Let $\{f_k\}$ be a sequence of measurable functions defined on a measurable set $E$ with $m(E)<\infty$. Suppose that for each $x$ in $E$, there exist a constant $M_x$ such that $$\sup_k |f_k(x)|\leq M_x$$
Prove that for all $\varepsilon >0$ there is a measurable set $F \subset E$ and a finite $M$ such that $m(E \setminus F) < \varepsilon$ and $|f_k(x)| \leq M$ for all $k$ and all $x$ in $F$. (Hint: Consider the sets $E_n = \{x\in E: \sup_k |f_k(x)| \leq n \}$).
My proof. Let $f(x) = \sup_k |f_k(x)|$. Then $E_n = \{f \leq n\}$. Since $E_n$ is measurable and $m(E_n)<\infty$ there exist a closed set $F_n \subset E_n$ such that for all $\varepsilon>0$, $m(E_n \setminus F_n)<\varepsilon$ with $m(F_n) < m(E_n) < \infty$. But every closed set is measurable.
And i'm stuck here, any suggestion?
The sequence $(E_n)_{n \in \mathbb{N}}$ of measurable sets is increasing (i.e. $E_1 \subseteq E_2 \subseteq \dots$), satisfies
$$\bigcup_{n \in \mathbb{N}} E_n =E$$
and $m(E_n) \leq m(E)<\infty$ for all $n \in \mathbb{N}$. It follows directly from the continuity of the measure $m$ that $$\lim_{n \to \infty} m(E \backslash E_n) = 0.$$ For any fixed $\epsilon>0$, choose $n \in \mathbb{N}$ sufficiently large such that $m(E \backslash E_n)<\epsilon$. If we set $F := E_n$, the claim follows.