Measure and Integration: Recorded, Not Proved Here

For $E\subseteq \mathbb{R}$ put

${\lambda }^{\ast }(E):=\inf \{{\sum }_{k=1}^{\mathrm{\infty }}\mathrm{\mid }{I}_k\mathrm{\mid }:E\subseteq {\bigcup }_{k=1}^{\mathrm{\infty }}{I}_k,I_k\text{ open intervals}\},$

the Lebesgue outer measure of $E$. Call $E$ measurable when it satisfies the Caratheodory criterion

${\lambda }^{\ast }(A)={\lambda }^{\ast }(A\cap E)+{\lambda }^{\ast }(A\setminus E)\text{for every }A\subseteq \mathbb{R}\mathrm{.}$

Then the measurable sets form a $\sigma$-algebra $\mathcal{M}$ containing every open set, $\lambda :={\lambda }^{\ast }{\mathrm{\mid }}_{\mathcal{M}}$ is countably additive, $\lambda ([a,b])=b-a$, $\lambda$ is invariant under translation, and $\mathcal{M}$ is complete: every subset of a set of outer measure zero is measurable and null. The same construction in ${\mathbb{R}}^n$ with boxes in place of intervals produces ${\lambda }_n$.

A function $f:\mathbb{R}\to [-\mathrm{\infty },+\mathrm{\infty }]$ is measurable when $f^{-1}((c,+\mathrm{\infty }])\in \mathcal{M}$ for every $c\in \mathbb{R}$. For measurable $f\ge 0$ the Lebesgue integral is

$\int fd\lambda :=\sup \{{\sum }_{i=1}^m{c}_i\lambda (A_i):0\le {\sum }_{i=1}^m{c}_i{\mathbf{1}}_{A_i}\le f,A_i\in \mathcal{M}\text{ disjoint}\},$

and a measurable $f$ is Lebesgue integrable when $\int \mathrm{\mid }f\mathrm{\mid }d\lambda <\mathrm{\infty }$, in which case $\int fd\lambda :=\int f^{+}d\lambda -\int f^{-}d\lambda$. The integrable functions modulo equality almost everywhere form $L^1(\lambda )$. Finally, every Riemann integrable $f$ on $[a,b]$ is Lebesgue integrable there and the two integrals agree, so the Lebesgue integral extends the Riemann integral.

This library does not prove any of it. Everything above is recorded here with citations and used nowhere in a proof. The whole measure track is deferred.

What would prove it. The construction is standard and long: countable subadditivity of ${\lambda }^{\ast }$; Caratheodory's theorem that the sets satisfying the criterion form a $\sigma$-algebra on which an outer measure is countably additive; the fact that intervals satisfy the criterion, which is what gives $\lambda ([a,b])=b-a$ and forces the Borel sets into $\mathcal{M}$; approximation of a nonnegative measurable function from below by simple functions; and additivity of the integral, which is where the monotone convergence theorem enters. That is the opening chapter of any measure theory course and it is exactly the material this library has not built.

Which page it serves. It is the natural endpoint of the Riemann integral page and of the Cantor set, Baire and measure zero page: those pages prove Lebesgue's criterion for Riemann integrability using only the elementary covering notion of a null set, and the theorem above is what explains why that notion is the right one. It is also the base of everything else in this category.

What is available here without it. A great deal, and it should not be confused with what is deferred. The elementary notion "$E$ is null if for every $\epsilon >0$ it is covered by countably many intervals of total length below $\epsilon$", that is ${\lambda }^{\ast }(E)=0$ with no measurability theory attached, is in scope, and so is "almost everywhere" in that sense. Lebesgue's criterion for Riemann integrability, the vanishing of the Cantor function's derivative almost everywhere, Volterra's function, Jordan content and Jordan measurability all live inside the elementary theory. What is missing is the $\sigma$-algebra, the measure defined on it, and the integral.

Choice. The construction above is not free of choice, and the statement displayed above is not a theorem of ZF. What is choice-free is the definition of ${\lambda }^{\ast }$, its monotonicity, and its subadditivity over finitely many sets. What is not is countable subadditivity of ${\lambda }^{\ast }$, and with it the countable additivity of $\lambda$ asserted above and the statement "a countable union of null sets is null": each needs a countable choice principle (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$) ↗) to select one $\epsilon 2^{-n}$ cover per index. If ZF is consistent then ZF proves none of them, since in the Feferman-Levy model of ZF the set $\mathbb{R}$ is a countable union of countable sets, so $[0,1]$ there is a countable union of null sets while ${\lambda }^{\ast }([0,1])=1$. A measure track built here would have to keep the same ledger of choice principles that the rest of this library keeps.

Let $(X,\mathcal{A},\mu )$ be a measure space and let $f_1\le f_2\le f_3\le \cdots$ be measurable functions $X\to [0,+\mathrm{\infty }]$ with pointwise limit $f={\lim }_n{f}_n$. Then $f$ is measurable and

${\lim }_{n\to \mathrm{\infty }}{\int }_X{f}_{n}d\mu ={\int }_{X}fd\mu ,$

both sides being allowed the value $+\mathrm{\infty }$. Equivalently, for measurable $g_n:X\to [0,+\mathrm{\infty }]$,

${\int }_X{\sum }_{n=1}^{\mathrm{\infty }}{g}_{n}d\mu ={\sum }_{n=1}^{\mathrm{\infty }}{\int }_X{g}_{n}d\mu \mathrm{.}$

No integrability hypothesis and no dominating function are needed: monotonicity and nonnegativity are the whole hypothesis.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. The Lebesgue integral of a nonnegative measurable function as a supremum over simple minorants (Lebesgue measure and the Lebesgue integral ‡), plus continuity from below of the measure, that is $\mu ({\bigcup }_n{E}_n)={\lim }_n\mu (E_n)$ for an increasing sequence of measurable sets. The standard argument fixes a simple $0\le s\le f$ and $c\in (0,1)$, applies continuity from below to $E_n=\{f_n\ge cs\}$, and lets $c\to 1$. The theorem is then used immediately to prove that the integral is additive, so it is not a corollary of the basic theory but part of its foundation.

Which page it serves. The Riemann integral page, where the corresponding statement is false without extra hypotheses: an increasing sequence of Riemann integrable functions on $[0,1]$ with bounded integrals may converge pointwise to a bounded function that is not Riemann integrable. That failure is exactly the defect this theorem repairs, and it is the honest motivation for building a measure track at all.

A naming warning. "Monotone convergence theorem" names two unrelated results, and this library uses the phrase for both. The one stated here is Lebesgue's, sharpened by Beppo Levi in 1906, and it is about integrals of functions on a measure space. The other is the elementary theorem that a monotone sequence of reals converges if and only if it is bounded, proved in A monotone sequence converges if and only if it is bounded ↗. Nothing on this page bears on that sequence theorem, and neither result is a special case of the other. A reader who wants "bounded monotone sequences converge" is in the wrong item, and neither result is a special case of the other.

Interchange results that this library does prove. Uniform convergence permits interchange for the Riemann integral, and so does Arzela's bounded convergence theorem, which is a Riemann-level result and is in scope here even though its natural home is next to dominated convergence. The deferral is of the measure-theoretic statements, not of every interchange theorem.

Let $(X,\mathcal{A},\mu )$ be a measure space and let $f_n:X\to [0,+\mathrm{\infty }]$ be measurable. Then

${\int }_X{\mathrm{lim\hspace{0.17em}inf}}_{n\to \mathrm{\infty }}{f}_{n}d\mu \le {\mathrm{lim\hspace{0.17em}inf}}_{n\to \mathrm{\infty }}{\int }_X{f}_{n}d\mu \mathrm{.}$

The inequality can be strict. On $\mathbb{R}$ with Lebesgue measure, $f_n={\mathbf{1}}_{[n,n+1]}$ has ${\mathrm{lim\hspace{0.17em}inf}}_n{f}_n=0$ pointwise while $\int f_{n}d\lambda =1$ for every $n$, so the left side is $0$ and the right side is $1$: mass escapes to infinity. With $f_n=n{\mathbf{1}}_{(0,1\mathrm{/}n)}$ the same values arise with the mass escaping upward instead.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. One line from the monotone convergence theorem (Monotone convergence theorem (Beppo Levi) ‡): apply it to the increasing sequence $g_n:={\inf }_{k\ge n}{f}_k$, whose pointwise limit is ${\mathrm{lim\hspace{0.17em}inf}}_n{f}_n$, and use $g_n\le f_n$ to get $\int g_n\le \int f_n$. Fatou's lemma is thus not independent machinery; it is the convenient one-sided form of monotone convergence, and it is what the dominated convergence theorem is proved from.

