, pp. 685–686

Let $(X,\mathcal{M},\mu)$ be a measure space.

$L^p$ implies in measure. If $\|f_n - f\|_p \to 0$ for some $1 \le p < \infty$ , then $f_n \to f$ in measure. Folland (Folland (1999), Prop. 2.29, p. 61) states and proves the case $p = 1$ ; the general case is the same Chebyshev estimate.
Cauchy in measure is enough. If $\{f_n\}$ is Cauchy in measure, there is a measurable $f$ with $f_n \to f$ in measure, and some subsequence $\{f_{n_j}\}$ with $f_{n_j} \to f$ a.e. The limit is unique up to a null set: if also $f_n \to g$ in measure then $g = f$ a.e.
Consequently, if $f_n \to f$ in $L^1$ then some subsequence converges to $f$ a.e.
Egorov's theorem. If $\mu(X) < \infty$ and $f_n \to f$ a.e., then for every $\varepsilon > 0$ there is a measurable $E$ with $\mu(E) < \varepsilon$ such that $f_n \to f$ uniformly on $X \setminus E$ ; that is, a.e. convergence on a finite measure space is almost uniform convergence. Evans (Evans (2010), Thm 2, pp. 685–686) states it for Lebesgue measure on a measurable $A \subset \mathbb{R}^n$ with $|A| < \infty$ , with the conclusion written as $|A \setminus E| \le \varepsilon$ and $f_k \to f$ uniformly on $E$ .
Almost uniform implies both a.e. convergence and convergence in measure. With part 4 this gives: on a finite measure space, a.e. convergence implies convergence in measure.
Domination upgrades a.e. to $L^1$ . If $f_n \to f$ a.e. and $|f_n| \le g$ for all $n$ with $g \in L^1$ , then $f_n \to f$ in $L^1$ , by dominated convergence applied to $|f_n - f| \le 2g$ .

No implication other than those listed, and their compositions, is valid: the remark below gives four sequences which between them block every remaining implication, some of them only in combination with an implication already listed.

Idea of the proofOriginal

The three substantive statements have three different mechanisms, and it is worth separating them.

Convergence in measure is weak because it only asks that a bad set be small; Chebyshev's inequality converts an integral bound into exactly such a statement, which is why every $L^p$ convergence implies it. Recovering pointwise information from convergence in measure is possible only along a subsequence, and the device is to make the bad sets summable: if the $n$ th bad set has measure $2^{-n}$ , then the union of all bad sets from the $k$ th on is still small, and off that union the sequence is uniformly Cauchy by comparison with a geometric series. Egorov's theorem runs the opposite way, from pointwise to almost uniform, and the engine is continuity of the measure from above: the sets where some later term is still off by more than $1/k$ shrink to the empty set, so their measures shrink to zero — and this last step is exactly what fails without $\mu(X) < \infty$ . Choosing the tolerance $\varepsilon 2^{-k}$ at level $k$ and taking the union lets one error budget pay for all $k$ at once.

Proof(Part 1: Chebyshev's inequality)AdaptedAdapted from Folland (1999), Prop. 2.29, p. 61

Fix $\varepsilon > 0$ and put $E_{n,\varepsilon} = \{x : |f_n(x) - f(x)| \ge \varepsilon\}$ . On $E_{n,\varepsilon}$ the integrand $|f_n - f|^p$ is at least $\varepsilon^p$ , so restricting the integral to that set and bounding below,

\|f_n - f\|_p^p = \int_X |f_n - f|^p \,d\mu \;\ge\; \int_{E_{n,\varepsilon}} |f_n - f|^p \,d\mu \;\ge\; \varepsilon^p\, \mu(E_{n,\varepsilon}) ,

using monotonicity of the integral at both steps. Hence $\mu(E_{n,\varepsilon}) \le \varepsilon^{-p}\|f_n - f\|_p^p \to 0$ . As $\varepsilon > 0$ was arbitrary, $f_n \to f$ in measure. $\square$

Proof(Part 2: from Cauchy in measure to an a.e. limit)AdaptedAdapted from Folland (1999), Thm 2.30, pp. 61–62

Step 1: choose a fast subsequence. Being Cauchy in measure, the sequence admits indices $n_1 < n_2 < \cdots$ such that, writing $g_j = f_{n_j}$ and

E_j = \{x : |g_j(x) - g_{j+1}(x)| \ge 2^{-j}\} ,

we have $\mu(E_j) \le 2^{-j}$ . Indeed, for each $j$ apply the Cauchy condition with $\varepsilon = 2^{-j}$ to obtain an index beyond which all pairs have bad set of measure at most $2^{-j}$ .

Step 2: a tail of bad sets is small. Let $F_k = \bigcup_{j \ge k} E_j$ . Countable subadditivity gives $\mu(F_k) \le \sum_{j\ge k} 2^{-j} = 2^{1-k}$ .

Step 3: off $F_k$ the subsequence is uniformly Cauchy. Let $x \notin F_k$ and $i > j \ge k$ . Then $x \notin E_l$ for every $l \ge k$ , so telescoping and summing a geometric series,

|g_j(x) - g_i(x)| \le \sum_{l=j}^{i-1} |g_l(x) - g_{l+1}(x)| < \sum_{l=j}^{\infty} 2^{-l} = 2^{1-j} .

The bound $2^{1-j}$ does not depend on $x$ or on $i$ , so $\{g_j\}$ is uniformly Cauchy on $X \setminus F_k$ and converges there.

Step 4: the limit function. Let $F = \bigcap_k F_k$ . Since $\mu(F) \le \mu(F_k) \le 2^{1-k}$ for every $k$ , $\mu(F) = 0$ . Define $f(x) = \lim_j g_j(x)$ for $x \notin F$ and $f(x) = 0$ for $x \in F$ ; this is a pointwise limit of measurable functions off a measurable set, hence measurable. Every $x \notin F$ lies outside some $F_k$ , so $g_j \to f$ a.e.

Step 5: the subsequence converges in measure. Letting $i \to \infty$ in step 3 gives $|g_j(x) - f(x)| \le 2^{1-j}$ for all $x \notin F_k$ and all $j \ge k$ . Hence for $\varepsilon>0$ , as soon as $2^{1-j} < \varepsilon$ and $j \ge k$ , the set where $|g_j - f| \ge \varepsilon$ is contained in $F_k$ , of measure at most $2^{1-k}$ . Taking $k$ large makes this arbitrarily small, so $g_j \to f$ in measure.

\{|f_n - f| \ge \varepsilon\} \subseteq \{|f_n - g_j| \ge \tfrac{\varepsilon}{2}\} \cup \{|g_j - f| \ge \tfrac{\varepsilon}{2}\} .

Given $\delta > 0$ , choose $j$ with $\mu(\{|g_j - f| \ge \varepsilon/2\}) < \delta/2$ and with $n_j$ large enough that the Cauchy condition makes $\mu(\{|f_n - g_j|\ge \varepsilon/2\}) < \delta/2$ for all large $n$ . Then $\mu(\{|f_n - f|\ge\varepsilon\}) < \delta$ .

Step 7: uniqueness of the limit. If also $f_n \to g$ in measure, the same two-set inclusion applied to $f$ , $f_n$ , $g$ shows that $\mu(\{|f - g| \ge \varepsilon\})$ is smaller than any positive number, hence zero, for every $\varepsilon > 0$ . Taking $\varepsilon = 1/m$ and using countable subadditivity on $\{|f-g| > 0\} = \bigcup_m \{|f-g| \ge 1/m\}$ gives $f = g$ a.e. $\square$

Proof(Part 4: Egorov's theorem)AdaptedAdapted from Folland (1999), Thm 2.33, p. 62

First reduce to convergence at every point. Let $Z = \{x : f_n(x) \not\to f(x)\}$ , a measurable set with $\mu(Z) = 0$ , and replace $f$ and every $f_n$ by $0$ on $Z$ . The modified sequence converges to the modified limit at every point of $X$ , and the modification changes neither $\mu$ of any set nor the difference $f_n - f$ off $Z$ . So if the theorem is proved for the modified sequence, giving a measurable $E$ with $\mu(E) < \varepsilon$ and uniform convergence off $E$ , then the original sequence converges uniformly off $E \cup Z$ , which is measurable with $\mu(E \cup Z) \le \mu(E) + \mu(Z) = \mu(E) < \varepsilon$ and so is an admissible exceptional set. We may therefore assume $f_n \to f$ at every point of $X$ .

Step 1: the sets that measure "still off by $1/k$ somewhere later". For $k, n \in \mathbb{N}$ put

E_n(k) = \bigcup_{m \ge n} \Bigl\{ x : |f_m(x) - f(x)| > \tfrac{1}{k} \Bigr\} .

For fixed $k$ these decrease as $n$ increases, because the union is over fewer terms.

Step 2: they shrink to nothing. If $x \in \bigcap_{n} E_n(k)$ then $|f_m(x) - f(x)| > 1/k$ for infinitely many $m$ , contradicting $f_m(x) \to f(x)$ . So $\bigcap_n E_n(k) = \varnothing$ . Here $\mu(X) < \infty$ enters: continuity of a measure from above applies to a decreasing sequence only when some member has finite measure, and then it gives $\mu(E_n(k)) \to \mu(\varnothing) = 0$ as $n \to \infty$ .

Step 3: spend the budget geometrically. Let $\varepsilon > 0$ . For each $k$ use step 2 to choose $n_k$ with $\mu\bigl(E_{n_k}(k)\bigr) < \varepsilon 2^{-k}$ , and set $E = \bigcup_{k\ge1} E_{n_k}(k)$ . Countable subadditivity gives $\mu(E) < \varepsilon \sum_{k\ge1} 2^{-k} = \varepsilon$ .

Step 4: uniformity off $E$ . Let $x \notin E$ and let $k$ be given. Then $x \notin E_{n_k}(k)$ , which by the definition in step 1 says precisely that $|f_m(x) - f(x)| \le 1/k$ for every $m \ge n_k$ . The index $n_k$ depends on $k$ but not on $x$ , so $\sup_{x \notin E} |f_m(x) - f(x)| \le 1/k$ for $m \ge n_k$ : the convergence on $X\setminus E$ is uniform. $\square$

Proof(Part 5: what almost uniform convergence gives)AdaptedAdapted from Folland (1999), the remark following Thm 2.33 and Exercise 39, pp. 62–63

Suppose $f_n \to f$ almost uniformly.

A.e. convergence. For each $m$ pick $E_m$ with $\mu(E_m) < 1/m$ and $f_n \to f$ uniformly, hence pointwise, on $X\setminus E_m$ . Let $N = \bigcap_m E_m$ . Then $\mu(N) \le \mu(E_m) < 1/m$ for every $m$ , so $\mu(N) = 0$ ; and every $x \notin N$ lies outside some $E_m$ , where convergence holds. So $f_n \to f$ off the null set $N$ .

Convergence in measure. Let $\varepsilon, \delta > 0$ . Choose $E$ with $\mu(E) < \delta$ and $f_n \to f$ uniformly on $X \setminus E$ , and then $N_0$ with $\sup_{X\setminus E}|f_n - f| < \varepsilon$ for $n \ge N_0$ . For such $n$ the set $\{|f_n - f| \ge \varepsilon\}$ is contained in $E$ , so its measure is below $\delta$ . As $\delta$ was arbitrary, $\mu(\{|f_n-f|\ge\varepsilon\}) \to 0$ .

