I'm trying to solve the following exercise.
[Cohn, 2.3.5] Let $(X, \mathscr{A}, \mu)$ be a measure space, and $f: X \to [-\infty, +\infty]$ an $\mathscr{A}$-measurable function whose integral exists and is not equal to $-\infty$. Show that if $g: X \to [-\infty, +\infty]$ is $\mathscr{A}$-measurable and satisfies $f \leq g$ for $\mu$-almost every $x \in X$, then the integral of $g$ exists and satisfies $\int g \, d\mu \geq \int f \, d\mu$.
Here's what I did.
Suppose that $S$ and $T$ are $[0, +\infty]$-valued $\mathscr{A}$-measurable functions on $X$ such that $S \leq T$ for $\mu$-almost every $x$, and let $A = \{x : S(x) = T(x)\}$. Write $\tilde{T} = \mathbf{1}_{A} T +h$, where $h(x) = +\infty$ on $A^c$, and else $h(x) = 0$. It's clear that $\int h = 0$ (take $h_n = n \mathbb{1}_{A^c}$. the functions $h_n$ are simple, $h_n \uparrow h$, and thus $\int h = \lim (n \mu(A^c)) = 0$.). Additionally, $\int S \leq \int \tilde{T}$, since $S \leq \tilde{T}$ everywhere and any simple function $\phi$ satisfying $\phi \leq S$ also satisfies $\phi \leq \tilde{T}$. By linearity, $\int \tilde{T} = \int_A T = \int T$, with the last equality due to $0 \leq \mathbf{1}_{A^c} T \leq h$.
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?