Proof of The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere

Each result cited is universally quantified over the data appearing in its own statement, and is applied here to the data named in the statement above. The claims are proved in the order 1, 2, 3, 6, 5, 4, 7; no claim is used before it has been proved. We use countable choice, to select a measurable null superset for each member of a sequence of null sets.

Throughout we use repeatedly that $\mathcal{F}$ contains the countable intersection of any sequence of its members, since $\bigcap_{m}A_{m}=X\setminus\bigcup_{m}(X\setminus A_{m})$ and $\mathcal{F}$ is closed under complements and countable unions by Sigma-Algebra and Measurable Space.

Claim 1. Let $N\subseteq N'$ with $N'$ null. By Null Set of a Measure there is $B\in\mathcal{F}$ with $N'\subseteq B$ and $\mu(B)=0$; then $N\subseteq B$, so $N$ is null. Next let $(N_{m})_{m\in\mathbb{N}}$ be a sequence of null sets, and choose for each $m$ a set $B_{m}\in\mathcal{F}$ with $N_{m}\subseteq B_{m}$ and $\mu(B_{m})=0$. Put $B=\bigcup_{m}B_{m}\in\mathcal{F}$. By claim 4 of Basic Properties of a Measure, $\mu(B)\le\sum_{m}\mu(B_{m})$, and the latter sum is $0$: its partial sums are all $0$, hence bounded above, and their least upper bound is $0$, which is the sum by the convention of Measure, Measure Space, and Probability Measure. So $\mu(B)=0$, and $\bigcup_{m}N_{m}\subseteq B$ is null.

For the consequence, let $N_{m}$ be the set of points at which $Q_{m}$ fails; it is null by hypothesis, by A Property Holding Almost Everywhere. The set of points at which $Q_{m}$ fails for at least one $m$ is $\bigcup_{m}N_{m}$, which is null by the previous paragraph.

Claim 2. Choose $B\in\mathcal{F}$ with $N\subseteq B$ and $\mu(B)=0$. Since $X\setminus B\subseteq X\setminus N$, the hypothesis gives $f(x)=0$ for every $x\in X\setminus B$. Let $s$ be a nonnegative simple function with $s(x)\le f(x)$ for every $x$, and let $s=\sum_{i=1}^{r}c_{i}\mathbf{1}_{A_{i}}$ be its standard representation as in Simple Function and Its Integral, so that the $A_{i}=s^{-1}(\{c_{i}\})$ are members of $\mathcal{F}$. Fix $i$. If $c_{i}\ne 0$ then $c_{i}>0$, because $s$ is nonnegative; for $x\in A_{i}$ we get $0<c_{i}=s(x)\le f(x)$, so $f(x)\ne 0$ and therefore $x\in B$. Thus $A_{i}\subseteq B$, and $\mu(A_{i})\le\mu(B)=0$ by claim 2 of Basic Properties of a Measure, so $c_{i}\mu(A_{i})=0$. If $c_{i}=0$ then $c_{i}\mu(A_{i})=0$ as well, by the convention $0\cdot\infty=0$ of Measure, Measure Space, and Probability Measure when $\mu(A_{i})=\infty$ and trivially otherwise. Hence $\int_{X}s\,d\mu=\sum_{i=1}^{r}c_{i}\mu(A_{i})=0$.

So every nonnegative simple minorant of $f$ has integral $0$, and the set of these integrals is $\{0\}$, whose least upper bound is $0$. By Lebesgue Integral of a Nonnegative Measurable Function, $\int_{X}f\,d\mu=0$.

Claim 3. Let $u,v:X\to[0,\infty]$ be measurable with $u\le v$ almost everywhere, let $N$ be the set where $v(x)<u(x)$, which is null, and choose $B\in\mathcal{F}$ with $N\subseteq B$ and $\mu(B)=0$.

Define $\tilde u,\tilde v,w:X\to[0,\infty]$ by

and $\tilde v$ from $v$ in the same way as $\tilde u$ from $u$. These are measurable: for a real $c<0$ each of the sets $\{\tilde u>c\}$, $\{w>c\}$ is $X$, while for $0\le c$ we have $\{\tilde u>c\}=\{u>c\}\cap(X\setminus B)$ and $\{w>c\}=\{u>c\}\cap B$, both members of $\mathcal{F}$.