Which page it serves. It is the tool that makes the limit theorems on the sequences and series pages usable for integrals, and it is the standard route to dominated convergence (Dominated convergence theorem ‡). Its two escaping mass examples are the sharpest available answer to the question "why is a dominating function needed", which the Riemann integral page can pose but not answer.

Let $(X,\mathcal{A},\mu )$ be a measure space, let $f_n:X\to \mathbb{R}$ be measurable with $f_n\to f$ pointwise almost everywhere, and suppose there is a single $g\in L^1(\mu )$ with $\mathrm{\mid }{f}_n\mathrm{\mid }\le g$ almost everywhere for every $n$. Then $f$ and every $f_n$ are integrable,

${\lim }_{n\to \mathrm{\infty }}{\int }_X\mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu =0,\text{hence}{\lim }_{n\to \mathrm{\infty }}{\int }_X{f}_{n}d\mu ={\int }_{X}fd\mu \mathrm{.}$

The domination hypothesis cannot be dropped: $f_n=n{\mathbf{1}}_{(0,1\mathrm{/}n)}$ on $[0,1]$ converges pointwise to $0$ while $\int f_{n}d\lambda =1$, and the least function dominating all $f_n$ is $1\mathrm{/}x$, which is not integrable on $(0,1)$.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. Fatou's lemma (Fatou's lemma ‡) applied to the nonnegative sequences $g+f_n$ and $g-f_n$, whose liminfs are $g+f$ and $g-f$; the two resulting inequalities squeeze $\int f_n$ to $\int f$. The convergence in $L^1$ comes from the same argument applied to $2g-\mathrm{\mid }{f}_n-f\mathrm{\mid }\ge 0$. So the three convergence theorems form one block: the monotone convergence theorem is proved from the construction of the integral, Fatou from monotone convergence, and this from Fatou.

Which page it serves. It is the endpoint of the Riemann integral page and of the uniform convergence page. Uniform convergence on a bounded interval is what this library can offer for interchanging a limit and an integral, and it is a much heavier hypothesis than pointwise convergence with a dominating function.

The Riemann-level substitute that is in scope. Arzela's bounded convergence theorem, that a uniformly bounded sequence of Riemann integrable functions on $[a,b]$ converging pointwise to a Riemann integrable limit may be integrated term by term, is a theorem about the Riemann integral and is not deferred. It needs the limit function's Riemann integrability as a hypothesis, which is exactly the weakness the Lebesgue theory removes.

Call a family $\mathcal{V}$ of nondegenerate closed intervals a Vitali cover (a fine cover) of $E\subseteq \mathbb{R}$ when for every $x\in E$ and every $\epsilon >0$ there is $I\in \mathcal{V}$ with $x\in I$ and $0<\mathrm{\mid }I\mathrm{\mid }<\epsilon$.

Vitali covering theorem. If ${\lambda }^{\ast }(E)<\mathrm{\infty }$ and $\mathcal{V}$ is a Vitali cover of $E$, then for every $\epsilon >0$ there are finitely many pairwise disjoint $I_1,\dots ,I_N\in \mathcal{V}$ with

${\lambda }^{\ast }(E\setminus {\bigcup }_{k=1}^N{I}_k)<\epsilon ,$

and there is a countable pairwise disjoint family $\{I_k{\}}_{k\ge 1}\subseteq \mathcal{V}$ with ${\lambda }^{\ast }(E\setminus {\bigcup }_{k\ge 1}{I}_k)=0$. The finiteness of ${\lambda }^{\ast }(E)$ may be dropped for the countable form, since $\mathbb{R}$ is a countable union of bounded pieces. The same statement holds in ${\mathbb{R}}^n$ for closed balls, and the underlying combinatorial device is the $5r$ covering lemma: from any family of balls with uniformly bounded radii one may extract a disjoint subfamily $\mathcal{D}$ such that the balls of $\mathcal{D}$ dilated by the factor $5$ cover the union of the original family.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. Lebesgue outer measure and the measurability theory of Lebesgue measure and the Lebesgue integral ‡, together with a greedy selection: repeatedly choose an interval of nearly maximal length among those disjoint from the ones already chosen, and estimate the leftover by the $5r$ lemma. The selection is a recursion over a countable index set with an explicit rule, so it costs only dependent choice, not the full Axiom of Choice.

Which page it serves. It is the covering machinery behind almost everywhere differentiability: the Lebesgue differentiation theorems (Lebesgue's differentiation theorem for monotone functions ‡, Lebesgue differentiation theorem for $L^1$ functions ‡) and, through them, the sharp fundamental theorem of calculus (The sharp fundamental theorem of calculus (absolute continuity) ‡). Its natural home is the monotone functions and discontinuities page, which can prove that a monotone function has at most countably many discontinuities but cannot reach differentiability almost everywhere.

Relation to the elementary covering arguments already in scope. The Heine-Borel and nested interval arguments this library uses, and the elementary covering definition of a null set, are all finite or countable covering statements about total length. Vitali's theorem is the first covering result that needs the measure itself rather than just total length, which is why it is here and they are not. The weaker form sufficient for null sets is recorded separately as Mini-Vitali covering theorem ‡.

Work with the covering relation $\mathcal{H}=\{([u,v],w):u<v,u\le w\le v\}$ and the premeasure $\mathrm{\ell }([u,v],w)=v-u$. A family $\beta \subseteq \mathcal{H}$ is a full cover of $E$ when for each $x\in E$ there is $\delta >0$ with $([u,v],x)\in \beta$ whenever $u\le x\le v$ and $v-u<\delta$, and a fine cover of $E$ when for each $x\in E$ and each $\delta >0$ some $([u,v],x)\in \beta$ has $v-u<\delta$. Write $V(\mathrm{\ell },\beta )$ for the supremum of ${\sum }_i(v_i-u_i)$ over packings, that is over finite or countable subfamilies of $\beta$ with pairwise nonoverlapping intervals.

Mini-Vitali covering theorem. For $E\subseteq \mathbb{R}$ the following are equivalent, and each says that $E$ has Lebesgue measure zero:

1. for every $\epsilon >0$ there is an open $G\supseteq E$ with $\lambda (G)<\epsilon$;
2. for every $\epsilon >0$ there is a full cover $\beta$ of $E$ with $V(\mathrm{\ell },\beta )<\epsilon$;
3. for every $\epsilon >0$ there is a fine cover $\beta$ of $E$ with $V(\mathrm{\ell },\beta )<\epsilon$.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. Two elementary covering lemmas about finite families of compact intervals, then a compactness argument; Bruckner, Bruckner and Thomson give it in Section 3.10 as the cheap precursor to the full theorem. It is genuinely easier than Vitali covering theorem ‡, because it is the same assertion restricted to null sets: the full theorem says ${\mathrm{\ell }}^{\circ }={\mathrm{\ell }}^{\bullet }={\lambda }^{\ast }$ for the measures generated by covers and by packings, and the mini version says only that the three vanish together.

Which page it serves. The same pages as the full theorem, and it is the piece that a first pass at a measure track should prove first: it already yields the Lebesgue differentiation theorem for monotone functions (Lebesgue's differentiation theorem for monotone functions ‡) by the growth lemma route, without the full covering theorem.

