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

In the setting of Measure Spaces and the Lebesgue Integral: Standing Notation, let $(X,\mathcal{F},\mu)$ be a measure space. Then the following hold.

1. (Null sets)¶ Every subset of a null set is null, and the union of the members of a sequence $(N_{m})_{m\in\mathbb{N}}$ of null sets is null. Consequently, if for each $m\in\mathbb{N}$ a property $Q_{m}$ of points of $X$ holds almost everywhere, then the set of points at which $Q_{m}$ fails for at least one $m$ is null; that is, almost every point of $X$ satisfies $Q_{m}$ for every $m$ simultaneously.

2. (Integrals over null sets)¶ Let $f:X\to[0,\infty]$ be measurable and let $N$ be a null set such that $f(x)=0$ for every $x\in X\setminus N$. Then

3. (Almost-everywhere comparison)¶ Let $u,v:X\to[0,\infty]$ be measurable. If $u(x)\le v(x)$ for almost every $x$, then $\int_{X}u\,d\mu\le\int_{X}v\,d\mu$; if $u(x)=v(x)$ for almost every $x$, then $\int_{X}u\,d\mu=\int_{X}v\,d\mu$. If $f,g:X\to\mathbb{R}$ are measurable, $f$ is integrable and $f(x)=g(x)$ for almost every $x$, then $g$ is integrable and $\int_{X}f\,d\mu=\int_{X}g\,d\mu$.

4. (Vanishing integral)¶ Let $f:X\to[0,\infty]$ be measurable. Then $\int_{X}f\,d\mu=0$ if and only if $f(x)=0$ for almost every $x$.

5. (Finiteness almost everywhere)¶ Let $f:X\to[0,\infty]$ be measurable with $\int_{X}f\,d\mu<\infty$. Then the set $\{x\in X:f(x)=\infty\}$ belongs to $\mathcal{F}$ and has measure $0$; in particular $f(x)<\infty$ for almost every $x$.

6. (Markov's inequality)¶ Let $f:X\to[0,\infty]$ be measurable and let $t$ be a positive real number. Then the set $A_{t}=\{x\in X:t\le f(x)\}$ belongs to $\mathcal{F}$ and

the product being formed in $[0,\infty]$.

7. (Dominated convergence almost everywhere)¶ Let $f$ and $f_{m}$, for $m\in\mathbb{N}$, be measurable maps from $X$ to $\mathbb{R}$, and let $g:X\to\mathbb{R}$ be integrable. Suppose that for almost every $x\in X$ the sequence $(f_{m}(x))_{m\in\mathbb{N}}$ converges to $f(x)$, and that for each $m\in\mathbb{N}$ one has $|f_{m}(x)|\le g(x)$ for almost every $x$. Then $f$ and every $f_{m}$ are integrable,