Beppo Levi's Monotone Convergence Theorem

Question

Let $(X, \cal{A}, \mu)$ be a measure space, $f : X → [0, ∞]$ and $(f_n)_{n∈\mathbb{N}}$ be a sequence of $\cal{A}$-measurable functions, where $f_n : X → [0, ∞]$, $n ∈ \mathbb{N}$, s.t. $f_n → f$ and $∀n ∈ \mathbb{N}$, $f_n ≤ f_{n+1}$ $\mu$-a.e. Show that: $$ \int{fd\mu}=\lim_{n→\infty}\int{f_nd\mu} $$ It seems very much with the Beppo Levi's theorem, but with the difference that we restrict $f_n$ and now is $\mu$-almost everywhere. Could you help me prove this, please?

Answer 1

Let $A_n:=\{x\in X\mid f_{n+1}(x)<f_n(x)\}$ and let $A=\bigcup_{n=1}^{\infty}A_n$.

Be aware that $A$ is a measurable set with $\mu(A)=0$.

If for a fixed $x_0\in X$ the sequence $(f_n(x_0))n$ is not monotonically increasing then some $n$ must exist with $f_{n+1}(x_0)<f_n(x_0)$.

Equivalently we could say that in that case $x_0\in A$.

Then we can conclude that the functions $f_n\mathbf1_{A^{\complement}}$ are monotonically increasing with $f_n\mathbf1_{A^{\complement}}\to f\mathbf1_{A^{\complement}}$ and by Monotone Convergence Theorem we conclude that:$$\int_{A^{\complement}}fd\mu=\lim_{n\to\infty}\int_{A^{\complement}}f_nd\mu$$

But $\mu(A)=0$ tells us that $\int_{A^{\complement}}fd\mu=\int fd\mu$ and $\int_{A^{\complement}}f_nd\mu=\int f_nd\mu$ for every $n$ so that also:$$\int fd\mu=\lim_{n\to\infty}\int f_nd\mu$$