Let $m$ be Lebesgue measure on $\mathbb R$ and $\mathcal{F}$ a countable set of Lebesgue measurable functions defined on Lebesgue measurable set $E\subset\mathbb R$ with $m(E)<\infty$. Assume that for each $x\in E$, the set $\{|f(x)|:f\in\mathcal F\}$ is a bounded subset of $\mathbb R$. Show that for all $\epsilon>0$, there is a closed set $F\subset E$ and a number $M<\infty$ such that $m(E\setminus F)<\epsilon$ and $|f(x)|\le M$ for all $x\in F$ and all $f\in\mathcal F$.
My attempt:
By Lusin's theorem, for each $f_n\in\mathcal F$, there exists a closed subset of $F_n$ such that $f_n$ is continuous on $F_n$ with $m(E\setminus {F_n})<\epsilon/2^n$. Then for every $f\in\mathcal F$, $f$ is continuous on the closed set $F=\bigcap_{n=1}^\infty F_n$ with $m(E\setminus F )<\epsilon$.
Now how to prove that it is uniformly bounded on $F$?
Best Answer
Hint: Let $G(x) = \sup_{f \in \mathcal F} |f(x)|$, and $B_N = \{x \in E: G(x) > N\}$. Show that $\lim_{N \to \infty} m(B_N) = 0$. Then after discarding a set of arbitrarily small measure, $\mathcal F$ is uniformly bounded on what remains.