Why it is deferred at all, given that the elementary null sets are in scope. The statement is about Lebesgue measure zero, which this library does have in the elementary covering sense, but its content is the equivalence with the full and fine cover formulations, and those are the language of the measure track. Nothing on the topology of $\mathbb{R}$ page or the Riemann integral page currently needs them, so there is no loss; a future measure page will state and prove this before anything else.

Let $F:[a,b]\to \mathbb{R}$ be monotone. Then $F$ is differentiable at almost every point of $[a,b]$, the derivative $F^{\mathrm{\prime }}$ is measurable, and if $F$ is increasing then

${\int }_a^b{F}^{\mathrm{\prime }}d\lambda \le F(b)-F(a),$

with equality precisely when $F$ is absolutely continuous. The same conclusion holds for every $F$ of bounded variation, since such an $F$ is a difference of two increasing functions.

The inequality is genuinely an inequality. The Cantor function $c:[0,1]\to [0,1]$ is continuous and increasing with $c^{\mathrm{\prime }}=0$ almost everywhere, so ${\int }_0^1{c}^{\mathrm{\prime }}d\lambda =0<1=c(1)-c(0)$.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. The Vitali covering theorem (Vitali covering theorem ‡) applied to the sets where the upper and lower Dini derivates differ, or the rising sun lemma route, or the mini-Vitali route through growth lemmas; all three need Lebesgue outer measure as a measure, not just the elementary null sets. The inequality ${\int }_a^b{F}^{\mathrm{\prime }}\le F(b)-F(a)$ then follows from Fatou's lemma (Fatou's lemma ‡) applied to the difference quotients $n(F(x+1\mathrm{/}n)-F(x))$.

The equality case. Equality ${\int }_a^b{F}^{\mathrm{\prime }}d\lambda =F(b)-F(a)$ holds for an increasing $F$ exactly when $F$ is absolutely continuous in the sense of Absolutely continuous functions ‡, which is the content of The sharp fundamental theorem of calculus (absolute continuity) ‡. The gap between the two sides is carried by the singular part of $F$, and the Cantor function is the case where the singular part is everything.

Which page it serves. The monotone functions and discontinuities page, which proves that a monotone function has at most countably many discontinuities and therefore is continuous almost everywhere, and then stops. Differentiability almost everywhere is the next statement in every classical treatment and cannot be reached from the elementary theory. It is also what allows the Cantor function counterexample to be stated at full strength on the Cantor set page: not merely "$c^{\mathrm{\prime }}=0$ off a null set", which is elementary, but "$c$ is one of the monotone functions to which Lebesgue's theorem applies, and it saturates the inequality in the wrong direction".

A naming warning. The phrase "Lebesgue differentiation theorem" is used for two different results. In the classical one-variable literature, including Thomson's article cited above, it names the theorem stated here, that monotone functions are almost everywhere differentiable. In the $L^1$ and harmonic analysis literature it names the averaging statement recorded separately as Lebesgue differentiation theorem for $L^1$ functions ‡. Neither is proved here, and this library keeps them under distinct names.

Let $f\in L^1({\mathbb{R}}^n)$. Then for almost every $x\in {\mathbb{R}}^n$,

${\lim }_{r\to 0^{+}}\frac{1}{{\lambda }_n(B(x,r))}{\int }_{B(x,r)}fd{\lambda }_n=f(x),$

and indeed almost every $x$ is a Lebesgue point, meaning

${\lim }_{r\to 0^{+}}\frac{1}{{\lambda }_n(B(x,r))}{\int }_{B(x,r)}\mathrm{\mid }f(y)-f(x)\mathrm{\mid }d{\lambda }_n(y)=0.$

In dimension one this says that for $f\in L^1[a,b]$ the indefinite integral $F(x)={\int }_a^{x}fd\lambda$ satisfies $F^{\mathrm{\prime }}=f$ almost everywhere, which is the Lebesgue form of the first fundamental theorem of calculus. A special case is the Lebesgue density theorem: a measurable set $E$ has density $1$ at almost every point of $E$ and density $0$ at almost every point of its complement.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. The Hardy-Littlewood maximal inequality, itself proved from the $5r$ covering lemma inside Vitali covering theorem ‡, plus density of the continuous functions in $L^1$, plus the dominated convergence theorem (Dominated convergence theorem ‡). Every ingredient is measure-theoretic.

Which page it serves. The fundamental theorems of calculus page. That page proves that $F(x)={\int }_a^{x}f$ is differentiable with $F^{\mathrm{\prime }}=f$ at every point where $f$ is continuous, which is as far as the Riemann theory reaches. The theorem above removes continuity entirely and replaces "at every point of continuity" by "at almost every point", and together with The sharp fundamental theorem of calculus (absolute continuity) ‡ it closes the subject.

Why it is stated separately from the monotone case. See the naming warning in Lebesgue's differentiation theorem for monotone functions ‡. The two results are close relatives, each provable from Vitali-type covering arguments, but they are different statements about different objects, and conflating them is a common source of confusion about what the fundamental theorem of calculus actually says.

A function $F:[a,b]\to \mathbb{R}$ is absolutely continuous when for every $\epsilon >0$ there is $\delta >0$ such that for every finite family of pairwise disjoint subintervals $(a_1,b_1),\dots ,(a_N,b_N)$ of $[a,b]$,

${\sum }_{k=1}^N(b_k-a_k)<\delta ⟹{\sum }_{k=1}^N\mathrm{\mid }F(b_k)-F(a_k)\mathrm{\mid }<\epsilon \mathrm{.}$

Write $AC[a,b]$ for the class of such $F$. The inclusions

$\text{Lipschitz}⊊AC[a,b]⊊\{\text{continuous and of bounded variation}\}⊊\{\text{continuous}\}$

are all strict: $x\mapsto \sqrt{x}$ on $[0,1]$ is absolutely continuous but not Lipschitz; the Cantor function is continuous, increasing and of bounded variation but not absolutely continuous; and $x\mapsto x\sin (1\mathrm{/}x)$ on $(0,1]$, extended by $0$, is continuous but not of bounded variation. $AC[a,b]$ is a vector space, closed under products, and $F\in AC[a,b]$ has Luzin's property (N): it maps null sets to null sets.

Partly proved elsewhere in this library. The elementary definition is Absolute continuity on a compact interval ↗. The hierarchy $C^1\subseteq \mathrm{L}\mathrm{i}\mathrm{p}\mathrm{s}\mathrm{c}\mathrm{h}\mathrm{i}\mathrm{t}\mathrm{z}\subseteq AC\subseteq C\cap BV$ and strictness witnesses are proved in $C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation ↗ and its companion examples. Closure under vector-space operations and products, Luzin's property (N), the sharp Lebesgue-integral FTC and the Banach–Zarecki characterisation remain unproved.

What would prove it. Absolute continuity implies bounded variation by a covering argument on $[a,b]$; property (N) follows from the definition applied to a cover of the null set by intervals of small total length; strictness of the inclusion at the Cantor function needs that $c$ maps a null set onto a set of measure $1$, and that is a genuine measure statement. The real theorem about the class is The sharp fundamental theorem of calculus (absolute continuity) ‡, and its characterisation without integrals is Banach-Zarecki theorem ‡.

Which page it serves. The bounded variation and Riemann-Stieltjes page, which builds the theory of $BV[a,b]$, the Jordan decomposition into a difference of increasing functions, and the total variation function. Absolute continuity is the next class in that hierarchy and the one for which the Newton-Leibniz formula is exactly true, so the page can define it, place it in the hierarchy with the counterexamples above, and then must stop.

What is in scope on that page. The whole $BV$ theory, the Jordan decomposition, the Cantor function as a continuous increasing non-constant function with derivative zero off a null set, and the elementary statement that absolutely continuous implies uniformly continuous and of bounded variation. What is not in scope is any statement whose formulation needs the Lebesgue integral.

Let $F:[a,b]\to \mathbb{R}$. The following are equivalent.

