Vtyjkyuk

I'm trying to solve the following exercise.

[Cohn, 2.3.5] Let $(X, mathscrA, mu)$ be a measure space, and $f: X to [-infty, +infty]$ an $mathscrA$-measurable function whose integral exists and is not equal to $-infty$. Show that if $g: X to [-infty, +infty]$ is $mathscrA$-measurable and satisfies $f leq g$ for $mu$-almost every $x in X$, then the integral of $g$ exists and satisfies $int g , dmu geq int f , dmu$.

Here's what I did.

Suppose that $S$ and $T$ are $[0, +infty]$-valued $mathscrA$-measurable functions on $X$ such that $S leq T$ for $mu$-almost every $x$, and let $A = x : S(x) = T(x)$. Write $tildeT = mathbf1_A T +h$, where $h(x) = +infty$ on $A^c$, and else $h(x) = 0$. It's clear that $int h = 0$ (take $h_n = n mathbb1_A^c$. the functions $h_n$ are simple, $h_n uparrow h$, and thus $int h = lim (n mu(A^c)) = 0$.). Additionally, $int S leq int tildeT$, since $S leq tildeT$ everywhere and any simple function $phi$ satisfying $phi leq S$ also satisfies $phi leq tildeT$. By linearity, $int tildeT = int_A T = int T$, with the last equality due to $0 leq mathbf1_A^c T leq h$.

Now decompose $f =f^+ - f^-$and $g = g^+ - g^-$ as is standard. At once, $g^+ geq f^+$ and $g^- leq f^-$ almost everywhere hold. Thus, by the argument above, one obtains $int g^+ geq int f^+$ and $int g^- leq int f^- < +infty$ (the last inequality follows because if $int f^+ < infty$, then $int f^- < +infty$, otherwise $int f = -infty$). This implies $int g$ exists and further,
$$
int g = int g^+ - int g^- geq int f^+ - int f^- = int f.
$$
This is the desired result.

Does that look fine?