Let $(f_n)_{n = 1}^{\infty}$ be a sequence of measurable functions defined on a measurable set $E$, $m(E) < \infty.$ Suppose that $(\forall x \in E)(\exists M_x \in \mathbb{R})$ such that $\sup_n \mid f_n(x) \mid \le M_x.$ Prove that $\forall \varepsilon > 0$ there exists $F$, a measurable subset of $E$, $m(E \setminus F) < \varepsilon$ and $M = M(F, \varepsilon)$ such that $\mid f_n(x) \mid \le M$ for all $n \in \mathbb{N}$ and every $x \in F.$
I can't think what would the set $F$ look like. Something hints me that I should have used the fact that $f_n$ are measurable and $[0, M_x]$ are closed, but I can't figure it out.
Any help is appreciated.
For $j\in \Bbb N$ let $E(j)=\{x\in E:\sup_{n\in \Bbb N}|f_n(x)|\le j\}.$
For $j,n\in \Bbb N$ the set $E(n,j)=\{x\in E: |f_n(x)|\le j\}$ is measurable, so $E(j)=\cap_{n\in \Bbb N}E(n,j)$ is measurable.
Let $F(1)=E(1)$ and $F(j+1)=E(j+1)$ \ $E(j).$
Now $E(j)\subset E(j+1).$ So $\{F(j):j\in \Bbb N\}$ is a pair-wise disjoint family of measurable sets, and $\cup_{j\in \Bbb N}F(j)=E.$ So $$m(E)=\sum_{j\in \Bbb N}m(F(j))=\sum_{j=1}^{\infty}m(F(j))=\lim_{k\to \infty}\sum_{j=1}^km(F(j)).$$ Now $m(E)<\infty.$ So, given $\varepsilon >0,$ take $k\in \Bbb N$ large enough that $m(E)-\sum_{j=1}^km(F(j))<\varepsilon.$ Then $m(E \setminus E(k))=m(E)-\sum_{j=1}^km(F(j))<\varepsilon\;$ and by the def'n of $E(k),$ we are done.