Combining this with part 4: on a space of finite measure, a.e. convergence implies almost uniform convergence, which implies convergence in measure. $\square$

Remark(Where the hypotheses are used, and the standard counterexamples)AdaptedAdapted from Folland (1999), the four sequences on p. 61 and the discussion on pp. 61–62

Folland (Folland (1999), §2.4, p. 61) introduces four sequences on $\mathbb{R}$ with Lebesgue measure which between them block every implication not listed above. In our notation they are

\text{(i) } f_n = \tfrac{1}{n}\chi_{(0,n)}, \quad \text{(ii) } f_n = \chi_{(n,n+1)}, \quad \text{(iii) } f_n = n\,\chi_{[0,1/n]},

and (iv) the "typewriter" sequence: writing $n = 2^k + j$ with $0 \le j < 2^k$ , set $f_n = \chi_{[j2^{-k},\,(j+1)2^{-k}]}$ , so that the indicator of a dyadic interval of length $2^{-k}$ sweeps across $[0,1]$ and then halves its width.

$L^1$ is not implied by any pointwise mode. In (i) the convergence to $0$ is uniform, in (ii) it is pointwise everywhere, in (iii) it is a.e.; yet $\int |f_n| = 1$ in all three cases, so none converges in $L^1$ . Part 6 is therefore not a triviality: the hypothesis that supplies what is missing is the integrable dominating function, and in (i), (ii), (iii) the smallest dominating function is $\sup_n f_n$ , which is comparable to $1/x$ for large $x$ in (i), equal to $1$ on $(1,\infty)$ in (ii), and comparable to $1/x$ near $0$ in (iii) — none of them integrable.
Convergence in measure does not imply $L^1$ convergence. In (i) the set $\{f_n \ge \varepsilon\}$ equals $(0,n)$ while $1/n \ge \varepsilon$ and is empty once $n > 1/\varepsilon$ , so its measure is eventually $0$ ; in (iii) it is $[0,1/n]$ , of measure $1/n$ . Both sequences therefore converge to $0$ in measure, and neither converges in $L^1$ .
A.e. convergence does not imply convergence in measure when $\mu(X) = \infty$ . In (ii), $f_n \to 0$ at every point, but $\mu(\{f_n \ge \tfrac12\}) = 1$ for every $n$ , so the sequence is not even Cauchy in measure. This is exactly the failure of step 2 of the Egorov proof: the sets $E_n(k)$ there have infinite measure and continuity from above does not apply. It is why Egorov and part 5 carry the hypothesis $\mu(X) < \infty$ , and why in probability the corresponding arrow holds with no extra assumption.
Convergence in measure (even in $L^1$ ) does not imply a.e. convergence. The typewriter sequence (iv) has $\int|f_n| = 2^{-k}$ for $2^k \le n < 2^{k+1}$ , so $f_n \to 0$ in $L^1$ and hence in measure; but for every $x \in [0,1]$ there are infinitely many $n$ with $f_n(x) = 1$ and infinitely many with $f_n(x) = 0$ , so $f_n(x)$ converges at no point of $[0,1]$ . Part 3 is the best that can be said: along the subsequence $n = 2^k$ (say) convergence to $0$ is a.e.
Egorov's conclusion cannot be improved to "off a null set". The witness is $f_n(x) = x^n$ on $[0,1]$ , which converges to $0$ at every point of $[0,1)$ and so a.e. Let $N$ be any null set. A null set contains no interval, so $N$ cannot swallow $(1-\delta,1)$ for any $\delta > 0$ ; hence $[0,1]\setminus N$ contains points arbitrarily close to $1$ , and since $x^n \to 1$ as $x \uparrow 1$ for each fixed $n$ , $\sup_{[0,1]\setminus N}|f_n| = 1$ for every $n$ . Convergence off a null set is therefore never uniform here. Deleting a set of small positive measure does work: off $(1-\delta,1]$ the supremum is $(1-\delta)^n \to 0$ . So the exceptional set in Egorov's theorem has small, not zero, measure, and must genuinely depend on $\varepsilon$ . (Sequence (iv) is no help with this point: it converges nowhere on $[0,1]$ , so it does not satisfy Egorov's hypothesis at all.)
Egorov's theorem needs the limit to be finite a.e., implicitly: the proof measures $|f_m - f|$ , which requires $f$ to be a complex-valued (hence finite) function, as the statement assumes.

Method(Deciding how a given sequence converges)Original

Find the pointwise limit first. For each fixed $x$ , compute $\lim_n f_n(x)$ . If it exists off a null set, that function is the only possible limit in any of the four modes, by part 2 (uniqueness) and part 3.
Test convergence in measure by computing $\mu(\{|f_n - f| \ge \varepsilon\})$ explicitly as a function of $n$ and $\varepsilon$ . For sequences built from indicators this is just the length of an interval.
Test $L^p$ convergence by computing $\|f_n - f\|_p$ . This may hold for some $p$ and fail for others, so do it for the $p$ you need; failure for one $p$ says nothing about another.
If $\mu(X) < \infty$ and the pointwise limit exists a.e., convergence in measure is automatic by part 5, and only the $L^p$ question remains. Look for an integrable dominant to settle it by part 6; if none exists, the mass is escaping, either to infinity or into a spike.
If you need pointwise information but have only $L^p$ or in-measure convergence, pass to a subsequence by part 3 and continue there; this is the standard move in existence proofs for differential equations.
If you need uniformity and have a.e. convergence on a finite measure space, use Egorov: fix the budget $\varepsilon$ first, get a set $E$ with $\mu(E) < \varepsilon$ , argue uniformly on $X\setminus E$ , and bound whatever happens on $E$ by $\varepsilon$ times a crude estimate.

Example(A one-parameter family that separates the modes)Original

On $X = (0,1)$ with Lebesgue measure, and for a fixed real $\alpha > 0$ , let

f_n = n^{\alpha}\,\chi_{(0,1/n)} , \qquad n = 1, 2, \dots

Pointwise limit. For fixed $x \in (0,1)$ , $f_n(x) = 0$ as soon as $n > 1/x$ . So $f_n \to 0$ everywhere on $(0,1)$ , hence a.e.

In measure. For $0 < \varepsilon < n^{\alpha}$ the set $\{f_n \ge \varepsilon\}$ is $(0,1/n)$ , of measure $1/n$ ; for $\varepsilon \ge n^{\alpha}$ it is empty. Either way $\mu(\{f_n \ge \varepsilon\}) \le 1/n \to 0$ , so $f_n \to 0$ in measure for every $\alpha$ . (This also follows from part 5, since $\mu(X) = 1 < \infty$ .)

In $L^p$ . Directly,

\|f_n\|_p^p = \int_0^{1/n} n^{\alpha p}\,dx = n^{\alpha p - 1} , \qquad\text{so}\qquad \|f_n\|_p = n^{\alpha - 1/p} .

Hence $f_n \to 0$ in $L^p$ exactly when $\alpha < 1/p$ , that is when $p < 1/\alpha$ ; at $p = 1/\alpha$ the norm is identically $1$ , and for $p > 1/\alpha$ it diverges.

Numbers. Take $\alpha = \tfrac12$ and $n = 10^4$ . Then

\|f_n\|_1 = n^{-1/2} = 0.01 , \qquad \|f_n\|_2 = n^{0} = 1 , \qquad \|f_n\|_3 = n^{1/6} = 10^{2/3} = 4.64158883\ldots,

while $\mu(\{f_n \ge \varepsilon\}) = 10^{-4}$ for every $\varepsilon \le 100$ . One sequence, one limit, and three different verdicts: convergence in $L^1$ , no convergence in $L^2$ (the norm sits at $1$ forever), divergence in $L^3$ — with a.e. and in-measure convergence throughout. The smallest dominating function is $\sup_n f_n$ , which at $x$ equals $(\lceil 1/x\rceil - 1)^{1/2}$ , since the terms that are non-zero at $x$ are those with $n < 1/x$ ; this sits just below $x^{-1/2}$ and is comparable to it, as $(1/x - 1)^{1/2} \le (\lceil 1/x\rceil - 1)^{1/2} < x^{-1/2}$ . Now $x^{-1/2} \in L^1(0,1)$ , with integral $2$ , so part 6 applies and accounts for the $L^1$ convergence. It says nothing about $L^2$ , being stated for $L^1$ only; and the natural way to borrow it — apply part 6 to $|f_n|^2$ , which converges to $0$ a.e. and is dominated by $x^{-1}$ — fails at exactly the right place, because $x^{-1} \notin L^1(0,1)$ . There is in any case no $L^2$ convergence to explain.

Example(Egorov's theorem made explicit)Original

On $X = [0,1]$ with Lebesgue measure take $f_n(x) = x^n$ and $f = 0$ .

A.e. but not uniform. $f_n(x) \to 0$ for every $x \in [0,1)$ and $f_n(1) = 1$ for all $n$ , so $f_n \to 0$ a.e. (the exceptional set $\{1\}$ is null). The convergence is not uniform: $\sup_{[0,1]} |f_n| = 1$ for every $n$ .

In measure. For $0 < \varepsilon < 1$ , $\{x : x^n \ge \varepsilon\} = [\varepsilon^{1/n}, 1]$ , of measure $1 - \varepsilon^{1/n} \to 0$ . At $\varepsilon = \tfrac12$ and $n = 10$ this measure is $1 - 2^{-1/10} = 0.0669670\ldots$

Egorov with $\varepsilon = 2\times 10^{-2}$ . Take $E = (0.99,\,1]$ , so $\mu(E) = 0.01 < \varepsilon$ , as part 4 requires with its strict inequality. On $[0,0.99]$ ,

\sup_{0 \le x \le 0.99} |f_n(x)| = 0.99^{\,n} ,

which tends to $0$ , so the convergence is uniform there, as Egorov promises. The rate is explicit: $0.99^n < 10^{-3}$ requires $n > \log(10^{-3})/\log(0.99) = 687.3158\ldots$ , so $n \ge 688$ . Checking the two neighbouring values, $0.99^{687} = 1.00318\times 10^{-3}$ and $0.99^{688} = 9.93148\times 10^{-4}$ , which confirms $n = 688$ as the first index at which the uniform error on $X \setminus E$ drops below $10^{-3}$ .

Why the set cannot be null. Any $E$ with $\mu(E) = 0$ leaves points of $[0,1]$ arbitrarily close to $1$ in $X \setminus E$ , and at such a point $x$ we have $f_n(x) = x^n$ close to $1$ for the first several $n$ ; taking a sequence of such points shows $\sup_{X\setminus E}|f_n| = 1$ for every $n$ . Shrinking $\varepsilon$ forces $E$ to reach closer to $1$ and the index $n_k$ in the proof to grow — here, halving $\varepsilon$ to $10^{-2}$ means taking $E = (0.995,\,1]$ , of measure $0.005 < \varepsilon$ , in place of $(0.99,\,1]$ , and pushes the index for error $10^{-3}$ to $n \ge 1379$ , since $0.995^{1378} = 1.00047\times10^{-3}$ and $0.995^{1379} = 9.95470\times10^{-4}$ .

ApplicationOriginal

Existence proofs for differential equations. A typical argument produces approximate solutions $u_n$ bounded in some $L^p$ , extracts a subsequence converging in a weak or $L^p$ sense, and then needs pointwise convergence to pass to the limit inside a nonlinear term $F(u_n)$ . Part 3 supplies exactly that: $L^p$ convergence yields an a.e. convergent subsequence, and continuity of $F$ then gives $F(u_{n_j}) \to F(u)$ a.e., after which dominated convergence or Fatou's lemma closes the argument. This is why Egorov's theorem appears in the appendix of a partial differential equations text at all; see weak solutions and Sobolev spaces.

Laws of large numbers. The weak law is convergence in probability of the sample mean, the strong law is almost sure convergence; the chart says the strong law implies the weak one and not conversely, which is precisely why the two are separate theorems with separate hypotheses. See the law of large numbers.