By construction $u=\tilde u+w$ pointwise, and $w$ vanishes on $X\setminus B$, so $\int_{X}w\,d\mu=0$ by claim 2. By the additivity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral,

and in the same way $\int_{X}v\,d\mu=\int_{X}\tilde v\,d\mu$. Moreover $\tilde u(x)\le\tilde v(x)$ for every $x\in X$: on $B$ both values are $0$, and off $B$ the point $x$ lies outside $N$, so $u(x)\le v(x)$. The monotonicity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral gives $\int_{X}\tilde u\,d\mu\le\int_{X}\tilde v\,d\mu$, and hence $\int_{X}u\,d\mu\le\int_{X}v\,d\mu$.

If $u=v$ almost everywhere, then $u\le v$ almost everywhere and $v\le u$ almost everywhere, so each integral is at most the other and they are equal.

Finally let $f,g:X\to\mathbb{R}$ be measurable with $f$ integrable and $f=g$ almost everywhere. The set where $|f|$ and $|g|$ differ is contained in the set where $f$ and $g$ differ, hence is null by claim 1, so $|f|=|g|$ almost everywhere and $\int_{X}|g|\,d\mu=\int_{X}|f|\,d\mu<\infty$ by the part already proved; thus $g$ is integrable, by the criterion in Measure Spaces and the Lebesgue Integral: Standing Notation §integral. The positive and negative parts $f^{+},f^{-},g^{+},g^{-}$ of Integrable Function and the Lebesgue Integral are measurable, and the sets where $f^{+}$ and $g^{+}$, respectively $f^{-}$ and $g^{-}$, differ are again contained in the set where $f$ and $g$ differ. Hence $\int_{X}f^{+}\,d\mu=\int_{X}g^{+}\,d\mu$ and $\int_{X}f^{-}\,d\mu=\int_{X}g^{-}\,d\mu$, and subtracting gives $\int_{X}f\,d\mu=\int_{X}g\,d\mu$ by Integrable Function and the Lebesgue Integral.

Claim 6. We first show $A_{t}\in\mathcal{F}$. For $m\in\mathbb{N}$ the set $\{x:f(x)>t-\tfrac{1}{m}\}$ belongs to $\mathcal{F}$ by measurability, and

Indeed, if $t\le f(x)$ then $t-\tfrac1m<t\le f(x)$ for every $m$; conversely, if $f(x)<t$ then $f(x)$ is a real number and $t-f(x)$ is positive, so by The Archimedean Property of the Real Numbers there is $m$ with $\tfrac1m<t-f(x)$, that is $f(x)<t-\tfrac1m$, so $x$ is not in the intersection. Hence $A_{t}\in\mathcal{F}$.

The function $t\,\mathbf{1}_{A_{t}}$ is a nonnegative simple function, and $t\,\mathbf{1}_{A_{t}}(x)\le f(x)$ for every $x$: at $x\in A_{t}$ the left side is $t\le f(x)$, and at $x\notin A_{t}$ it is $0\le f(x)$. By the monotonicity and the homogeneity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral together with The Integral of an Indicator Function is the Measure of the Set,

Claim 5. Write $I=\int_{X}f\,d\mu$, a nonnegative real number by hypothesis, and $A=\{x\in X:f(x)=\infty\}$. As in claim 6, $A=\bigcap_{m\in\mathbb{N}}\{x:f(x)>m\}$: if $f(x)=\infty$ then $f(x)>m$ for every $m$, and if $f(x)$ is real then The Archimedean Property of the Real Numbers provides $m$ with $f(x)<m$. Hence $A\in\mathcal{F}$.

For every $m\in\mathbb{N}$ the nonnegative simple function $m\,\mathbf{1}_{A}$ satisfies $m\,\mathbf{1}_{A}\le f$ pointwise, so exactly as in claim 6 we get $m\,\mu(A)\le I$. If $\mu(A)=\infty$ this reads $\infty\le I$, contradicting $I<\infty$. If $\mu(A)$ is a positive real number $\alpha$, then $m\le I/\alpha$ for every $m\in\mathbb{N}$, contradicting The Archimedean Property of the Real Numbers. Hence $\mu(A)=0$. Since $f(x)<\infty$ for every $x\in X\setminus A$ and $A$ is a null set, $f$ is finite almost everywhere.

