# A countable family of point-wise bounded functions defined on finite measurable set is actually uniformly bounded on a closed subset

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 $$F$$?

## 1 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.