1. $F$ is absolutely continuous on $[a,b]$.
2. $F$ is differentiable almost everywhere on $[a,b]$, $F^{\mathrm{\prime }}\in L^1[a,b]$, and

$F(x)=F(a)+{\int }_a^x{F}^{\mathrm{\prime }}d\lambda \text{for every }x\in [a,b]\mathrm{.}$

3. There exists $g\in L^1[a,b]$ with $F(x)=F(a)+{\int }_a^{x}gd\lambda$ for every $x\in [a,b]$; and then $g=F^{\mathrm{\prime }}$ almost everywhere.

In particular, for $F\in AC[a,b]$ the Newton-Leibniz formula $F(b)-F(a)={\int }_a^b{F}^{\mathrm{\prime }}d\lambda$ holds, and $AC[a,b]$ is exactly the class of functions for which it holds in this sense.

The identity must be required at every $x$, not only at $x=b$. The endpoint identity alone does not characterise absolute continuity. Let $c$ be the Cantor function and set $F(x)=c(2x)$ on $[0,1\mathrm{/}2]$ and $F(x)=c(2-2x)$ on $[1\mathrm{/}2,1]$. Then $F$ is continuous, $F^{\mathrm{\prime }}=0$ almost everywhere, $F^{\mathrm{\prime }}\in L^1$, and

${\int }_0^1{F}^{\mathrm{\prime }}d\lambda =0=F(1)-F(0),$

yet $F$ is not absolutely continuous, since it is not constant while carrying all its variation on a null set. The identity fails at $x=1\mathrm{/}2$, where the left side is $1$ and the right side is $0$.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. The implication from 3 to 1 is absolute continuity of the indefinite Lebesgue integral, which follows from the dominated convergence theorem (Dominated convergence theorem ‡). The implication from 3 to $g=F^{\mathrm{\prime }}$ almost everywhere is the Lebesgue differentiation theorem (Lebesgue differentiation theorem for $L^1$ functions ‡). The hard direction, from 1 to 2, uses that $F\in AC$ is of bounded variation, hence differentiable almost everywhere by Lebesgue's differentiation theorem for monotone functions ‡, that $F^{\mathrm{\prime }}\in L^1$ with ${\int }_a^x{F}^{\mathrm{\prime }}\le F(x)-F(a)$ for increasing $F$, and then a Vitali covering argument to upgrade the inequality to equality using the $\delta$ from absolute continuity. Every step is measure-theoretic.

Which page it serves. This is the natural endpoint of the fundamental theorems of calculus page, and the reason that page's results are called the working FTC rather than the FTC. That page proves: if $f$ is Riemann integrable on $[a,b]$ and $F$ is any antiderivative of $f$ on $[a,b]$, then ${\int }_a^{b}f=F(b)-F(a)$; and if $f$ is continuous then $x\mapsto {\int }_a^{x}f$ is an antiderivative. Both statements carry hypotheses that the theorem above deletes. The library states the sharp version here so that no reader concludes the working FTC is the last word, and so that the counterexamples on that page (a derivative that is not Riemann integrable, the Cantor function, Volterra's function) have a stated theorem to be counterexamples to.

What this page's other items add. Banach-Zarecki theorem ‡ characterises the same class without mentioning an integral at all, and Henstock-Kurzweil versus Lebesgue: $f$ is Lebesgue integrable iff $f$ and $\mathrm{\mid }f\mathrm{\mid }$ are both HK integrable ‡ records the integral for which the Newton-Leibniz formula holds for every everywhere-differentiable $F$, with no integrability hypothesis on $F^{\mathrm{\prime }}$ whatsoever.

Let $F:[a,b]\to \mathbb{R}$. Then $F$ is absolutely continuous on $[a,b]$ if and only if all three of the following hold:

1. $F$ is continuous on $[a,b]$;
2. $F$ is of bounded variation on $[a,b]$;
3. $F$ has Luzin's property (N): $\lambda (F(E))=0$ for every $E\subseteq [a,b]$ with $\lambda (E)=0$.

None of the three may be dropped. The Cantor function satisfies 1 and 2 and fails 3, since it maps the Cantor set, a null set, onto $[0,1]$. The function $x\mapsto x\sin (1\mathrm{/}x)$ on $(0,1]$ extended by $0$ satisfies 1 and 3 and fails 2. A jump function satisfies 2 and 3 and fails 1.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. The forward direction is a covering estimate directly from the definition of absolute continuity (Absolutely continuous functions ‡). The converse uses the Vitali covering theorem (Vitali covering theorem ‡) together with the Banach indicatrix formula, which expresses the total variation of a continuous function as the integral over $\mathbb{R}$ of the number of preimages, and hence needs the Lebesgue integral (Lebesgue measure and the Lebesgue integral ‡).

Which page it serves. The bounded variation and Riemann-Stieltjes page. That page has both the continuity and the bounded variation hypotheses available as proved notions, and property (N) can be stated with the elementary covering notion of a null set, so the statement is fully intelligible there. Only the proof is out of reach.

Why it is worth recording even unproved. It is the answer to the question that the sharp fundamental theorem of calculus (The sharp fundamental theorem of calculus (absolute continuity) ‡) leaves open, namely what absolute continuity is intrinsically, with no integral in sight. The pair of statements together explains the Cantor function completely: it fails Newton-Leibniz because it fails property (N), and it fails property (N) because it moves a null set onto a set of full measure.

Let $(X,\mathcal{A},\mu )$ be a measure space with $\mu (X)<\mathrm{\infty }$, and let $f_n,f:X\to \mathbb{R}$ be measurable with $f_n\to f$ pointwise almost everywhere. Then for every $\epsilon >0$ there is a measurable $E\subseteq X$ with

$\mu (X\setminus E)<\epsilon \text{and}{f}_n\to f\text{ uniformly on }E\mathrm{.}$

The finiteness of $\mu (X)$ is essential: on $\mathbb{R}$ with Lebesgue measure, $f_n={\mathbf{1}}_{[n,n+1]}\to 0$ pointwise, but on any set whose complement has finite measure the convergence is not uniform. The conclusion cannot be improved to $\epsilon =0$: convergence is in general not uniform off a null set, as $f_n=x^n$ on $[0,1]$ shows.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. Countable subadditivity and continuity from above of a finite measure, applied to the sets $E_{n,k}={\bigcup }_{m\ge n}\{\mathrm{\mid }{f}_m-f\mathrm{\mid }\ge 1\mathrm{/}k\}$: each decreases in $n$ to a null set, so some $E_{n_k,k}$ has measure below $\epsilon 2^{-k}$, and the complement of their union works. The argument is short but is entirely about the measure, which this library does not have (Lebesgue measure and the Lebesgue integral ‡). Continuity from above is where $\mu (X)<\mathrm{\infty }$ is spent.

Which page it serves. The uniform convergence page. That page proves what uniform convergence buys, and it can exhibit sequences that converge pointwise but not uniformly; Egorov is the theorem that says the failure is always confined to a small set, and it is the standard bridge from pointwise hypotheses to uniform conclusions. It is also the usual route to Lusin's theorem (Lusin's theorem ‡).

Let $E\subseteq {\mathbb{R}}^n$ be measurable with ${\lambda }_n(E)<\mathrm{\infty }$ and let $f:E\to \mathbb{R}$ be measurable. Then for every $\epsilon >0$ there is a closed set $K\subseteq E$ with

${\lambda }_n(E\setminus K)<\epsilon \text{and}f{\mathrm{\mid }}_K\text{ continuous}\mathrm{.}$

$K$ may be taken compact when $E$ is bounded, and $f{\mathrm{\mid }}_K$ extends to a continuous function on all of ${\mathbb{R}}^n$ by Tietze. The assertion is about the restriction $f{\mathrm{\mid }}_K$ and not about continuity of $f$ at the points of $K$: the Dirichlet function ${\mathbf{1}}_{\mathbb{Q}}$ is nowhere continuous on $\mathbb{R}$, yet its restriction to the closed set $\mathbb{R}\setminus U$, where $U$ is an open set of measure below $\epsilon$ containing $\mathbb{Q}$, is identically zero and so continuous.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. Regularity of Lebesgue measure, which supplies closed sets from inside and open sets from outside, applied to the preimages of a countable base of intervals, plus Egorov's theorem (Egorov's theorem ‡) in the version where a measurable function is an almost everywhere limit of simple functions. Both ingredients are measure-theoretic (Lebesgue measure and the Lebesgue integral ‡).