Monte Carlo estimates. An estimator that converges in $L^2$ has a variance that goes to zero and so comes with an error bar; one that converges only in probability has a guarantee of the form "the chance of being off by more than $\varepsilon$ is small", with no control of how bad the rare failures are. Sequence (iii) of the remark, $f_n = n\chi_{[0,1/n]}$ , is the model of an estimator that is almost always right and occasionally catastrophically wrong: the error vanishes off a set of measure $1/n$ , so the estimator converges in measure (in probability), but on that shrinking set it takes the value $n$ , and $\|f_n\|_2^2 = n^2 \cdot n^{-1} = n$ diverges. That is why heavy-tailed sampling problems resist mean-square error analysis: the rare failures, invisible to convergence in probability, dominate the variance. See Monte Carlo methods.

Signal processing and the choice of error norm. A reconstruction may converge in mean square while failing to converge at any single instant (the typewriter sequence), or converge at every instant while the $L^1$ mass of the error stays constant (sequence (i), a plateau of height $1/n$ spreading over $(0,n)$ ; here the energy $\|f_n\|_2^2 = 1/n$ does go to zero). The spike (iii) converges at every instant except $x = 0$ and keeps its $L^1$ mass too, but its energy $\|f_n\|_2^2 = n$ diverges — the same $L^1$ bookkeeping, opposite verdicts in mean square. Deciding which failure is acceptable is a modelling decision, and the four modes are the vocabulary for stating it.

Approximation of measurable functions. Egorov's theorem is one of the two standard bridges from measurable to continuous behaviour, the other being Lusin's theorem, and it is what makes "a measurable function is nearly continuous, a convergent sequence of measurable functions is nearly uniformly convergent" a precise statement rather than a slogan.

L^p spaces

Theorem

L^p spaces

Adapted

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Establishes the two inequalities that make $\|\cdot\|_p$ a norm, proves that $L^p$ is complete for $1 \le p \le \infty$ (Riesz-Fischer), and identifies the simple and continuous functions as dense subspaces, so that every $L^p$ function can be approached by something concrete.

Prerequisites: Dominated convergence, Monotone convergence and Fatou's lemma, Modes of convergence, The Lebesgue integral, Measurable functions, Lebesgue measure, Sigma-algebras and measures

Definition(

L^p

L^\infty

and conjugate exponents)AdaptedAdapted from Folland (1999), §6.1, pp. 181–184

Fix a measure space $(X,\mathcal{M},\mu)$ . For measurable $f : X \to \mathbb{C}$ and $0 < p < \infty$ put

\|f\|_p = \Bigl[\int_X |f|^p \,d\mu\Bigr]^{1/p} \in [0,\infty] , \qquad L^p(X,\mathcal{M},\mu) = \{\, f \text{ measurable} : \|f\|_p < \infty \,\} ,

and for $p = \infty$ put

\|f\|_\infty = \inf\{\, a > 0 : \mu(\{x : |f(x)| > a\}) = 0 \,\} , \qquad L^\infty = \{\, f : \|f\|_\infty < \infty \,\},

with $\inf\varnothing = \infty$ . The number $\|f\|_\infty$ is the essential supremum of $|f|$ , and the infimum is attained, because $\{|f| > \|f\|_\infty\}$ is the countable union of the null sets $\{|f| > \|f\|_\infty + 1/n\}$ . As with $L^1$ , two functions equal a.e. are the same element of $L^p$ ; without this identification $\|f\|_p = 0$ would not force $f = 0$ .

$L^p$ is a vector space: $|f+g|^p \le [2\max(|f|,|g|)]^p \le 2^p(|f|^p + |g|^p)$ .

For $1 < p < \infty$ the conjugate exponent is the number $q = p/(p-1)$ , equivalently $p^{-1} + q^{-1} = 1$ . Following Folland (Folland (1999), §6.1, p. 184) we also call $1$ and $\infty$ conjugate to each other, which is the formal reading of $1^{-1} + \infty^{-1} = 1$ .

When $\mu$ is counting measure on a set $A$ one writes $\ell^p(A)$ , and $\ell^p$ for $A = \mathbb{N}$ ; then $\|a\|_p = (\sum_n |a_n|^p)^{1/p}$ and every statement below is a statement about series.

Theorem(Hölder's inequality)AdaptedAdapted from Folland (1999), Thm 6.2 with (6.3), p. 182, and Thm 6.8(a), p. 184; Reed & Simon (1980), Thm S.3, p. 348

Let $1 < p < \infty$ and let $q$ be its conjugate exponent. For all measurable $f, g$ on $X$ ,

\|fg\|_1 \;\le\; \|f\|_p \,\|g\|_q .

In particular if $f \in L^p$ and $g \in L^q$ then $fg \in L^1$ , and in that case equality holds if and only if $\alpha|f|^p = \beta|g|^q$ a.e. for some constants $\alpha,\beta$ with $\alpha\beta \ne 0$ . Reed and Simon (Reed & Simon (1980), Thm S.3, p. 348) state the same equality condition in the equivalent form: if neither $f$ nor $g$ vanishes a.e., equality holds exactly when $|g|$ is a.e. a constant multiple of $|f|^{p-1}$ .

The endpoint case is Folland's (Folland (1999), Thm 6.8(a), p. 184): $\|fg\|_1 \le \|f\|_1\|g\|_\infty$ , with equality for $f \in L^1$ , $g \in L^\infty$ exactly when $|g| = \|g\|_\infty$ a.e. on the set where $f \ne 0$ .

For $p = q = 2$ this is the Cauchy-Schwarz inequality of inner-product spaces.

Theorem(Minkowski's inequality, completeness, density)AdaptedAdapted from Folland (1999), Thm 6.5, Thm 6.6 and Prop. 6.7, p. 183, Thm 6.8(b)(d)(e), p. 184, and Prop. 7.9, p. 217; Reed & Simon (1980), Thm III.1(a)(b), p. 68

Let $(X,\mathcal{M},\mu)$ be a measure space.

Minkowski's inequality. If $1 \le p < \infty$ and $f, g \in L^p$ then $\|f+g\|_p \le \|f\|_p + \|g\|_p$ . The same holds for $p = \infty$ . Hence $\|\cdot\|_p$ is a norm on $L^p$ for every $1 \le p \le \infty$ , and $L^p$ is a normed vector space (see normed and Banach spaces).
Riesz-Fischer: completeness. For $1 \le p < \infty$ , $L^p$ is complete, hence a Banach space; and so is $L^\infty$ . Equivalently, every $\|\cdot\|_p$ -Cauchy sequence of $L^p$ functions converges in $L^p$ to an $L^p$ function.
Density of simple functions. For $1 \le p < \infty$ , the simple functions $f = \sum_{j=1}^n a_j \chi_{E_j}$ with $\mu(E_j) < \infty$ for every $j$ are dense in $L^p$ . In $L^\infty$ the simple functions are dense without the finite-measure restriction.
Density of continuous functions. If $\mu$ is a Radon measure on a locally compact Hausdorff space $X$ — Lebesgue measure on $\mathbb{R}^n$ is one — then $C_c(X)$ , the continuous functions of compact support, is dense in $L^p(\mu)$ for $1 \le p < \infty$ .

Part 4 fails at $p = \infty$ : a uniform limit of continuous functions is continuous, so the closure of $C_c(\mathbb{R})$ in $L^\infty(\mathbb{R})$ contains no discontinuous function.

Note(Three variants in the library)AdaptedAdapted from Reed & Simon (1980), Thm III.1(c), p. 68; Grimmett & Stirzaker (2020), note (b), p. 353 and Thms (6), (7) and (8), p. 364; Stakgold & Holst (2011), §4.3, Examples 6 and 7, pp. 239–240, and Exercise 4.3.6, p. 244

Three exponents. Reed and Simon (Reed & Simon (1980), Thm III.1(c), p. 68) state Hölder in the more flexible form: if $p,q,r \ge 1$ satisfy $p^{-1} + q^{-1} = r^{-1}$ and $f \in L^p$ , $g \in L^q$ , then $fg \in L^r$ with $\|fg\|_r \le \|f\|_p\|g\|_q$ . Taking $r = 1$ recovers the statement above; taking $r = 2$ is what one needs to multiply two functions and stay in $L^2$ . They also record the dual description of the norm, $\|f\|_p = \sup\{\|fg\|_1 : \|g\|_q = 1\}$ , from which Minkowski follows in two lines by taking suprema in $\|(f+g)h\|_1 \le \|fh\|_1 + \|gh\|_1$ ; that is the route of their Corollary 2, and it is different from the route proved below.
Random variables. Grimmett and Stirzaker (Grimmett & Stirzaker (2020), Thms (6) and (7), p. 364) state the same two inequalities as $\|XY\|_1 \le \|X\|_p\|Y\|_q$ and $\|X+Y\|_p \le \|X\|_p + \|Y\|_p$ with $\|X\|_p = (\mathbb{E}|X^p|)^{1/p}$ , the case of a probability measure. They add the $C_p$ inequality $\mathbb{E}(|X+Y|^p) \le C_p(\mathbb{E}|X^p| + \mathbb{E}|Y^p|)$ with $C_p = 1$ for $0 < p \le 1$ and $C_p = 2^{p-1}$ for $p > 1$ , which is the substitute for Minkowski in the range $p<1$ where the triangle inequality is unavailable.
$L^2$ as a completion. Stakgold and Holst (Stakgold & Holst (2011), §4.3, Examples 6 and 7, pp. 239–240) make completeness concrete the other way round: the continuous functions on $[a,b]$ under the mean-square norm are not complete, as the sequence $u_k(x) = \tfrac12 + \pi^{-1}\arctan(kx)$ on $(-1,1)$ shows — it is Cauchy in that norm and its limit $\tfrac12 + \tfrac12\operatorname{sgn} x$ is discontinuous — and they define $L_2(a,b)$ as the completion. Their Exercise 4.3.6 lists dense subspaces of $L_2(a,b)$ : polynomials, continuous functions, differentiable functions, twice-differentiable functions vanishing at the endpoints, and the test functions $C_0^\infty(a,b)$ .

Idea of the proofOriginal

The three results have three different engines.

Hölder is convexity in disguise. The inequality $a^{\lambda}b^{1-\lambda} \le \lambda a + (1-\lambda) b$ says that the logarithm is concave; applied pointwise with $a = |f|^p$ , $b = |g|^q$ and $\lambda = 1/p$ it turns a product into a sum, which integrates. Normalising both norms to $1$ beforehand is what makes the right-hand side integrate to $p^{-1}+q^{-1} = 1$ exactly.

Minkowski is Hölder applied once, after the trick of writing $|f+g|^p = |f+g|\,|f+g|^{p-1}$ and noticing that the conjugate exponent sends $p-1$ back to $p$ . The whole content of "the triangle inequality holds exactly for $p \ge 1$ " is the convexity of $t \mapsto t^p$ .

Completeness is where the convergence theorems are used, and the route is to replace a Cauchy sequence by a series: a Cauchy sequence has a subsequence whose consecutive differences have summable norms, and such a series can be handled with monotone convergence (to build a dominating function) and then dominated convergence (to get convergence of the partial sums in norm). Density is the simple-function approximation of a measurable function, upgraded from pointwise to norm convergence by dominated convergence once more, and then, for continuity, one indicator at a time by squeezing a continuous function between a compact set and an open set.

Proof(Hölder)AdaptedAdapted from Folland (1999), Lemma 6.1 and Thm 6.2, p. 182

Step 1: a pointwise inequality between a product and a sum. For $a, b \ge 0$ and $0 < \lambda < 1$ ,

a^{\lambda}b^{1-\lambda} \le \lambda a + (1-\lambda) b , \qquad\text{with equality iff } a = b .

