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$\bf{\text{Background:}}$

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.

roo
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  • Isn't it enough to observe that $f = f \chi_\Omega$? – Chris Janjigian Sep 04 '13 at 21:06
  • That seems to work. But then $\sigma$-finiteness has not been used? – roo Sep 04 '13 at 21:12
  • Or does $\sigma$-finiteness imply that $\Omega$ is measurable? – roo Sep 04 '13 at 21:13
  • The fact that $\Sigma$ is a $\sigma-$ algebra implies that $\Omega$ is measurable (it's part of the definition). This result is true for any measurable map between measure spaces. – Chris Janjigian Sep 04 '13 at 21:14
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    Maybe the problem statement should be: $f$ is measurable iff $\chi_Ef$ is measurable for each $E\in\Sigma$ such that $\mu(E)<\infty$? Then $\sigma-finitness is really needed. – Etienne Sep 04 '13 at 21:24
  • You are correct Etienne, sorry for the omission. – roo Sep 04 '13 at 21:37
  • Is it a diagonal argument like the one I am trying to apply? I still am stuck in the same place. – roo Sep 04 '13 at 21:47
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    Let $A_n = F_n \setminus F_{n-1}$. Then $\mu(A_n) < \infty$, $\Omega = \bigcup A_n$, and the $A_n$ are disjoint. For each $n$, let $(g_k^n){k \in \mathbb{N}}$ a sequence of simple functions converging (a.e.) to $f\chi{A_n}$ vanishing outside $A_n$. Let $f_m = \sum_{n=1}^m g_m^n$. – Daniel Fischer Sep 04 '13 at 22:44
  • Thanks! This has been bugging me for 2 days now. :) – roo Sep 04 '13 at 22:44

1 Answers1

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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.

Jonathan Y.
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    @DanielFischer, apologies for missing your comment. – Jonathan Y. Sep 04 '13 at 22:54
  • I tried to skip your technique of calculating the limit of $\chi_{F_{n}}h_{m}$ for $m\to\infty$, and tried to take $h_{m}$'s limit directly and see why you needed to restrict.

    Thanks for the details!

     – roo Sep 06 '13 at 05:49