Proof question: Sequences of measurable functions $f_n$, such that for almost all $x$, set $f_n(x)$ is bounded...

Let $g:=\sup_{n}|f_{n}(x)|$, which is measurable (but possibly takes value $\infty$). Let $A=\{x\in X\mid\{f_{n}(x)\mid n\in\mathbb{N}\}\mbox{ is bounded\}}.$ By assumption, $\mu(X\setminus A)=0$. Let $B=\{x\in X\mid g(x)\in\mathbb{R}\}$ and $B_{n}=\{x\in X\mid g(x)\leq n\}$. Clearly $B_{1}\subseteq B_{2}\subseteq\ldots$ and $B=\cup_{n}B_{n}$. By continuity of measure, $\mu(B)=\lim_{n}\mu(B_{n})$. Note that $A=B$ and $\mu(X)<\infty$, so we have that $0=\mu(A^{c})=\mu(B^{c})=\mu(X)-\mu(B)=\mu(X)-\lim_{n}(B_{n})$. Given $\varepsilon>0$, we can choose $n$ such that $\mu(X)-\mu(B_{n})<\varepsilon$. Put $c=n$ and $E=B_{n}$. Now, for each $x\in E$, we have that $g(x)\leq n\Rightarrow|f_{k}(x)|\leq c$ for all $k$. Moreover, $\mu(X\setminus E)<\varepsilon$.