If $b = 0$ both sides are what they are and the claim is immediate; otherwise divide by $b$ and set $t = a/b$ , reducing the claim to $t^{\lambda} \le \lambda t + (1-\lambda)$ . The function $t \mapsto t^{\lambda} - \lambda t$ has derivative $\lambda(t^{\lambda-1} - 1)$ , positive for $t<1$ and negative for $t>1$ , so its maximum over $t>0$ is its value $1-\lambda$ at $t = 1$ , attained only there. That is the claim, with equality only at $t=1$ , that is $a=b$ .

Step 2: reduce to normalised functions. The inequality is trivial if $\|f\|_p$ or $\|g\|_q$ is $0$ , since then $f$ or $g$ vanishes a.e. and so does $fg$ ; and it is trivial if either norm is infinite. So assume both are finite and non-zero. Replacing $f$ by $af$ and $g$ by $bg$ multiplies both sides by $|ab|$ , so it costs nothing to assume $\|f\|_p = \|g\|_q = 1$ , and the claim becomes $\|fg\|_1 \le 1$ .

Step 3: apply step 1 pointwise. Fix $x$ and take $a = |f(x)|^p$ , $b = |g(x)|^q$ , $\lambda = 1/p$ , so that $1-\lambda = 1/q$ and $a^\lambda b^{1-\lambda} = |f(x)|\,|g(x)|$ :

|f(x)g(x)| \le \frac{|f(x)|^p}{p} + \frac{|g(x)|^q}{q} .

Step 4: integrate. By monotonicity and linearity of the integral,

\|fg\|_1 \le \frac{1}{p}\int |f|^p + \frac{1}{q}\int|g|^q = \frac1p + \frac1q = 1 = \|f\|_p\|g\|_q .

Step 5: equality. The inequality in step 4 is strict unless the pointwise inequality of step 3 is an equality a.e., which by step 1 happens exactly when $|f|^p = |g|^q$ a.e. Undoing the normalisation of step 2 turns this into $\alpha|f|^p = \beta|g|^q$ a.e. for constants with $\alpha\beta \ne 0$ . $\square$

Proof(Minkowski)AdaptedAdapted from Folland (1999), Thm 6.5, p. 183

Let $1 \le p < \infty$ and $f, g \in L^p$ . For $p = 1$ the claim is $\int|f+g| \le \int|f| + \int|g|$ , which is the pointwise triangle inequality integrated; and if $f+g = 0$ a.e. there is nothing to prove. So let $p > 1$ with conjugate exponent $q$ , and note $(p-1)q = p$ .

Step 1: split off one factor. Pointwise, $|f+g|^p = |f+g|\,|f+g|^{p-1} \le \bigl(|f| + |g|\bigr)|f+g|^{p-1}$ .

Step 2: Hölder on each term. Integrating and applying Hölder to each of the two products,

\int |f+g|^p \;\le\; \bigl\| |f+g|^{p-1} \bigr\|_q \bigl( \|f\|_p + \|g\|_p \bigr) .

Step 3: identify the leftover factor. Because $(p-1)q = p$ ,

\bigl\||f+g|^{p-1}\bigr\|_q = \Bigl[\int |f+g|^{(p-1)q}\Bigr]^{1/q} = \Bigl[\int|f+g|^p\Bigr]^{1/q} ,

so with $A = \int |f+g|^p$ the last two steps read $A \le A^{1/q}\bigl(\|f\|_p + \|g\|_p\bigr)$ .

Step 4: divide. $A$ is finite, because $L^p$ is a vector space, and we may assume $A > 0$ . Dividing by $A^{1/q}$ and using $1 - 1/q = 1/p$ gives $A^{1/p} = \|f+g\|_p \le \|f\|_p + \|g\|_p$ . The division is the step that needs $A < \infty$ , and that in turn is where $f, g \in L^p$ is used rather than mere measurability.

For $p = \infty$ : $|f| \le \|f\|_\infty$ and $|g| \le \|g\|_\infty$ a.e., so $|f+g| \le \|f\|_\infty + \|g\|_\infty$ off the union of two null sets, and the essential supremum obeys the same bound. $\square$

Proof(Riesz-Fischer: completeness)AdaptedAdapted from Folland (1999), Thm 6.6, p. 183

Let $1 \le p < \infty$ .

Step 1: reduce a Cauchy sequence to a series. Let $\{h_n\}$ be Cauchy in $L^p$ . Choose indices $n_1 < n_2 < \cdots$ with $\|h_m - h_{n_k}\|_p \le 2^{-k}$ for every $m \ge n_k$ , and set $f_1 = h_{n_1}$ and $f_{k+1} = h_{n_{k+1}} - h_{n_k}$ . Then $\sum_{k\ge1} \|f_k\|_p \le \|h_{n_1}\|_p + \sum_{k\ge1} 2^{-k} < \infty$ , and the series telescopes: $\sum_{k=1}^{j} f_k = h_{n_j}$ . So if every absolutely convergent series in $L^p$ converges in $L^p$ , the subsequence $\{h_{n_j}\}$ converges in $L^p$ , and a Cauchy sequence with a convergent subsequence converges to the same limit by the triangle inequality. It therefore suffices to prove the statement for series.

Step 2: a dominating function. Let $\{f_k\} \subset L^p$ with $B = \sum_{k=1}^\infty \|f_k\|_p < \infty$ . Put $G_n = \sum_{k=1}^n |f_k|$ and $G = \sum_{k=1}^\infty |f_k| \in [0,\infty]$ . Minkowski gives $\|G_n\|_p \le \sum_{k=1}^n\|f_k\|_p \le B$ for every $n$ . The functions $G_n^p$ increase pointwise to $G^p$ , so monotone convergence gives $\int G^p = \lim_n \int G_n^p \le B^p$ . Hence $G \in L^p$ and in particular $G < \infty$ a.e.

Step 3: the sum exists. Where $G(x) < \infty$ the series $\sum_k f_k(x)$ converges absolutely, hence converges; call its sum $F(x)$ , and set $F = 0$ on the null set where $G = \infty$ . Then $|F| \le G$ , so $F \in L^p$ .

Step 4: convergence in norm. For each $n$ , $\bigl|F - \sum_{k=1}^n f_k\bigr| \le |F| + G_n \le 2G$ , so $\bigl|F - \sum_{k=1}^n f_k\bigr|^p \le (2G)^p \in L^1$ by step 2. The left side tends to $0$ pointwise a.e. by step 3, so dominated convergence gives

\Bigl\|F - \sum_{k=1}^n f_k\Bigr\|_p^p = \int \Bigl|F - \sum_{k=1}^n f_k\Bigr|^p \longrightarrow 0 .

Thus $\sum_k f_k$ converges to $F$ in $L^p$ , and by step 1, $L^p$ is complete.

Step 5: the case $p = \infty$ . If $\{h_n\}$ is Cauchy in $L^\infty$ , let $N$ be the union over all $m,n$ of the null sets where $|h_n - h_m| > \|h_n-h_m\|_\infty$ ; $N$ is null, being a countable union. Off $N$ the sequence is uniformly Cauchy, hence converges uniformly to some bounded $h$ ; setting $h = 0$ on $N$ gives $h \in L^\infty$ and $\|h_n - h\|_\infty \to 0$ . $\square$

Proof(Density of simple and of continuous functions)AdaptedAdapted from Folland (1999), Prop. 6.7, p. 183, and Prop. 7.9, p. 217

Simple functions. Let $1 \le p < \infty$ and $f \in L^p$ . By the simple-function approximation theorem of measurable functions there are simple $f_n$ with $f_n \to f$ pointwise and $|f_n| \le |f|$ . Each $f_n$ lies in $L^p$ , and $|f_n - f|^p \le (|f_n|+|f|)^p \le 2^p|f|^p$ , which is integrable; since $|f_n - f|^p \to 0$ pointwise, dominated convergence gives $\|f_n - f\|_p \to 0$ . Finally, if $f_n = \sum_j a_j\chi_{E_j}$ with the $E_j$ disjoint and the $a_j$ non-zero, then $\sum_j |a_j|^p\mu(E_j) = \int|f_n|^p < \infty$ forces $\mu(E_j) < \infty$ for every $j$ . Finiteness of $p$ is used in invoking dominated convergence on $|f_n-f|^p$ ; at $p=\infty$ the approximation is instead uniform and needs no convergence theorem.

Continuous functions. By the previous paragraph it is enough to approximate a single $\chi_E$ with $\mu(E) < \infty$ . Let $\varepsilon > 0$ . Regularity of the Radon measure $\mu$ supplies a compact $K \subseteq E$ and an open $U \supseteq E$ with $\mu(U \setminus K) < \varepsilon$ , and Urysohn's lemma supplies $\varphi \in C_c(X)$ with $\chi_K \le \varphi \le \chi_U$ . The functions $\chi_E$ and $\varphi$ both lie between $0$ and $1$ and agree outside $U\setminus K$ , so $|\chi_E - \varphi|^p \le \chi_{U\setminus K}$ and

\|\chi_E - \varphi\|_p \le \mu(U\setminus K)^{1/p} < \varepsilon^{1/p} .

Since $\varepsilon$ was arbitrary and $p < \infty$ , $\chi_E$ lies in the $L^p$ -closure of $C_c(X)$ . $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Folland (1999), §6.1, pp. 181–185

$p \ge 1$ is exactly what the triangle inequality needs. Folland (Folland (1999), §6.1, p. 182) shows the failure below $1$ explicitly: for $0<p<1$ and $a,b>0$ one has $a^p + b^p > (a+b)^p$ , so if $E$ and $F$ are disjoint of positive finite measure and $a = \mu(E)^{1/p}$ , $b = \mu(F)^{1/p}$ , then $\|\chi_E + \chi_F\|_p = (a^p+b^p)^{1/p} > a + b = \|\chi_E\|_p + \|\chi_F\|_p$ . For $p<1$ the quantity $\|\cdot\|_p$ is therefore not a norm, and Grimmett and Stirzaker's $C_p$ inequality is what survives.
Conjugacy $p^{-1}+q^{-1}=1$ is used twice: once in step 3 of the Hölder proof, to make $\lambda$ and $1-\lambda$ the two reciprocals, and once in step 3 of the Minkowski proof, where $(p-1)q = p$ is the only reason the leftover factor is a power of the quantity being estimated. Neither works for a mismatched pair.
Finiteness of $A = \int|f+g|^p$ is what allows the division in Minkowski step 4. It is guaranteed by $f, g \in L^p$ together with the estimate $|f+g|^p \le 2^p(|f|^p+|g|^p)$ of the definition block, and the proof would break for merely measurable $f, g$ .
Completeness needs both convergence theorems, monotone convergence to build the dominating $G$ and dominated convergence to finish, and so it is a genuinely Lebesgue-theoretic fact. With the Riemann integral it is false: Stakgold and Holst's (Stakgold & Holst (2011), §4.3, Example 6, pp. 239–240) arctangent sequence is Cauchy in the mean-square norm with no continuous limit. Completeness is also what makes the $L^p$ limit in the modes of convergence usable without knowing the limit in advance.
$\sigma$ -finiteness is not needed anywhere above, and neither is any topology except in part 4, where a Radon measure on a locally compact Hausdorff space is assumed so that regularity and Urysohn's lemma are available.
There is no inclusion between $L^p$ and $L^q$ in general. Folland's examples on $(0,\infty)$ with $f_a(x) = x^{-a}$ , $a>0$ , isolate the two obstructions: $f_a\chi_{(0,1)} \in L^p$ iff $p < a^{-1}$ , so a function can be too singular at a point, while $f_a\chi_{(1,\infty)} \in L^p$ iff $p > a^{-1}$ , so it can decay too slowly at infinity. Raising $p$ helps against slow decay and hurts against singularities. Inclusions do hold when one obstruction is excluded — on a finite measure space $L^q \subseteq L^p$ for $p<q$ — and Folland's Prop. 6.10 gives the interpolation bound $\|f\|_q \le \|f\|_p^{\lambda}\|f\|_r^{1-\lambda}$ for $p<q<r$ with $q^{-1} = \lambda p^{-1} + (1-\lambda)r^{-1}$ .
Equality in Hölder is a rigidity statement, not a technicality: it says the inequality is sharp and identifies the extremisers, which is what makes the duality formula $\|f\|_p = \sup\{\|fg\|_1 : \|g\|_q = 1\}$ of Reed and Simon true with a supremum that is attained.