Which page it serves. The continuity page and the uniform convergence page, where the question "how badly can a function fail to be continuous" is answered only for specific examples (Dirichlet, Thomae, Volterra). Lusin's theorem is the general answer: measurability is exactly continuity after deleting a set of arbitrarily small measure, which is Littlewood's second principle. It also belongs beside the Riesz-Markov-Kakutani theorem (Riesz-Markov-Kakutani representation theorem ‡), since both express that continuous functions are dense in the measurable world.

Let $(X,\mathcal{A},\mu )$ be a measure space and for measurable $f$ put $\mathrm{\parallel }f{\mathrm{\parallel }}_p:=({\int }_X\mathrm{\mid }f{\mathrm{\mid }}^{p}d\mu {)}^{1\mathrm{/}p}$ for $1\le p<\mathrm{\infty }$, and $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}:={\mathrm{ess\hspace{0.17em}sup}}_X\mathrm{\mid }f\mathrm{\mid }$.

Holder. If $1\le p,q\le \mathrm{\infty }$ with $\frac{1}{p}+\frac{1}{q}=1$, then for all measurable $f,g$,

${\int }_X\mathrm{\mid }fg\mathrm{\mid }d\mu \le \mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }g{\mathrm{\parallel }}_q\mathrm{.}$

For $1<p<\mathrm{\infty }$ equality holds exactly when $\mathrm{\mid }f{\mathrm{\mid }}^p$ and $\mathrm{\mid }g{\mathrm{\mid }}^q$ are proportional almost everywhere. The case $p=q=2$ is the Cauchy-Schwarz inequality for the integral.

Minkowski. If $1\le p\le \mathrm{\infty }$, then for all measurable $f,g$,

$\mathrm{\parallel }f+g{\mathrm{\parallel }}_p\le \mathrm{\parallel }f{\mathrm{\parallel }}_p+\mathrm{\parallel }g{\mathrm{\parallel }}_p\mathrm{.}$

Hence $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_p$ is a seminorm on the measurable functions with finite $p$-th moment, and a norm on the quotient by equality almost everywhere. For $0<p<1$ Minkowski's inequality reverses on nonnegative functions and $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_p$ is not a norm.

Not proved in this library. The inequalities are recorded here in their integral form only, and used in no proof here.

What would prove it. Young's inequality $ab\le a^p\mathrm{/}p+b^q\mathrm{/}q$, or the concavity of the logarithm, plus the monotonicity and additivity of the Lebesgue integral (Lebesgue measure and the Lebesgue integral ‡). No convergence theorem is needed; the proofs are the finite-sum proofs with sums replaced by integrals.

Which page it serves. The roots and rational powers page already proves the weighted arithmetic-geometric mean inequality and the finite forms of Holder and Minkowski for finite sequences of reals with rational exponents, and the ${\mathbb{R}}^n$ as a normed space page uses those to get the $p$-norms on ${\mathbb{R}}^n$. The integral form is the same statement for the counting measure replaced by an arbitrary measure, and it is exactly the step that needs the integral.

What is genuinely missing here, and what is not. The inequality itself is not deep and its finite version is in scope. What is deferred is the setting: the statement quantifies over measurable functions and uses $\int$ and $\mathrm{ess\hspace{0.17em}sup}$, none of which this library defines. The moment the integral exists, these two lines follow at once, which is why they are recorded together rather than as separate results.

Let $(X,\mathcal{A},\mu )$ be a measure space and $1\le p\le \mathrm{\infty }$. Write $L^p(\mu )$ for the space of measurable $f$ with $\mathrm{\parallel }f{\mathrm{\parallel }}_p<\mathrm{\infty }$, modulo equality almost everywhere.

Riesz-Fischer theorem. $(L^p(\mu ),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_p)$ is a Banach space: every $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_p$-Cauchy sequence converges in $L^p(\mu )$. Moreover, if $f_n\to f$ in $L^p(\mu )$ then some subsequence converges to $f$ pointwise almost everywhere.

For $p=2$ this makes $L^2(\mu )$ a Hilbert space, and the classical 1907 form of the theorem is the surjectivity half of the correspondence with ${\mathrm{\ell }}^2$: for an orthonormal system $(e_n)$ in $L^2$ and any $(c_n)\in {\mathrm{\ell }}^2$ the series ${\sum }_n{c}_n{e}_n$ converges in $L^2$, so every square-summable sequence is the sequence of Fourier coefficients of an $L^2$ function.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. Minkowski's inequality (Holder and Minkowski inequalities in integral form ‡) to know $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_p$ is a norm, then the standard criterion that a normed space is complete when every absolutely convergent series converges, applied through the monotone convergence theorem (Monotone convergence theorem (Beppo Levi) ‡) to ${\sum }_n\mathrm{\parallel }{f}_{n+1}-f_n{\mathrm{\parallel }}_p<\mathrm{\infty }$, with the dominated convergence theorem (Dominated convergence theorem ‡) identifying the limit. The subsequence statement falls out of the same construction. The case $p=\mathrm{\infty }$ is separate and easier, since a countable union of null sets is again null.

Which page it serves. The completeness page, which proves $\mathbb{R}$ and ${\mathbb{R}}^n$ complete and the bounded functions with the sup metric complete, and the function-space page, which proves $C([0,1],\mathbb{R})$ complete for the uniform metric. The $L^1$ metric on $C[a,b]$ is not built here at all, which is the honest motivation for the Lebesgue theory: the completion of $C[a,b]$ under the $L^1$ metric exists abstractly, and the theorem above says it is a space of functions, namely $L^1[a,b]$. Without measure theory the completion stays an abstract object with no description.

Naming. The name covers both the completeness theorem for $L^p$ and the 1907 result about Fourier coefficients in $L^2$, which were proved independently by F. Riesz and E. Fischer. This library keeps both under this id because they are the same theorem in the case $p=2$.

Let $(X,\mathcal{A},\mu )$ be a $\sigma$-finite measure space whose $\sigma$-algebra is countably generated modulo null sets, for instance ${\mathbb{R}}^n$ with Lebesgue measure.

Separability for finite $p$. For $1\le p<\mathrm{\infty }$ the space $L^p(\mu )$ is separable: it has a countable dense subset. On ${\mathbb{R}}^n$ one may take finite rational linear combinations of indicators of boxes with rational vertices, or the continuous functions of compact support with rational data.

Failure at $p=\mathrm{\infty }$. $L^{\mathrm{\infty }}[0,1]$ is not separable. The family $\{{\mathbf{1}}_{[0,t]}:t\in [0,1]\}$ is uncountable and satisfies $\mathrm{\parallel }{\mathbf{1}}_{[0,s]}-{\mathbf{1}}_{[0,t]}{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ for $s\mathrm{\ne }t$, so the open balls of radius $\frac{1}{2}$ around its members are pairwise disjoint and any dense set must meet each of them. The same argument shows ${\mathrm{\ell }}^{\mathrm{\infty }}$ is not separable.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. For the positive half: approximation of an $L^p$ function by simple functions, then of a measurable set by a finite union of boxes up to small measure, then of the coefficients by rationals, with the three errors summed by Minkowski's inequality (Holder and Minkowski inequalities in integral form ‡). For the negative half: nothing beyond the displayed computation, once $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\mathrm{\infty }}$ and the notion of a countable set (Finite, countably infinite, countable, uncountable ↗) are available.

