Proof of Monotone Convergence Theorem

Step 0 (continuity from below). Let $(B_m)_{m\in\mathbb{N}}$ be a nondecreasing sequence in $\mathcal{F}$ with union $B$. Set $D_1=B_1$ and $D_m=B_m\setminus B_{m-1}$ for $m\ge 2$; the $D_m$ are pairwise disjoint members of $\mathcal{F}$ (Sigma-Algebra and Measurable Space) with $\bigcup_{m\le k}D_m=B_k$ and $\bigcup_m D_m=B$. By countable additivity and finite additivity (the finite case of Measure, Measure Space, and Probability Measure, obtained by appending copies of $\varnothing$),

Step 1 ($f$ is measurable and $\sup_m\int f_m\le\int f$). For $a\in\mathbb{R}$, $\{f>a\}=\bigcup_m\{f_m>a\}\in\mathcal{F}$, since $f(x)>a$ holds exactly when some $f_m(x)>a$; so $f$ is measurable in the sense of Lebesgue Integral of a Nonnegative Measurable Function. Monotonicity of the nonnegative integral is immediate from that definition: if $g\le h$ pointwise, every simple minorant of $g$ is a simple minorant of $h$, so $\int g\,d\mu\le\int h\,d\mu$. Since $f_m\le f$ for every $m$, we get $\sup_m\int_X f_m\,d\mu\le\int_X f\,d\mu$.

Step 2 (reverse inequality). Let $s$ be a nonnegative simple function with $s\le f$, with standard representation $s=\sum_{i=1}^{r}c_i\mathbf{1}_{A_i}$, and let $c\in(0,1)$. For $m\in\mathbb{N}$ put

using $\{f_m\ge a\}=\bigcap_{j\in\mathbb{N}}\{f_m>a-1/j\}\in\mathcal{F}$. The sets $E_m$ are nondecreasing because the $f_m$ are, and $\bigcup_m E_m=X$: if $s(x)=0$ then $x\in E_1$; if $s(x)>0$ then $f(x)\ge s(x)>c\,s(x)$, so $f_m(x)\ge c\,s(x)$ for some $m$ by definition of the pointwise supremum. The function $c\,s\,\mathbf{1}_{E_m}$ is simple, satisfies $c\,s\,\mathbf{1}_{E_m}\le f_m$ pointwise, and its integral is $c\sum_i c_i\,\mu(A_i\cap E_m)$ (its standard representation refines the sets $A_i\cap E_m$; the value follows from Simple Function and Its Integral). Hence by Lebesgue Integral of a Nonnegative Measurable Function,

Letting $m\to\infty$ and applying Step 0 to the nondecreasing sequences $(A_i\cap E_m)_m$ with union $A_i$,

Taking the supremum over $c\in(0,1)$ and then over all simple $s\le f$ gives $\sup_m\int_X f_m\,d\mu\ge\int_X f\,d\mu$ by Lebesgue Integral of a Nonnegative Measurable Function. With Step 1, equality holds; since the sequence of integrals is nondecreasing (Step 1), the supremum is also its limit when finite, and the convergence statement holds in $[0,\infty]$ as claimed. $\blacksquare$