Claim 4. Suppose first that $f=0$ almost everywhere, and let $N$ be the null set of points where $f(x)\ne 0$. Then $f$ vanishes on $X\setminus N$, so $\int_{X}f\,d\mu=0$ by claim 2.

Conversely suppose $\int_{X}f\,d\mu=0$. For $m\in\mathbb{N}$ put $E_{m}=\{x\in X:\tfrac1m\le f(x)\}$, which belongs to $\mathcal{F}$ by claim 6 applied with the positive number $\tfrac1m$, and satisfies $\tfrac1m\mu(E_{m})\le 0$ by that claim. Were $\mu(E_{m})$ nonzero, $\tfrac1m\mu(E_{m})$ would be positive, so $\mu(E_{m})=0$. Now

since $f$ is nonnegative, so $f(x)\ne0$ means $0<f(x)$: if $f(x)=\infty$ then $x\in E_{1}$, and if $f(x)$ is a positive real then The Archimedean Property of the Real Numbers gives $m$ with $\tfrac1m<f(x)$; the reverse inclusion is immediate. By claim 4 of Basic Properties of a Measure the union has measure at most $\sum_{m}\mu(E_{m})=0$, so $\{f\ne 0\}$ is a measurable set of measure $0$ and hence null, that is, $f=0$ almost everywhere.

Claim 7. Let $N_{0}$ be the set of points at which $(f_{m}(x))_{m}$ fails to converge to $f(x)$, and for $m\in\mathbb{N}$ let $N_{m}$ be the set of points at which $|f_{m}(x)|\le g(x)$ fails; all are null by hypothesis. By claim 1 their union $N$ is null, and we choose $B\in\mathcal{F}$ with $N\subseteq B$ and $\mu(B)=0$.

Set $\tilde f_{m}=f_{m}\,\mathbf{1}_{X\setminus B}$, $\tilde f=f\,\mathbf{1}_{X\setminus B}$ and $\tilde g=g\,\mathbf{1}_{X\setminus B}$, products of real-valued maps, which are measurable by claims 1 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, the indicator being measurable because $X\setminus B\in\mathcal{F}$. For $x\in B$ all these functions take the value $0$, and for $x\notin B$ the point $x$ lies outside $N$, so $(f_{m}(x))_{m}$ converges to $f(x)$ and $|f_{m}(x)|\le g(x)$ for every $m$. Hence, for every $x\in X$, the sequence $(\tilde f_{m}(x))_{m}$ converges to $\tilde f(x)$, and $|\tilde f_{m}(x)|\le\tilde g(x)$ for every $m$.

Since $\tilde g$ and $g$ differ only on $B$, they agree almost everywhere, so $\tilde g$ is integrable by claim 3. The hypotheses of Dominated Convergence Theorem are therefore met by $(\tilde f_{m})$, $\tilde f$ and $\tilde g$, and it yields that $\tilde f$ is integrable, that $\int_{X}|\tilde f_{m}-\tilde f|\,d\mu$ converges to $0$, and that $\int_{X}\tilde f_{m}\,d\mu$ converges to $\int_{X}\tilde f\,d\mu$. Each $\tilde f_{m}$ is integrable too, since $|\tilde f_{m}|\le\tilde g$ everywhere gives $\int_{X}|\tilde f_{m}|\,d\mu\le\int_{X}\tilde g\,d\mu<\infty$ by claim 1 of Linearity and Monotonicity of the Lebesgue Integral.

Finally $f$ and $\tilde f$ agree off $B$, hence almost everywhere, and likewise $f_{m}$ and $\tilde f_{m}$; so by claim 3 the functions $f$ and every $f_{m}$ are integrable, with $\int_{X}f\,d\mu=\int_{X}\tilde f\,d\mu$ and $\int_{X}f_{m}\,d\mu=\int_{X}\tilde f_{m}\,d\mu$. The nonnegative measurable functions $|f_{m}-f|$ and $|\tilde f_{m}-\tilde f|$ agree off $B$, hence almost everywhere, so their integrals agree by claim 3. Substituting these equalities into the two limit statements obtained from Dominated Convergence Theorem gives the two assertions of the claim.