Which page it serves. Separability is defined later in Separability: the existence of an at most countable dense subset ↗. The pair of statements above is the standard first example of a natural Banach space that is not separable, and it is also why $L^{\mathrm{\infty }}$ behaves differently from every $L^p$ with $p$ finite. It is not separable, so it has no countable dense subset to run approximation arguments on, and the duality stops at it. For $\sigma$-finite $\mu$ the space $L^{\mathrm{\infty }}(\mu )$ is itself the dual of $L^1(\mu )$; what has no counterpart is the return trip, since the dual of $L^{\mathrm{\infty }}(\mu )$ is not $L^1(\mu )$ but a space of bounded finitely additive set functions. So the identification $L^p(\mu {)}^{\ast }=L^q(\mu )$ that holds for $1<p<\mathrm{\infty }$ stops here.

Hypotheses worth stating carefully. Separability of $L^p$ is a property of the measure space and not of $p$ alone: $L^p(\mu )$ for the counting measure on an uncountable set is not separable for any $p$, since the indicators of singletons are pairwise at distance $2^{1\mathrm{/}p}$. The clean statement is the one above, with $\sigma$-finiteness and a countably generated $\sigma$-algebra as hypotheses.

Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be $\sigma$-finite measure spaces and let $f$ be measurable for the product $\sigma$-algebra $\mathcal{A}\otimes \mathcal{B}$.

Tonelli. If $f\ge 0$ then $x\mapsto {\int }_{Y}f(x,y)d\nu (y)$ and $y\mapsto {\int }_{X}f(x,y)d\mu (x)$ are measurable and

${\int }_X\text{\hspace{0.17em}⁣}\left({\int }_{Y}fd\nu \right)d\mu ={\int }_Y\text{\hspace{0.17em}⁣}\left({\int }_{X}fd\mu \right)d\nu ={\int }_{X\times Y}fd(\mu \otimes \nu ),$

all three possibly $+\mathrm{\infty }$.

Fubini. If $f\in L^1(\mu \otimes \nu )$ then $f(x,\cdot )\in L^1(\nu )$ for $\mu$-almost every $x$, $f(\cdot ,y)\in L^1(\mu )$ for $\nu$-almost every $y$, the two almost everywhere defined iterated integrals are integrable, and the same chain of equalities holds.

In practice the two are used together: Tonelli applied to $\mathrm{\mid }f\mathrm{\mid }$ establishes the integrability hypothesis of Fubini, which is then applied to $f$. Neither theorem asserts that equality of the two iterated integrals implies integrability, and neither dispenses with $\sigma$-finiteness.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. Construction of the product measure by Caratheodory extension from measurable rectangles, uniqueness of that extension on $\sigma$-finite spaces by a Dynkin system argument, the fact that sections of a product-measurable set are measurable, and then the monotone convergence theorem (Monotone convergence theorem (Beppo Levi) ‡) to pass from indicators to simple functions to nonnegative measurable functions. $\sigma$-finiteness is used twice: once to make the product measure unique, once to make the section function measurable.

Which page it serves. The Fubini and change of variables page of the multivariable track. That page proves the Fubini theorem for the Riemann integral on a box, where the hypothesis is continuity or Riemann integrability of $f$ together with existence of the inner integrals, and where the counterexamples are about existence rather than about measurability. The Lebesgue version is what makes the theorem usable for the functions that actually arise, and its two failure modes are recorded here as Failure of Tonelli without $\sigma$-finiteness: the diagonal under Lebesgue times counting measure ‡ and Sierpinski's example under the continuum hypothesis ‡.

Two hypotheses that are often dropped and should not be. The first is $\sigma$-finiteness, whose failure is the diagonal example. The second is product measurability of $f$: equality of iterated integrals for a function that is not measurable on the product is not asserted by anything above, and Sierpinski's example shows the iterated integrals can then exist and differ.

Let $X=[0,1]$ with Lebesgue measure $\lambda$ and let $Y=[0,1]$ with counting measure $\kappa$, so that $\kappa (A)$ is the number of elements of $A$ when $A$ is finite and $+\mathrm{\infty }$ otherwise. Let

$f={\mathbf{1}}_{\mathrm{\Delta }},\mathrm{\Delta }=\{(x,y)\in [0,1{]}^2:x=y\},$

the indicator of the diagonal. Then $f$ is nonnegative and measurable for the product $\sigma$-algebra, and

${\int }_X({\int }_{Y}f(x,y)d\kappa (y))d\lambda (x)={\int }_0^{1}1d\lambda =1,$

${\int }_Y({\int }_{X}f(x,y)d\lambda (x))d\kappa (y)={\int }_{Y}0d\kappa =0.$

The two iterated integrals are $1$ and $0$. Tonelli's theorem does not apply because $\kappa$ is not $\sigma$-finite on the uncountable set $[0,1]$, and no hypothesis on $f$ can repair this: $f$ is an indicator of a closed set, as good as a function can be.

Not proved in this library. The computation is a two-line consequence of the definitions of the two integrals, but both integrals belong to the deferred measure track (Lebesgue measure and the Lebesgue integral ‡), so it is recorded here rather than proved.

What would prove it. Only the definitions: the inner integral against counting measure of ${\mathbf{1}}_{\{x\}}$ is $1$ for each fixed $x$, and the inner integral against Lebesgue measure of the indicator of a single point is $0$ for each fixed $y$. The measurability of $\mathrm{\Delta }$ in the product $\sigma$-algebra is the one point needing care, and it follows since $\mathrm{\Delta }$ is closed and the product $\sigma$-algebra here contains the Borel sets of the square.

Which page it serves. The Fubini and change of variables page, as the counterexample that shows why the Lebesgue statement carries a $\sigma$-finiteness hypothesis while the Riemann statement on a box does not need one. It is the cheapest possible witness: no choice, no pathological set, no continuum hypothesis, just a measure that is too large.

Contrast with the other failure recorded here. In this example $f$ is perfectly measurable and the measures are at fault. In Sierpinski's example under the continuum hypothesis ‡ the measures are the best possible and the function is at fault. Between them they show that both hypotheses of Fubini-Tonelli theorem and the $\sigma$-finiteness hypothesis ‡ are needed.

Assume the continuum hypothesis. Then there is a set $E\subseteq [0,1{]}^2$ such that

• for every $x\in [0,1]$ the vertical section $E_x=\{y:(x,y)\in E\}$ has countable complement in $[0,1]$, and
• for every $y\in [0,1]$ the horizontal section $E^y=\{x:(x,y)\in E\}$ is countable.

Consequently every section is Lebesgue measurable, $\lambda (E_x)=1$ for every $x$ and $\lambda (E^y)=0$ for every $y$, so both iterated integrals of ${\mathbf{1}}_E$ exist and

${\int }_0^1\text{\hspace{0.17em}⁣}({\int }_0^1{\mathbf{1}}_E(x,y)dy)dx=1,{\int }_0^1\text{\hspace{0.17em}⁣}({\int }_0^1{\mathbf{1}}_E(x,y)dx)dy=0.$

By Fubini's theorem $E$ is therefore not measurable in $[0,1{]}^2$. Sierpinski's construction uses a well ordering of $[0,1]$ in order type ${\omega }_1$, which is where the continuum hypothesis is spent, and sets $E=\{(x,y):y⪯̸x\}$ for that well ordering $⪯$.

Not proved in this library. It is recorded with a citation to Sierpinski's 1920 paper and used in no proof here.

What would prove it. A well ordering of $[0,1]$ of order type ${\omega }_1$, which under the continuum hypothesis (The continuum hypothesis, and what this page does not prove ↗) exists by the well ordering theorem, plus the observation that each initial segment of such a well ordering is countable. The measure theory needed is then only that countable sets are null and that Fubini's theorem (Fubini-Tonelli theorem and the $\sigma$-finiteness hypothesis ‡) forbids the displayed pair of values for a measurable set.

Which page it serves. The Fubini and change of variables page, beside Failure of Tonelli without $\sigma$-finiteness: the diagonal under Lebesgue times counting measure ‡. The two together answer the question "which hypothesis of Fubini's theorem is doing the work": the diagonal example kills $\sigma$-finiteness, this one kills the idea that measurability of every section is enough.