Method(Using the

L^p

toolkit)Original

To show $f \in L^p$ , split the domain at the places where $f$ is singular and where the domain is unbounded, and compare with a power $x^{-a}$ on each piece; the criterion is $ap < 1$ near a singularity and $ap > 1$ near infinity.
To bound $\int |fg|$ , choose a conjugate pair $(p,q)$ for which both $\|f\|_p$ and $\|g\|_q$ are finite, and apply Hölder. If several choices work, the bound depends on the choice and may be optimised over $p$ , as in the example below.
To bound a product of three or more factors, use the three-exponent form of Reed and Simon (Reed & Simon (1980), Thm III.1(c), p. 68) repeatedly, or Hölder with exponents $p_1,\dots,p_k$ satisfying $\sum_j p_j^{-1} = 1$ .
To construct an element of $L^p$ as a limit, verify that the candidate sequence is Cauchy in $\|\cdot\|_p$ and invoke completeness; this is how solutions of differential equations, Fourier series and conditional expectations are produced without a formula.
To prove an identity or inequality for all $f \in L^p$ , prove it for simple functions or for $C_c$ functions, check that both sides are continuous in $\|\cdot\|_p$ , and pass to the limit by density. This is the standard proof pattern for translation continuity, for the Plancherel theorem and for the boundedness of an operator defined at first only on nice functions.
Never use density at $p = \infty$ without checking: continuous functions are not dense there, and neither are the compactly supported ones.

Example(Hölder with an optimised exponent)Original

Estimate, and then compare with the true value of,

I = \int_0^1 x^{-1/3}(1-x)^{-1/2}\,dx .

Step 1: the two factors separately. Let $f(x) = x^{-1/3}$ and $g(x) = (1-x)^{-1/2}$ on $(0,1)$ . For $p < 3$ ,

\|f\|_p = \Bigl[\int_0^1 x^{-p/3}dx\Bigr]^{1/p} = \Bigl[\frac{3}{3-p}\Bigr]^{1/p} ,

since $\int_0^1 x^{-s}dx = 1/(1-s)$ for $s<1$ . For $q<2$ , likewise $\|g\|_q = [2/(2-q)]^{1/q}$ .

Step 2: the Hölder bound as a function of $p$ . With $q = p/(p-1)$ the condition $q<2$ reads $p>2$ , and $2-q = (p-2)/(p-1)$ , so the bound $I \le \|f\|_p\|g\|_q$ becomes

B(p) = \Bigl[\frac{3}{3-p}\Bigr]^{1/p}\Bigl[\frac{2(p-1)}{p-2}\Bigr]^{(p-1)/p} , \qquad 2 < p < 3 .

Both brackets blow up at the ends of the interval, so $B$ has an interior minimum.

Step 3: the optimal exponent. Try to make the two brackets equal. Setting $3/(3-p) = 2(p-1)/(p-2)$ gives $3(p-2) = 2(p-1)(3-p)$ , that is $3p - 6 = -2p^2 + 8p - 6$ , so $2p^2 = 5p$ and $p = 5/2$ . Then $q = 5/3$ , both brackets equal $6$ , and

B(5/2) = 6^{2/5}\cdot 6^{3/5} = 6^{2/5 + 3/5} = 6 .

Numerically $\|f\|_{5/2} = 6^{2/5} = 2.0476725\ldots$ and $\|g\|_{5/3} = 6^{3/5} = 2.9301561\ldots$ , whose product is $6.0000000$ . Neighbouring exponents are worse: $B(2.45) = 6.020374$ and $B(2.55) = 6.019925$ , so $p = 5/2$ is the minimiser and the best bound this method gives is

I \le 6 .

Step 4: the true value. The integral is a Beta function, $I = B\bigl(\tfrac23,\tfrac12\bigr) = \Gamma(\tfrac23)\Gamma(\tfrac12)/\Gamma(\tfrac76)$ (see the beta function), and numerically

I = 2.5871095592\ldots

so the optimised Hölder bound overshoots by a factor $6/2.58711 = 2.3192$ . That is typical: Hölder is sharp only when $\alpha|f|^p = \beta|g|^q$ a.e., which here would require $x^{-5/6}$ to be proportional to $(1-x)^{-5/6}$ , and it is not. The example shows both the strength of the method — a finite bound with no special function needed — and the price of its generality.

Example(Density made quantitative)Original

Take $X = (0,1)$ with Lebesgue measure and $f(x) = x^{-1/3}$ . Then $\|f\|_2 = [\int_0^1 x^{-2/3}dx]^{1/2} = \sqrt{3} = 1.7320508$ , so $f \in L^2(0,1)$ although it is unbounded.

Truncation. For $N \ge 1$ let $f_N = \min(f, N)$ , that is $f_N(x) = N$ for $0 < x \le N^{-3}$ and $f_N(x) = x^{-1/3}$ for $N^{-3} < x < 1$ . The two functions differ only on $(0,N^{-3}]$ , and there

\|f - f_N\|_2^2 = \int_0^{N^{-3}}\bigl(x^{-1/3} - N\bigr)^2 dx = \int_0^{N^{-3}} x^{-2/3}dx - 2N\int_0^{N^{-3}} x^{-1/3}dx + N^2 N^{-3} .

The three terms are $3N^{-1}$ , $-2N\cdot\tfrac32 N^{-2} = -3N^{-1}$ and $N^{-1}$ , so

\|f - f_N\|_2^2 = \frac{1}{N} , \qquad \|f-f_N\|_2 = N^{-1/2} .

At $N = 100$ the error is exactly $0.1$ , against $\|f\|_2 = 1.7320508$ : a relative error of $5.77\%$ . Numerical integration of $\int_0^{10^{-6}}(x^{-1/3}-100)^2dx$ returns $1.0000\times 10^{-2}$ , confirming the algebra.

From bounded to simple. $f_N$ is bounded by $N$ , so the dyadic simple functions $s_m = 2^{-m}\lfloor 2^m f_N\rfloor$ satisfy $0 \le f_N - s_m \le 2^{-m}$ everywhere, whence $\|f_N - s_m\|_2 \le 2^{-m}$ because the measure of $(0,1)$ is $1$ . Each $s_m$ takes finitely many values, each on a set of finite measure. Combining, $\|f - s_m\|_2 \le N^{-1/2} + 2^{-m}$ : with $N = 10^4$ and $m = 7$ this is $0.01 + 0.0078125 = 0.0178125$ , that is a simple function whose $L^2$ distance from $f$ is at most $0.0178$ , about $1.03\%$ of $\|f\|_2$ . Density is not an abstract statement; the approximants are explicit and the error is computable.

ApplicationOriginal

Quantum mechanics. States are unit vectors of $L^2(\mathbb{R}^3)$ , and completeness is what makes that space a Hilbert space, so that orthonormal bases exist, projections onto closed subspaces exist, and a formally convergent eigenfunction expansion actually converges to a state. Cauchy-Schwarz, the $p=q=2$ case of Hölder, is the inequality behind the uncertainty principle. Density of $C_c^\infty$ is why one may verify an operator identity on smooth compactly supported wavefunctions and conclude it on all states.

Partial differential equations. Energy estimates live in $L^2$ , Sobolev spaces are built from $L^p$ norms of derivatives, and existence proofs almost always produce a solution as the limit of a Cauchy sequence whose limit exists only because $L^p$ is complete. Hölder's inequality with a carefully chosen exponent pair is the standard device for absorbing a nonlinear term into a linear one; see weak solutions and Sobolev spaces.

Probability and statistics. $L^2$ of a probability space is the space of finite-variance random variables, with the inner product given by covariance; the conditional expectation is the orthogonal projection onto a sub- $\sigma$ -algebra, and it exists because $L^2$ is complete. Hölder with $p=q=2$ gives $|\operatorname{Cov}(X,Y)| \le \sigma_X\sigma_Y$ , that is $|\rho| \le 1$ for the correlation coefficient, and the equality case says $|\rho|=1$ exactly for an affine relation. See probability inequalities.

Signal processing. The energy of a signal is its squared $L^2$ norm, and Plancherel's theorem is the statement that the Fourier transform is an isometry of $L^2$ ; it is proved by establishing the identity on a dense class and extending by continuity, which is exactly step 5 of the method above. Power constraints are $L^2$ bounds, peak constraints are $L^\infty$ bounds, and the lack of inclusion between the two is why both are specified in practice.

Interpolation and scaling arguments. The bound $\|f\|_q \le \|f\|_p^\lambda\|f\|_r^{1-\lambda}$ is the prototype of the interpolation inequalities used to trade regularity against integrability throughout fluid mechanics and kinetic theory, and the $x^{-a}$ examples are the scaling calculations that tell one in advance which exponent can possibly work.

Theorem · advanced

The Radon-Nikodym theorem

Adapted

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Gives the exact condition under which one measure is an integral of a density against another, constructs that density as the Radon-Nikodym derivative $d\nu/d\mu$ , and separates off the singular part of a measure that has no density at all.

Prerequisites: L^p spaces, Monotone convergence and Fatou's lemma, The Lebesgue integral, Measurable functions, Sigma-algebras and measures, The Riesz representation theorem

Definition(Absolute continuity, singularity and densities)AdaptedAdapted from Folland (1999), §3.1, pp. 86–87, and §3.2, pp. 88–89; Reed & Simon (1980), §I.4, p. 22

Throughout, $(X,\mathcal{M})$ is a measurable space and all measures are defined on $\mathcal{M}$ . Recall from sigma-algebras and measures that a measure $\mu$ is $\sigma$ -finite if $X$ is a countable union of sets of finite measure.

Let $\mu$ and $\nu$ be positive measures.

$\nu$ is absolutely continuous with respect to $\mu$ , written $\nu \ll \mu$ , if $\mu(E) = 0$ forces $\nu(E) = 0$ . Every $\mu$ -null set is then $\nu$ -null, so " $\mu$ -almost everywhere" implies " $\nu$ -almost everywhere".
$\nu$ and $\mu$ are mutually singular, written $\nu \perp \mu$ , if $X$ splits as a disjoint union $X = B \sqcup C$ with $\mu(B) = 0$ and $\nu(C) = 0$ . Informally the two measures live on disjoint sets. Reed and Simon (Reed & Simon (1980), §I.4, p. 22) phrase it as a single set $B$ with $\mu(B) = 0$ and $\nu(X \setminus B) = 0$ , which is the same condition with $C = X \setminus B$ .

The two relations are opposites rather than variants: if $\nu \ll \mu$ and $\nu \perp \mu$ then $\nu = 0$ , because with $B, C$ as above $\mu(B) = 0$ gives $\nu(B) = 0$ by absolute continuity, while $\nu(C) = 0$ by singularity, so $\nu(X) = 0$ .

