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

Let $(f_n)$ be a sequence of measurable functions in a finite measure space $X$. Suppose that for almost every $x$, $\{f_n(x)\}$ is a bounded set. Then for any $\epsilon > 0$ there exist a positive number $c$ and a measurable set $E$ with $\mu(X-E) < \epsilon$, such that $|f_n(x)| \leq c$ for all $x \in E, n = 1, 2, \ldots$.

Attempt at proof
First, define $g(x) := \sup_n |f_n(x)|$. From Theorem A ($f$ measurable $\implies |f|$ measurable) and Theorem B ($f_n$ measurable for all $n$ $\implies$ $\sup_n f_n$ measurable), it follows that $g$ is measurable. Thus, $\{ x : g(x) \leq c \}$ is measurable for all $c \in \mathbb{R}$. Given $\epsilon > 0$, let $E := \{ x : g(x) \leq c \}$ for some $c$ such that $\mu(X \setminus E) < \epsilon$. (???). Thus $|f_n(x)| \leq c$ for all $x \in E, n = 1, 2, \ldots$.
Questions
Is this a proof of the statement? In particular, at the part marked with (???), I am wondering if the existence of $c$ needs elaboration, and if so, how can I do it?

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, $\mu(X\setminus E)<\varepsilon$.