What it costs, exactly. More than ZFC: the statement is conditional on the continuum hypothesis, which is independent of ZFC, so this is not a ZFC counterexample but a consistency result. The dependence is an equivalence and not merely a sufficient condition. A set $E$ with the two section properties above splits the square into $E$, whose horizontal sections are countable, and its complement, whose vertical sections are countable; and by Sierpinski's decomposition theorem such a splitting exists, for the square or equally for the plane, if and only if the continuum hypothesis holds. So under the negation of the continuum hypothesis this construction is unavailable outright.

What is not thereby restored. The conclusion, unlike the construction, does not go away. Martin's axiom also implies that there is a function on the unit square whose two iterated integrals are defined and unequal, so Martin's axiom together with the negation of the continuum hypothesis still produces one, by a different route. What is consistent with ZFC, by a theorem of Friedman (1980), is the opposite: there are models of ZFC in which no such function exists at all, so that whenever both iterated integrals of a function on the unit square exist they agree. The strong Fubini statement for non-measurable functions is therefore independent of ZFC, and this item is one half of that independence. The ordinals and transfinite recursion machinery the construction uses is itself only ordinary ZFC, and this library plans that page; what is deferred is the independence apparatus and the measure theory, not the well ordering.

Let $V\subseteq [0,1]$ contain exactly one element of each coset of $\mathbb{Q}$ in $\mathbb{R}$, that is, one representative of each class of the equivalence relation $x\sim y⟺x-y\in \mathbb{Q}$. Such a $V$ exists by the Axiom of Choice. Then $V$ is not Lebesgue measurable.

Indeed, enumerate the rationals of $[-1,1]$ as $q_1,q_2,\dots$ and put $V_k=V+q_k$. The $V_k$ are pairwise disjoint by the choice of one representative per class, and

$[0,1]\subseteq {\bigcup }_{k\ge 1}{V}_k\subseteq [-1,2]\mathrm{.}$

If $V$ were measurable then so would each $V_k$ be, with $\lambda (V_k)=\lambda (V)$ by translation invariance, and countable additivity would give $1\le {\sum }_k\lambda (V)\le 3$. The sum is $0$ if $\lambda (V)=0$ and $+\mathrm{\infty }$ otherwise, so neither case is possible.

The construction needs more than ZF + DC. By Solovay's theorem, if ZFC together with "there exists an inaccessible cardinal" is consistent, then there is a model of ZF + DC in which every subset of $\mathbb{R}$ is Lebesgue measurable. On that hypothesis ZF + DC does not prove that a Vitali set exists. The large-cardinal assumption is part of the claim and not a technicality: it is what Solovay's construction consumes, and by Shelah it cannot be removed.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. The countable additivity and translation invariance of Lebesgue measure (Lebesgue measure and the Lebesgue integral ‡), plus a choice function on the family of cosets (The Axiom of Choice ↗). The argument itself is three lines; the measure it argues about is what this library lacks.

Which page it serves. The order, Zorn and the Axiom of Choice page and its examples page, where the cost of choice is tracked result by result, and the Cantor set, Baire and measure zero page, which shows how badly a set can behave while remaining elementary. The Vitali set is the standard answer to "why is Lebesgue measure not defined on all of $\mathcal{P}(\mathbb{R})$", and it cannot be stated at all until a measure exists.

Where it sits in the choice ledger. It needs a choice principle strictly beyond dependent choice, which is a strong statement in a library that otherwise tracks choice carefully (The choice ledger: what costs the Axiom of Choice and what does not ↗). The relevant facts, all external here, are Solovay's model of ZF + DC in which every set of reals is Lebesgue measurable and has the Baire property, and Shelah's refinement, which shows that the inaccessible cardinal is genuinely needed for the measurability half though not for the Baire property half. Weaker principles than full choice already suffice: the Boolean prime ideal theorem yields a non-measurable set, by way of the Hahn-Banach theorem.

Identify a subset of $\mathbb{N}$ with its characteristic function, so that $\mathcal{P}(\mathbb{N})$ becomes the Cantor space $\{0,1{\}}^{\mathbb{N}}$ with its product topology and its uniform product measure; equivalently, transport it to $[0,1]$ by binary expansions and use Lebesgue measure. Then:

no free ultrafilter on $\mathbb{N}$, read as a subset of $\{0,1{\}}^{\mathbb{N}}$, is measurable, and none has the Baire property.

The non-measurability is Sierpiński (1938), where it appears as the observation that a free ultrafilter yields a non-measurable set and hence a non-measurable additive function. The category form, that no such set has the Baire property, is the standard analogue and is recorded in the same place in the literature on weak choice principles.

The consequence usually wanted is negative: a free ultrafilter can never be exhibited by a construction that produces only measurable sets, or only sets with the Baire property. It is a precise sense in which such an object cannot be written down.

• Not proved in this library. The statement needs the product measure on $\{0,1{\}}^{\mathbb{N}}$, or Lebesgue measure on $[0,1]$, and a specialised topological zero-one law on a Polish space. The library now has the general Baire/category background, but not that zero-one-law argument or the measure and integration track.

• What would prove it. A free ultrafilter $\mathcal{U}$ is unchanged by altering finitely many coordinates, since it contains every cofinite set, so it is a tail event: the Kolmogorov zero-one law would force its measure to be $0$ or $1$, and the topological zero-one law would force it to be meagre or comeagre. Complementation $A\mapsto \mathbb{N}\setminus A$ is a measure-preserving homeomorphism of $\{0,1{\}}^{\mathbb{N}}$ that carries $\mathcal{U}$ exactly onto its own complement, because an ultrafilter contains exactly one of $A$ and $\mathbb{N}\setminus A$ (Characterisation of ultrafilters: every set or its complement ↗). A measurable $\mathcal{U}$ would therefore have measure $\frac{1}{2}$, and a $\mathcal{U}$ with the Baire property would be neither meagre nor comeagre; both contradict the zero-one laws. The two zero-one laws are the missing machinery, and they belong to the measure-theory and Baire-category tracks.

• Why it matters elsewhere. The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter ↗ produces an ultrafilter from Zorn's lemma with no description of it, and FALSE, once the ultrafilter lemma is available: every ultrafilter is principal ↗ uses that to produce a free ultrafilter on $\mathbb{N}$ and can say nothing further about it. This item is the sharp reason why nothing further can be said by the usual means: every free instance of Ultrafilter ↗ on $\mathbb{N}$ lies outside the measurable sets and outside the sets with the Baire property.

• No consistency hypothesis is needed. Unlike the independence results this library also records, this is an outright theorem: it says of any free ultrafilter that exists that it is non-measurable, and is vacuously true in a model with none.

Let $B\subseteq {\mathbb{R}}^3$ be a closed ball. There are a partition of $B$ into finitely many pieces $B=A_1\cup \cdots \cup A_n$ and rigid motions $g_1,\dots ,g_n$ of ${\mathbb{R}}^3$ such that $g_1(A_1),\dots ,g_n(A_n)$ partition two disjoint balls each congruent to $B$. Five pieces suffice, and four is impossible. More generally, any two bounded subsets of ${\mathbb{R}}^3$ with nonempty interior are equidecomposable.

The pieces are necessarily non-measurable, so the statement contradicts no theorem about volume; what it refutes is the existence of a finitely additive, isometry-invariant extension of Lebesgue measure to all subsets of ${\mathbb{R}}^3$. In ${\mathbb{R}}^1$ and ${\mathbb{R}}^2$ such extensions do exist, by a Banach construction using the amenability of the isometry groups there, so the paradox is a fact about $n\ge 3$, where the rotation group contains a free subgroup of rank two.