If $f : X \to [0,\infty]$ is measurable, the set function

\nu(E) \;=\; \int_E f \, d\mu \qquad (E \in \mathcal{M})

is a positive measure, by termwise integration of non-negative series (Monotone convergence and Fatou's lemma), and it satisfies $\nu \ll \mu$ , because $\mu(E) = 0$ makes the integral over $E$ vanish. We abbreviate this relation to

d\nu = f\,d\mu ,

and call $f$ a density of $\nu$ with respect to $\mu$ . When $\nu \ll \mu$ and such an $f$ exists it is unique up to $\mu$ -a.e. equality provided $\mu$ and $\nu$ are $\sigma$ -finite, which is the standing hypothesis of the theorem below and is what Step 9 uses; without it uniqueness fails, as $X = \{1\}$ with $\mu(\{1\}) = \nu(\{1\}) = \infty$ shows, where $f \equiv 1$ and $f \equiv 2$ are both densities. Under that hypothesis the resulting equivalence class is the Radon-Nikodym derivative, written $d\nu/d\mu$ . Following Folland (Folland (1999), §3.2, p. 91) we treat $d\nu/d\mu$ as a function, remembering that only its $\mu$ -a.e. equivalence class is determined.

For the signed version stated in the note below we also need Folland's term: a measurable $f : X \to [-\infty,\infty]$ is extended $\mu$ -integrable when $\int f^{+}\,d\mu < \infty$ or $\int f^{-}\,d\mu < \infty$ , so that $\int f\,d\mu$ is defined in $[-\infty,\infty]$ .

Theorem(Lebesgue-Radon-Nikodym)AdaptedAdapted from Folland (1999), Thm 3.8, p. 90; Reed & Simon (1980), Thms I.19 and I.20, p. 23

Let $\mu$ and $\nu$ be $\sigma$ -finite positive measures on $(X,\mathcal{M})$ .

(Lebesgue decomposition.) There are $\sigma$ -finite positive measures $\lambda$ and $\rho$ on $(X,\mathcal{M})$ with

\nu = \lambda + \rho , \qquad \lambda \perp \mu , \qquad \rho \ll \mu ,

and this pair is unique. 2. (Existence of a density.) There is a measurable $f : X \to [0,\infty]$ with $\rho(E) = \int_E f\,d\mu$ for every $E \in \mathcal{M}$ , and any two such functions agree $\mu$ -a.e. 3. (Radon-Nikodym.) If in addition $\nu \ll \mu$ , then $\lambda = 0$ and

\nu(E) = \int_E \frac{d\nu}{d\mu}\, d\mu \qquad \text{for all } E \in \mathcal{M} .

Conversely any $\nu$ of this form is absolutely continuous with respect to $\mu$ . So, for $\sigma$ -finite measures, having a density with respect to $\mu$ and being absolutely continuous with respect to $\mu$ are the same property.

Note(The version in the sources, and what we prove)AdaptedAdapted from Folland (1999), Thm 3.8, p. 90, Thm 3.4, p. 87, Thm 3.12, p. 93, and Exercise 14, p. 92; Reed & Simon (1980), Thms I.19 and I.20, p. 23

Folland (Folland (1999), Thm 3.8, p. 90) states the theorem with $\nu$ allowed to be a $\sigma$ -finite signed measure: $\lambda$ and $\rho$ are then $\sigma$ -finite signed measures and $f$ is extended $\mu$ -integrable with values in $\mathbb{R}$ . The general case follows from the positive case proved below by applying it separately to the two positive measures $\nu^{+}$ and $\nu^{-}$ of the Jordan decomposition $\nu = \nu^{+} - \nu^{-}$ (Folland (Folland (1999), Thm 3.4, p. 87)) and subtracting; Folland's Thm 3.12 (p. 93) does the same for complex $\nu$ . We state and prove the positive case, which is what the rest of this handbook uses.

Reed and Simon (Reed & Simon (1980), Thms I.19 and I.20, p. 23) split the result into two statements — the Radon-Nikodym theorem for $\nu \ll \mu$ , and the Lebesgue decomposition theorem — and state both for measures on a measure space without writing a $\sigma$ -finiteness hypothesis into the statement; their proof in the supplement (pp. 344–346) treats the finite case and leaves the passage to $\sigma$ -finite measures as a problem. Folland's Exercise 14 (p. 92) records that for part 3 alone the hypothesis on $\nu$ can be dropped: if $\mu$ is $\sigma$ -finite and $\nu$ is an arbitrary signed measure with $\nu \ll \mu$ , an extended $\mu$ -integrable density still exists. The $\sigma$ -finiteness of $\mu$ cannot be dropped; see the remark on hypotheses.

Idea of the proofOriginal

A measure $\nu$ fails to have a density with respect to $\mu$ only because of the part of $\nu$ that lives where $\mu$ has no mass, so the whole problem is to locate that part. The trick is to stop comparing $\nu$ with $\mu$ and compare both with their sum $\alpha = \mu + \nu$ , which dominates each of them. Against $\alpha$ nothing can blow up: $E \mapsto \nu(E)$ extends to a bounded linear functional on the Hilbert space $L^2(\alpha)$ , so the Riesz representation theorem hands us a single function $F$ with $d\nu = F\,d\alpha$ , and $0 \le F \le 1$ because $0 \le \nu \le \alpha$ . Subtracting, $d\mu = (1-F)\,d\alpha$ . Now read off the answer: on $\{F = 1\}$ the factor $1-F$ vanishes, so $\mu$ gives that set no mass and it carries the singular part of $\nu$ ; on $\{F < 1\}$ we may divide, and $F/(1-F)$ is the density we want.

ProofAdaptedAdapted from Reed & Simon (1980), Thm S.1 and its proof, pp. 344–346; Folland (1999), Thm 3.8, Case II, p. 90

The argument for finite measures is von Neumann's, which Reed and Simon (Reed & Simon (1980), pp. 344–346) give as an application of the Riesz lemma; the passage to $\sigma$ -finite measures follows Folland (Folland (1999), Thm 3.8, Case II, p. 90). Steps 1–7 assume $\mu(X) < \infty$ and $\nu(X) < \infty$ .

Step 1: a bounded functional on $L^2(\alpha)$ . Put $\alpha = \mu + \nu$ , a finite positive measure. Let $L^2(\alpha)$ be the real Hilbert space of square- $\alpha$ -integrable real functions with $(g,h) = \int gh \, d\alpha$ ; it is complete by Riesz-Fischer (L^p spaces). Define

L(g) = \int_X g \, d\nu \qquad (g \in L^2(\alpha)).

This is defined and bounded: $\nu \le \alpha$ as set functions, so $\int |g|\,d\nu \le \int |g| \, d\alpha$ , and Hölder's inequality with $p = q = 2$ applied to $|g| \cdot 1$ on the finite measure space $(X,\alpha)$ gives

|L(g)| \;\le\; \int_X |g| \, d\alpha \;\le\; \|g\|_{L^2(\alpha)}\,\alpha(X)^{1/2} .

Finiteness of $\alpha(X)$ , which is where both measures being finite is used, is what makes the constant $\alpha(X)^{1/2}$ available; $L$ is linear because the integral is.

Step 2: the representing function. By the Riesz representation theorem there is $F \in L^2(\alpha)$ with $L(g) = \int_X gF\,d\alpha$ for every $g \in L^2(\alpha)$ . Since $\alpha$ is finite, $\chi_E \in L^2(\alpha)$ for every $E \in \mathcal{M}$ , and taking $g = \chi_E$ gives

\nu(E) = \int_E F \, d\alpha \qquad (E \in \mathcal{M}). \tag{$*$}

Step 3: $0 \le F \le 1$ $\alpha$ -almost everywhere. Fix $\gamma > 1$ and let $E_\gamma = \{F \ge \gamma\}$ . On $E_\gamma$ the integrand in $(*)$ is at least $\gamma$ , so

\nu(E_\gamma) \;\ge\; \gamma\,\alpha(E_\gamma) \;=\; \gamma\bigl(\mu(E_\gamma) + \nu(E_\gamma)\bigr) \;\ge\; \gamma\,\nu(E_\gamma) .

Here $\nu(E_\gamma) \le \nu(X) < \infty$ , so this may be rearranged to $(\gamma - 1)\nu(E_\gamma) \le 0$ , and $\gamma > 1$ forces $\nu(E_\gamma) = 0$ ; feeding that back gives $\gamma\,\alpha(E_\gamma) \le 0$ , so $\alpha(E_\gamma) = 0$ . Letting $\gamma$ run through $1 + 1/n$ and using countable subadditivity, $\alpha(\{F > 1\}) = 0$ . In the other direction let $D_n = \{F \le -1/n\}$ ; then $(*)$ gives $0 \le \nu(D_n) \le -\alpha(D_n)/n \le 0$ , so $\alpha(D_n) = 0$ , and the union over $n$ gives $\alpha(\{F < 0\}) = 0$ . Changing $F$ on an $\alpha$ -null set changes no integral in $(*)$ , so we may and do assume $0 \le F(x) \le 1$ for every $x$ .

Step 4: the companion formula $d\mu = (1-F)\,d\alpha$ , and its use on functions. Subtracting $(*)$ from $\alpha(E) = \mu(E) + \nu(E)$ and using $\alpha(E) = \int_E 1 \, d\alpha$ ,

\mu(E) = \int_E (1 - F)\,d\alpha \qquad (E \in \mathcal{M}),

all three quantities being finite. We need this for integrands other than indicators: for every measurable $h : X \to [0,\infty]$ ,

\int_X h \, d\mu = \int_X h\,(1-F)\,d\alpha . \tag{$**$}

Indeed $(**)$ is the displayed formula when $h = \chi_E$ ; it extends to non-negative simple $h$ by linearity of the integral; and for general $h \ge 0$ choose simple $0 \le s_1 \le s_2 \le \cdots$ with $s_k \uparrow h$ pointwise (Measurable functions), note that $s_k(1-F) \uparrow h(1-F)$ because $1 - F \ge 0$ , and apply the monotone convergence theorem (Monotone convergence and Fatou's lemma) on both sides.

Step 5: the singular set. Let

B = \{x : F(x) = 1\}, \qquad A = X \setminus B = \{x : F(x) < 1\} .

Taking $h = \chi_B$ in $(**)$ gives $\mu(B) = \int_B (1-F)\,d\alpha = 0$ , since $1 - F$ vanishes identically on $B$ . So $B$ is $\mu$ -null.

Step 6: the density on the rest. Define

f = \frac{F}{1-F}\,\chi_A ,

a non-negative measurable function (the quotient is formed only where $1 - F > 0$ , and $f = 0$ on $B$ ). For $E \in \mathcal{M}$ apply $(**)$ to $h = \chi_E f$ :

\int_E f \, d\mu = \int_X \chi_E \frac{F}{1-F}\chi_A (1-F) \, d\alpha = \int_{E \cap A} F \, d\alpha = \nu(E \cap A) ,

the last equality by $(*)$ applied to $E \cap A$ . The cancellation of $1-F$ is legitimate because it happens only on $A$ , where $1 - F > 0$ .

Step 7: the decomposition, for finite measures. Set $\lambda(E) = \nu(E \cap B)$ and $\rho(E) = \nu(E \cap A)$ . Both are positive measures, and $\lambda + \rho = \nu$ because $A$ and $B$ partition $X$ . Since $\mu(B) = 0$ by Step 5 and $\lambda(A) = \nu(\varnothing) = 0$ , the splitting $X = B \sqcup A$ witnesses $\lambda \perp \mu$ . Step 6 says $\rho(E) = \int_E f \, d\mu$ , which makes $\rho \ll \mu$ . Finally, if $\nu \ll \mu$ then $\mu(B) = 0$ gives $\nu(B) = 0$ , hence $\lambda = 0$ and $\nu(E) = \int_E f\,d\mu$ for every $E$ : part 3.

