$\bf{\text{Background:}}$
Let $(\Omega,\Sigma,\mu)$ be a $\sigma$-finite measure space, and $X$ be a Banach space.
A function $f:\Omega\to X$ is simple if it assumes only finitely many values. That is, there exists subsets $E_{1}, … , E_{n}$ of $\Omega$ and scalars $x_{1}, … , x_{n}\in X$ such that $f = \sum_{i=1}^{n}\chi_{E_{i}}x_{i}$.
If the sets $E_{i}$ can be chosen from $\Sigma$, then $f$ is $\mu$-measurable simple.
A function $f:\Omega\to X$ is measurable if it is the limit of a sequence of $\mu$-measurable simple functions (almost everywhere).
$\bf{\text{Problem Statement:}}$
A function $f:\Omega\to X$ is measurable $\Leftrightarrow$ $\chi_{E}f$ is measurable for all $E\in \Sigma$ such that $\mu(E) < \infty$.
$\bf{\text{Partial Solution:}}$
The $(\Rightarrow)$ direction is easy. Write $f = \lim_{n\to\infty}f_{n}$ for $\mu$-measurable simple $f_{n}$. Then $\chi_{E}f_{n}$ is $\mu$-measurable simple and converges $\chi_{E}f$ almost everywhere.
The $(\Leftarrow)$ direction is giving me problems. I thought at first to let (as in the definition of $\sigma$-finite) $(E_{n})_{n=1}^{\infty}$ be a sequence in $\Sigma$ such that $\mu(E_{n}) < \infty$ and $\Omega = \cup_{n=1}^{\infty}E_{n}$, and defining
$F_{n} = \cup_{j=1}^{n}E_{j}$. Then for each $n\geq 1$, take a sequence $(g^{n}_{m})_{m=1}^{\infty}$ of $\mu$-measurable simple functions which converges to $\chi_{F_{n}}f$, by assumption. I thought I might be able to show that $g_{n}^{n}\to f$ almost everywhere, but I couldn't figure out the trick. My idea is below.
I claim that $\{\lambda\in \Omega : g^{n}_{n}(\lambda)\not\to f(\lambda)\}\subset\bigcup_{n=1}^{\infty}\{\lambda\in\Omega : g^{n}_{m}(\lambda)\not\to \chi_{F_{n}}(\lambda)f(\lambda)\}$, the latter set being $\mu$-measure $0$.
Is am I on the right track? I can't prove my claim.
Best Answer
A diagonalization argument would work, we just need to take care how to approximate our functions. For ease of notation, define $$\begin{align}H_1 &= F_1\\ H_{n+1} &= F_{n+1}\setminus F_n\end{align}$$ so that the $H_n$'s are pairwise disjoint and $\bigcup_{k=1}^n H_k = F_n$.
For all $n\geq 1$ we can take measurable simple functions $\{g^{(n)}_m\}_{m\geq 1}$ such that $g^{(n)}_m\mathop{\longrightarrow}_{m\to\infty} f^{(n)} := \chi_{H_n}f$. W.l.o.g. we can assume that $g^{(n)}_m$ is supported in $H_n$.
Define the functions $$h_n := \sum_{k=1}^n g^{(k)}_n.$$ Clearly, for all $n\geq 1$ we have $\chi_{F_n}h_m\mathop{\longrightarrow}_{m\to\infty} \sum_{k=1}^n f^{(k)}=\chi_{F_n} f$ almost everywhere (since for every sufficiently large $m$ we have $\chi_{F_n}h_m = \sum_{k=1}^n g^{(k)}_m$), and since $\Omega=\bigcup F_n$ we're done.