It needs choice beyond ZF + DC. If ZFC together with "there exists an inaccessible cardinal" is consistent, then ZF + DC does not prove the paradox: in Solovay's model of ZF + DC every set of reals is Lebesgue measurable, and the same holds in ${\mathbb{R}}^3$, while the pieces above are necessarily non-measurable, so no such decomposition exists there. That model is built by collapsing an inaccessible, so the conclusion drawn from it is conditional on that large-cardinal hypothesis and not merely on the consistency of ZF. The paradox does not need the full Axiom of Choice either: it follows from the Hahn-Banach theorem, hence from the Boolean prime ideal theorem, which the Axiom of Choice implies outright and which, if ZF is consistent, does not imply the Axiom of Choice in return.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. The free group of rank two inside $SO(3)$, its paradoxical decomposition, transfer of that decomposition to the sphere by choosing one point from each orbit (The Axiom of Choice ↗), the handling of the countably many fixed points, and the Banach-Schroder-Bernstein theorem for equidecomposability. The group theory is elementary and reachable; the statement is deferred because it is about measure, and because its interest is precisely that the pieces are not measurable (Lebesgue measure and the Lebesgue integral ‡, The Vitali set: a non-measurable subset of $\mathbb{R}$ ‡).

Which page it serves. The order, Zorn and the Axiom of Choice examples page, as the most dramatic consequence of choice, and any future measure page as the reason finite additivity plus isometry invariance cannot be had on all sets in dimension three. It is also the standard corrective to the belief that choice is harmless: it is a theorem of ZFC and cannot be blamed on a defect of the measure.

A frequent misstatement. The pieces are not "infinitely thin" or "of measure zero": they have no measure at all, and there are finitely many of them. The word paradox records a conflict with intuition, not an inconsistency.

Let $f:[a,b]\to \mathbb{R}$. Then $f$ is Lebesgue integrable on $[a,b]$ if and only if both $f$ and $\mathrm{\mid }f\mathrm{\mid }$ are Henstock-Kurzweil integrable on $[a,b]$, and in that case the two integrals of $f$ agree.

Equivalently: the Henstock-Kurzweil integral is a non-absolute integral, and $L^1[a,b]$ is exactly its absolutely integrable part. The inclusion is strict. The function $F(x)=x^2\sin (1\mathrm{/}{x}^2)$ for $x\mathrm{\ne }0$, $F(0)=0$, is differentiable everywhere on $[0,1]$ and $F^{\mathrm{\prime }}$ is HK integrable with ${\int }_0^1{F}^{\mathrm{\prime }}=F(1)-F(0)$, but $\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$ is not integrable in any sense, so $F^{\mathrm{\prime }}$ is not Lebesgue integrable. This is the point of the HK integral: it integrates every derivative, and the Newton-Leibniz formula ${\int }_a^b{F}^{\mathrm{\prime }}=F(b)-F(a)$ holds for every everywhere-differentiable $F$, with no hypothesis on $F^{\mathrm{\prime }}$ at all.

Not proved in this library. The comparison is recorded here; the HK integral itself is not deferred and is planned as ordinary content.

What would prove it. In one direction, a Lebesgue integrable $f$ is HK integrable with the same integral, and so is $\mathrm{\mid }f\mathrm{\mid }$, by the Vitali covering argument that produces gauges from measurable approximations. In the other, if $f$ and $\mathrm{\mid }f\mathrm{\mid }$ are both HK integrable then the indefinite HK integral of $\mathrm{\mid }f\mathrm{\mid }$ is absolutely continuous and monotone, and its derivative recovers $\mathrm{\mid }f\mathrm{\mid }$ almost everywhere, so $f\in L^1$. Both directions quantify over Lebesgue integrability (Lebesgue measure and the Lebesgue integral ‡) and use the differentiation theory (Lebesgue's differentiation theorem for monotone functions ‡, The sharp fundamental theorem of calculus (absolute continuity) ‡), which is why only the comparison is deferred.

Which page it serves. A Henstock-Kurzweil page in the integration track, which this library intends to build: the gauge integral needs only tagged partitions, a gauge $\delta :[a,b]\to (0,\mathrm{\infty })$, and Cousin's lemma, all of which are elementary and in scope. That page can prove the full Newton-Leibniz theorem for the HK integral, and then must record here what its relationship to $L^1$ is.

Why the comparison is the deferred part. The theorem is a statement about two integrals, one of which does not exist in this library. Stating it as a theorem would require the Lebesgue integral in the hypothesis and in the conclusion. The HK side loses nothing by the deferral: the improper integrals page and the fundamental theorems of calculus page can both use the gauge integral without mentioning measure at all.

There is a continuous $2\pi$-periodic function $f:\mathbb{R}\to \mathbb{R}$ whose Fourier series diverges at a point: writing

$S_{N}f(x)={\sum }_{\mathrm{\mid }n\mathrm{\mid }\le N}\widehat{f}(n)e^{inx},\widehat{f}(n)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(t)e^{-int}dt,$

there is $x_0$ with ${\mathrm{lim\hspace{0.17em}sup}}_N\mathrm{\mid }{S}_{N}f(x_0)\mathrm{\mid }=+\mathrm{\infty }$. du Bois-Reymond gave the first such example in 1873. The set of points of divergence can be taken to be any prescribed set of measure zero, and by Carleson's theorem it can be no larger than that: the Fourier series of a continuous, indeed of any $L^2$, function converges almost everywhere.

Not proved in this library, for now. Unlike most of this page, this result is probably reachable here, and it is recorded rather than proved only because the page that would carry it has not been written.

What would prove it. Two routes. The explicit one is du Bois-Reymond's lacunary construction, a sum of blocks of conjugate Dirichlet kernels with rapidly increasing frequencies, which is elementary but intricate. The soft one is the uniform boundedness principle applied to the functionals $f\mapsto S_{N}f(0)$ on $C(\mathbb{T})$, whose norms are the Lebesgue constants $\mathrm{\parallel }{D}_N{\mathrm{\parallel }}_{L^1}\sim (4\mathrm{/}{\pi }^2)\log N\to \mathrm{\infty }$; the Baire category theorem then gives a comeagre set of continuous functions whose Fourier series diverge at $0$. The Baire category theorem for complete metric spaces is in scope in this library, and uniform boundedness is flagged in the deferral list as borderline, since its proof is Baire plus linearity. So the soft route becomes available as soon as normed spaces and the Lebesgue constants exist.

Which page it serves. A Fourier series page in the analysis track, which is not yet planned, and the approximation and compactness page, where the positive results live: Fejer's theorem gives uniform convergence of the Cesaro means for every continuous function, and Weierstrass approximation follows. The correct reading of the pair is that summability, not convergence, is the right notion for continuous functions.

Attribution. The first published proof is du Bois-Reymond, Ueber die Fourierschen Reihen, Nachr. Kon. Ges. Wiss. Gottingen 21 (1873), 571-582, which is the paper the cited Wikipedia article names as the first proof that the Fourier series of a continuous function can diverge. The extended treatment is his 1876 Munich memoir, and 1876 is quoted for the result by part of the literature; the earlier date is used here because it is the date of the first publication.

Contrast. This is the continuous, pointwise, Baire-reachable failure. The $L^1$ failure recorded in Kolmogorov 1926: an $L^1$ function whose Fourier series diverges everywhere ‡ is of a different order: it is everywhere and it needs the Lebesgue theory even to state.

There is $f\in L^1(\mathbb{T})$ whose Fourier series diverges at every point: for every $x\in \mathbb{T}$,

${\mathrm{lim\hspace{0.17em}sup}}_{N\to \mathrm{\infty }}\mid S_{N}f(x)\mid =+\mathrm{\infty },S_{N}f(x)={\sum }_{\mathrm{\mid }n\mathrm{\mid }\le N}\widehat{f}(n)e^{inx}\mathrm{.}$

Kolmogorov proved almost everywhere divergence in 1923 and everywhere divergence in 1926. The result is sharp in the scale of $L^p$ spaces: by Carleson's theorem (1966) and Hunt's extension (1968), the Fourier series of a function in $L^p(\mathbb{T})$ with $1<p\le \mathrm{\infty }$ converges almost everywhere, so $p=1$ is exactly where everywhere divergence becomes possible.