Step 8: $\sigma$ -finite measures. Write $X = \bigcup_i P_i$ with $\mu(P_i) < \infty$ and $X = \bigcup_k Q_k$ with $\nu(Q_k) < \infty$ . The countably many sets $P_i \cap Q_k$ have both measures finite and cover $X$ ; disjointifying them (replace each by what it adds to the union of its predecessors, as in Sigma-algebras and measures) gives a countable partition $X = \bigsqcup_j X_j$ with $\mu(X_j) < \infty$ and $\nu(X_j) < \infty$ . Apply Steps 1–7 to the finite measures $\mu_j : E \mapsto \mu(E \cap X_j)$ and $\nu_j : E \mapsto \nu(E \cap X_j)$ , and write $\alpha_j = \mu_j + \nu_j$ . This produces a set $B_j = \{F_j = 1\}$ with $\mu_j(B_j) = 0$ and a measurable $f_j \ge 0$ with $\nu_j(E) = \nu_j(E \cap B_j) + \int_E f_j\,d\mu_j$ . Both $F_j$ and $f_j$ are determined only up to $\alpha_j$ -null sets, and $\alpha_j(X \setminus X_j) = 0$ , so we are free to normalise the representatives outside $X_j$ ; following Folland's (Folland (1999), Thm 3.8, Case II, p. 90) normalisation we choose $F_j = 0$ and hence $f_j = 0$ on $X \setminus X_j$ . This makes $B_j \subseteq X_j$ , so the $B_j$ are pairwise disjoint and $\mu(B_j) = \mu_j(B_j) = 0$ ; and since $f_j$ vanishes off $X_j$ , $\int_E f_j \, d\mu_j = \int_{E \cap X_j} f_j\,d\mu = \int_E f_j \, d\mu$ . Since also $\nu_j(E) = \nu(E \cap X_j)$ and $\nu_j(E \cap B_j) = \nu(E \cap B_j)$ , the display above becomes

\nu(E \cap X_j) = \nu(E \cap B_j) + \int_E f_j \, d\mu .

Put $B = \bigcup_j B_j$ , $\lambda(E) = \nu(E \cap B)$ and $f = \sum_j f_j$ . Then $\mu(B) = 0$ by countable subadditivity, so $\lambda \perp \mu$ ; the $B_j$ are disjoint with $B_j \subseteq X_j$ , so $\sum_j \nu(E \cap B_j) = \nu(E \cap B)$ . Summing the display over $j$ and interchanging sum and integral by termwise integration of non-negative series (Monotone convergence and Fatou's lemma) gives $\nu(E) = \lambda(E) + \int_E f \, d\mu$ for all $E$ . Setting $\rho(E) = \int_E f\,d\mu$ completes the existence proof; $\lambda$ and $\rho$ are $\sigma$ -finite because each is at most $\nu$ on each $X_j$ . If $\nu \ll \mu$ then $\nu(B) = 0$ and $\lambda = 0$ , as before.

Step 9: uniqueness. Suppose $\nu = \lambda + \rho = \lambda' + \rho'$ with $\lambda, \lambda' \perp \mu$ and $\rho, \rho' \ll \mu$ . Choose $\mu$ -null sets $C, C'$ carrying $\lambda, \lambda'$ , that is $\mu(C) = 0 = \lambda(X \setminus C)$ and likewise for $C'$ , and put $N = C \cup C'$ , so $\mu(N) = 0$ . For any $E$ , absolute continuity gives $\rho(E \cap N) = 0 = \rho'(E \cap N)$ , while $\lambda(E \setminus N) = 0 = \lambda'(E\setminus N)$ since $N \supseteq C, C'$ . Hence

\lambda(E) = \lambda(E \cap N) = \nu(E \cap N) = \lambda'(E \cap N) = \lambda'(E), \qquad \rho(E) = \rho(E \setminus N) = \nu(E \setminus N) = \rho'(E) ,

with no subtraction of possibly infinite quantities anywhere. For the density, suppose $\int_E f \, d\mu = \int_E g\,d\mu$ for all $E$ , with $f, g \ge 0$ . Using the partition $\{X_j\}$ of Step 8, refined so that in addition $\rho(X_j) < \infty$ , take $E = X_j \cap \{f > g\}$ : both integrals are finite and equal, so $\int_E (f - g)\,d\mu = 0$ with a strictly positive integrand, which forces $\mu(E) = 0$ . Exchanging $f$ and $g$ and taking the union over $j$ gives $f = g$ $\mu$ -a.e. $\square$

Theorem(Calculus of Radon-Nikodym derivatives)AdaptedAdapted from Folland (1999), Prop. 3.9 and Cor. 3.10, p. 91

Let $\lambda, \mu, \nu$ be $\sigma$ -finite positive measures on $(X,\mathcal{M})$ .

Change of variables. If $\nu \ll \mu$ then for every measurable $g \ge 0$ , and for every $g \in L^1(\nu)$ ,

\int_X g \, d\nu = \int_X g \,\frac{d\nu}{d\mu}\, d\mu .

Chain rule. If $\nu \ll \mu$ and $\mu \ll \lambda$ then $\nu \ll \lambda$ and

\frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu}\,\frac{d\mu}{d\lambda} \qquad \lambda\text{-a.e.}

Reciprocal. If $\mu \ll \lambda$ and $\lambda \ll \mu$ (the measures are then called equivalent) then $(d\lambda/d\mu)(d\mu/d\lambda) = 1$ almost everywhere with respect to either.

Proof(Calculus rules)AdaptedAdapted from Folland (1999), Prop. 3.9, p. 91

1. With $h = d\nu/d\mu$ , the identity $\int g\,d\nu = \int gh\,d\mu$ holds for $g = \chi_E$ by the definition of $h$ , hence for non-negative simple $g$ by linearity, hence for all measurable $g \ge 0$ by monotone convergence applied to a simple increasing approximation (Monotone convergence and Fatou's lemma), exactly as in Step 4 of the previous proof. For $g \in L^1(\nu)$ , split $g = g^{+} - g^{-}$ and subtract the two finite results.

2. $\nu \ll \lambda$ is immediate: $\lambda(E) = 0 \Rightarrow \mu(E) = 0 \Rightarrow \nu(E) = 0$ . For the formula, fix $E$ and apply part 1 to the pair $\mu \ll \lambda$ with the non-negative integrand $g = \chi_E \,(d\nu/d\mu)$ :

\nu(E) = \int_E \frac{d\nu}{d\mu}\,d\mu = \int_E \frac{d\nu}{d\mu}\,\frac{d\mu}{d\lambda}\,d\lambda .

Since also $\nu(E) = \int_E (d\nu/d\lambda)\,d\lambda$ for all $E$ , the $\lambda$ -a.e. uniqueness of a density (Step 9) gives the claim.

3. Apply part 2 with $\nu = \lambda$ , using that $d\lambda/d\lambda = 1$ by the same uniqueness statement: this gives $(d\lambda/d\mu)(d\mu/d\lambda) = 1$ $\lambda$ -a.e. The identity then also holds $\mu$ -a.e., because $\mu \ll \lambda$ makes every $\lambda$ -null set $\mu$ -null. $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Folland (1999), Thm 3.5 and the discussion on p. 89, p. 91, and Exercises 10 and 13, p. 92

$\sigma$ -finiteness of $\mu$ is essential, and it is used in Step 8. Without it there is no partition into pieces on which Step 1 can run, and the theorem is false. Folland's (Folland (1999), Exercise 13, p. 92) example: on $X = [0,1]$ with the Borel sets, let $m$ be Lebesgue measure and $\mu$ counting measure. The only $\mu$ -null set is $\varnothing$ , so $m \ll \mu$ trivially. But if $m(E) = \int_E f \, d\mu$ for all $E$ , then taking $E = \{x\}$ gives $f(x) = m(\{x\}) = 0$ for every $x$ , hence $f \equiv 0$ and $m \equiv 0$ , contradicting $m([0,1]) = 1$ . Folland adds that this $\mu$ also has no Lebesgue decomposition with respect to $m$ . Note that the failure is in the reference measure: $m$ here is finite, and $\mu$ is not $\sigma$ -finite because $[0,1]$ is uncountable.
$\sigma$ -finiteness of $\nu$ is used only for the decomposition. For part 3 alone it can be dropped, by Folland's Exercise 14 quoted in the note above.
Finiteness of $\nu$ is used in Step 3, to rearrange $\nu(E_\gamma) \ge \gamma \nu(E_\gamma)$ into $(\gamma-1)\nu(E_\gamma) \le 0$ . Had $\nu(E_\gamma)$ been infinite the inequality would have been vacuous and no bound on $F$ would follow. This is why the proof runs on finite pieces and only then is assembled.
Absolute continuity is used only in the last line of Steps 7 and 8. Everything before it holds for arbitrary $\sigma$ -finite $\nu$ , and that is the content of part 1: a measure always splits into a part with a density and a part concentrated where $\mu$ is blind.
What the singular part looks like. Take $\mu = m$ on $\mathbb{R}$ and $\nu = \delta_0$ , the point mass at $0$ . Then $\nu \perp \mu$ , with $B = \{0\}$ , and the theorem produces $f = 0$ : there is no density. Folland (Folland (1999), p. 91) notes that the nonexistent derivative $d\delta_0/dm$ is what physicists call the Dirac delta function. It is recovered as a limit of densities, or as a distribution rather than a function; see Delta-function identities and Test functions and distributions. The Radon-Nikodym theorem is thus a precise statement of why the delta function cannot be a function.
Absolute continuity as an $\varepsilon$ - $\delta$ statement needs $\nu$ finite. Folland's Thm 3.5 (Folland (1999), p. 89) says that for a finite signed $\nu$ , $\nu \ll \mu$ is equivalent to: for every $\varepsilon > 0$ there is $\delta > 0$ with $|\nu(E)| < \varepsilon$ whenever $\mu(E) < \delta$ . His Exercise 10 (p. 92) shows this fails for infinite $\nu$ : on $(0,1)$ take $d\nu = dx/x$ and $d\mu = dx$ . Then $\nu \ll \mu$ , but $\nu((0,\delta)) = \infty$ for every $\delta > 0$ .
The derivative is a class, not a function. $F$ in Step 2 is given by the Riesz theorem only as an element of $L^2(\alpha)$ , and Step 3 modified it on a null set. Correspondingly $d\nu/d\mu$ is determined only $\mu$ -a.e., which is exactly the ambiguity that Step 9 shows is the whole of the ambiguity. Any statement about the pointwise value of $d\nu/d\mu$ at a particular point is meaningless unless the point carries positive $\mu$ -measure.

Method(Finding and using a density)Original

Test absolute continuity. Identify the $\mu$ -null sets and check that $\nu$ vanishes on all of them. The usual obstruction is an atom: if $\nu(\{a\}) > 0$ while $\mu(\{a\}) = 0$ , then $\nu \not\ll \mu$ and the best available result is the decomposition, not a density.
Check $\sigma$ -finiteness of $\mu$ (and of $\nu$ if you want the decomposition). On $\mathbb{R}^n$ with Lebesgue measure this is automatic; with counting measure on an uncountable set it fails.
Guess the density. For $\mu = m$ on $\mathbb{R}$ , compute $G(b) = \nu((-\infty,b])$ and differentiate: if $G$ is an indefinite integral of a locally integrable $g$ , then $g$ is the candidate.
Verify it on a determining class. To confirm $\nu(E) = \int_E f\,dm$ for every Borel $E$ , it suffices to check it for $E = (a,b]$ . Both sides are then measures agreeing on the algebra of finite disjoint unions of such intervals, and if they are $\sigma$ -finite there the uniqueness clause of the extension theorem (Lebesgue measure) forces them to agree on all of $\mathcal{B}_{\mathbb{R}}$ .
Compute with it using the change-of-variables rule: every integral against $\nu$ becomes an integral against $\mu$ with an extra factor $d\nu/d\mu$ . This is the one formula that turns a statement about measures into a statement about functions.
Change reference measure with the chain rule, and invert with the reciprocal rule when the two measures are equivalent.
Compare several measures by dominating them all at once: for $\nu_1,\dots,\nu_n$ the measure $\mu = \sum_j \nu_j$ satisfies $\nu_j \ll \mu$ for every $j$ (Folland, Prop. 3.11 (Folland (1999), p. 91)), so all of them acquire densities simultaneously and their ratios are well defined. This is what makes a likelihood ratio meaningful whether the model is discrete or continuous.
If the test in step 1 fails, split instead: strip off the atoms, then the remaining singular part, and apply steps 3–6 to what is left.

Example(A measure with an atom: decomposition, density, and numbers)Original

Let $X = [0,2]$ with its Borel sets, let $\mu$ be Lebesgue measure on $X$ , and let

\nu(E) \;=\; \int_{E \cap [0,1]} 3x^2\,dx \;+\; \tfrac12\,\delta_1(E) ,

where $\delta_1$ is the point mass at $x = 1$ . Both are finite: $\mu(X) = 2$ and $\nu(X) = 1 + \tfrac12 = \tfrac32$ .

Step 1: $\nu \not\ll \mu$ . We have $\mu(\{1\}) = 0$ but $\nu(\{1\}) = \tfrac12 \neq 0$ . So no density exists, and we must decompose.

Step 2: the function $F$ of the proof. Here $\alpha = \mu + \nu$ acts as follows: on $[0,1)$ it has Lebesgue density $1 + 3x^2$ ; on $(1,2]$ it has Lebesgue density $1$ ; and $\alpha(\{1\}) = 0 + \tfrac12 = \tfrac12$ . Set

F(x) = \frac{3x^2}{1+3x^2} \ \ (0 \le x < 1), \qquad F(1) = 1, \qquad F(x) = 0 \ \ (1 < x \le 2) .

To check $(*)$ , split any Borel $E$ into three pieces. On $E \cap [0,1)$ ,

\int_{E\cap[0,1)} F \, d\alpha = \int_{E \cap [0,1)} \frac{3x^2}{1+3x^2}\,(1+3x^2)\,dx = \int_{E\cap[0,1)} 3x^2\,dx ,

which is $\nu(E \cap [0,1))$ . If $1 \in E$ , that piece contributes $F(1)\,\alpha(\{1\}) = 1 \cdot \tfrac12 = \tfrac12 = \nu(\{1\})$ . On $E \cap (1,2]$ both sides are $0$ . So $(*)$ holds, and $0 \le F \le 1$ as Step 3 of the proof predicts.

Step 3: read off the decomposition. The singular set is $B = \{F = 1\} = \{1\}$ , with $\mu(B) = 0$ , and

\lambda = \tfrac12\,\delta_1 , \qquad f(x) = \frac{F(x)}{1-F(x)}\chi_{A}(x) = \begin{cases} 3x^2, & 0 \le x < 1,\\[2pt] 0, & 1 \le x \le 2,\end{cases} \qquad \rho(E) = \int_E f \, d\mu .

The algebra behind the first line of the formula is $\bigl[3x^2/(1+3x^2)\bigr]\big/\bigl[1/(1+3x^2)\bigr] = 3x^2$ . As a numerical spot check at $x = \tfrac12$ : $F(\tfrac12) = \tfrac{3/4}{7/4} = \tfrac37$ , and $\tfrac{3/7}{4/7} = \tfrac34 = 3(\tfrac12)^2$ , as required.

Step 4: checkable numbers. From $\nu = \lambda + \rho$ ,

\nu\bigl([0,\tfrac12]\bigr) = \int_0^{1/2} 3x^2\,dx = \tfrac18 = 0.125 ,

\nu\bigl([\tfrac12,1]\bigr) = \int_{1/2}^{1} 3x^2\,dx + \tfrac12 = \tfrac78 + \tfrac12 = \tfrac{11}{8} = 1.375 .

The second of these exceeds the integral of the density over the same interval, $\int_{1/2}^{1} 3x^2\,dx = \tfrac78$ , by exactly $\tfrac12$ : that excess is the atom at $x = 1$ , equivalently the jump of $\tfrac12$ in $b \mapsto \nu([0,b])$ at $b = 1$ , and it is exactly what no density could produce. Finally the change-of-variables rule applied to the absolutely continuous part gives

\int_X x \, d\nu(x) = \int_0^1 x\cdot 3x^2\,dx + \tfrac12 \cdot 1 = \tfrac34 + \tfrac12 = \tfrac54 = 1.25 .

Each of these was confirmed symbolically: $\int_0^{1/2}3x^2dx = 1/8$ , $\int_{1/2}^{1}3x^2dx = 7/8$ , $\int_0^1 3x^3dx = 3/4$ .

Application(Densities and likelihood ratios in probability)Original

A random variable $X$ on $(\Omega,\mathcal{F},\mathbb{P})$ pushes $\mathbb{P}$ forward to its law $\mathbb{P}_X(E) = \mathbb{P}(X \in E)$ , a probability measure on $\mathbb{R}$ . Grimmett and Stirzaker (Grimmett & Stirzaker (2020), §2.3, Definition (2), p. 35) call $X$ continuous when its distribution function is an integral, $F(x) = \int_{-\infty}^{x} f(u)\,du$ , and call $f$ the density; in the language of this unit that is precisely the statement $\mathbb{P}_X \ll m$ , and $f = d\mathbb{P}_X/dm$ . Their discrete variables are the case $\mathbb{P}_X \ll$ counting measure on a countable set, with the mass function as density. They also note (p. 35) that a distribution may be neither — it may be singular, the Cantor distribution being the standard example — or a mixture of the three kinds; those are exactly the measures whose Lebesgue decomposition has a non-zero $\lambda$ . So the trichotomy that probability textbooks introduce by cases is one theorem.

The practical gain is that one formula covers all cases. Fix a dominating measure $\mu$ for a family of candidate laws (step 7 of the method) and write each as $p_\theta = dP_\theta/d\mu$ . Then $p_\theta$ is the likelihood, whether the model is discrete, continuous or mixed, and the ratio $dQ/dP$ of two laws is the likelihood ratio on which maximum likelihood and likelihood-ratio tests are built. A small example: let $P$ be the exponential law with rate $1$ and $Q$ the exponential law with rate $2$ on $(0,\infty)$ . Both are absolutely continuous with respect to $m$ , and because both densities $e^{-x}$ and $2e^{-2x}$ are strictly positive on $(0,\infty)$ the converse holds too: a set of positive Lebesgue measure has positive probability under each, so $m \ll P$ and $m \ll Q$ as well. Hence $P$ and $Q$ are each equivalent to $m$ , and therefore to each other; in particular $Q \ll P$ , which is what the chain rule needs. (Absolute continuity with respect to $m$ alone would not suffice: the uniform laws on $(0,1)$ and on $(1,2)$ are both $\ll m$ and yet mutually singular.) By the chain and reciprocal rules

\frac{dQ}{dP}(x) = \frac{2e^{-2x}}{e^{-x}} = 2e^{-x} .

Computing a $Q$ -probability as a $P$ -expectation then uses the change-of-variables rule:

Q(X > 1) = \int_{1}^{\infty} \frac{dQ}{dP}\,dP = \int_1^\infty 2e^{-x}\cdot e^{-x}\,dx = e^{-2} = 0.1353352832\ldots ,

which agrees with the direct computation $\int_1^\infty 2e^{-2x}dx = e^{-2}$ . Reweighting samples from one distribution to estimate probabilities under another is importance sampling, and the same identity, applied to the law of a diffusion, is the Girsanov change of measure behind risk-neutral pricing (The Ito integral and Ito's formula).

Finally, conditional expectation is a Radon-Nikodym derivative and not merely an analogue of one: given $Y \in L^1(\mathbb{P})$ and a sub- $\sigma$ -algebra $\mathcal{G} \subseteq \mathcal{F}$ , the set function $E \mapsto \int_E Y\,d\mathbb{P}$ on $\mathcal{G}$ is absolutely continuous with respect to $\mathbb{P}|_{\mathcal{G}}$ , and $\mathbb{E}[Y \mid \mathcal{G}]$ is defined as its density; Folland (Folland (1999), Exercise 17, p. 93) sets exactly this exercise. The a.e. uniqueness of Step 9 is the reason conditional expectation is well defined up to null sets and no better, and the tower property $\mathbb{E}\bigl[\mathbb{E}[Y \mid \mathcal{G}] \mid \mathcal{H}\bigr] = \mathbb{E}[Y \mid \mathcal{H}]$ for $\mathcal{H} \subseteq \mathcal{G}$ follows from that same uniqueness: both sides are densities on $\mathcal{H}$ of the one set function $E \mapsto \int_E Y\,d\mathbb{P}$ . (It is not an instance of the chain rule, which multiplies derivatives of three measures on a single $\sigma$ -algebra, whereas here one measure is restricted to successively smaller $\sigma$ -algebras.) Everything in Martingales rests on this.

Application(Spectral measures and the structure of

L^p

duality)AdaptedAdapted from Reed & Simon (1980), §VII.2, Thm VII.4 and the definitions on pp. 230–231; Reed & Simon (1980), Thm S.4, p. 350

Quantum mechanics. For a bounded self-adjoint operator $A$ on a Hilbert space $\mathcal{H}$ , each vector $\psi$ has a spectral measure $\mu_\psi$ on $\mathbb{R}$ (The spectral theorem), and Reed and Simon (Reed & Simon (1980), §VII.2, p. 230) apply this unit's decomposition, refined by a further splitting of the singular part, to every one of them (their Thm I.13): each $\mu_\psi$ splits uniquely into three pieces: a pure point part, a part that is absolutely continuous for Lebesgue measure, and a part that is both continuous and singular for it. Only the absolutely continuous / singular cut is the Lebesgue decomposition proved above; the splitting of the singular part into a pure point piece and a continuous singular piece is extra. Collecting the vectors by the type of their spectral measure gives

\mathcal{H} = \mathcal{H}_{\mathrm{pp}} \oplus \mathcal{H}_{\mathrm{ac}} \oplus \mathcal{H}_{\mathrm{sing}} ,

and Reed and Simon's Thm VII.4 states that all three subspaces are invariant under $A$ , that $A$ restricted to $\mathcal{H}_{\mathrm{pp}}$ has a complete set of eigenvectors, and that the restrictions to the other two summands have only absolutely continuous, respectively only continuous singular, spectral measures. The spectrum splits correspondingly into $\sigma_{\mathrm{pp}}$ , $\sigma_{\mathrm{ac}}$ and $\sigma_{\mathrm{sing}}$ (pp. 230–231). Physically, $\mathcal{H}_{\mathrm{pp}}$ holds the bound states of a Hamiltonian and $\mathcal{H}_{\mathrm{ac}}$ the scattering states, so this is the decomposition that separates the two regimes; the Radon-Nikodym derivative $d\mu_\psi/dm$ on the absolutely continuous part is the spectral density that appears in response functions and in the local density of states. Reed and Simon warn (p. 231) that $\sigma_{\mathrm{sing}}$ , the exotic middle case, need not be negligible: it can have positive Lebesgue measure.

Duality of $L^p$ spaces. The identification $(L^p)^* \cong L^q$ for $\sigma$ -finite measures is proved from this theorem. Given a positive bounded functional $L$ on $L^p$ , Reed and Simon (Reed & Simon (1980), Thm S.4, p. 350) define $\nu(E) = L(\chi_E)$ , verify countable additivity using monotone convergence, observe that $\mu(E) = 0$ makes $\chi_E$ the zero element of $L^p$ and hence $\nu(E) = 0$ , and then invoke the Radon-Nikodym theorem to produce the representing function $g$ with $L(h) = \int hg\,d\mu$ . The $q$ -integrability of $g$ is then extracted from the boundedness of $L$ . This is the main structural theorem about the spaces of L^p spaces, and the one that makes weak convergence and the direct method of variational calculus available.