Observe that if $f_n$ converges to $f$ almost everywhere then $f(x)=\limsup_{n\to \infty} f_n(x)$ almost everywhere. We know that $\limsup_{n\to \infty} f_n(x)$ is measurable since $\{f_n\}_{n\in \mathbb{N}}$ is a sequence of measurable functions.
Claim: Let $f:E\to \mathbb{R}$ and let $g:E\to\mathbb{R}$. If $f$ is measurable, and $f=g$ a.e., on $E$ then $g$ is measurable.
Proof of Claim: For $a\in \mathbb{R}$, let $A=\{x\in E: f(x)>a\}$ and $B=\{x\in E: g(x)>a\}$. Then $A$ is measurable and $A\setminus B\, , B\setminus A\subseteq \{x\in E: f(x)\neq g(x)\}$ and they have zero measures. Now observe that $B=(B\setminus A)\bigcup (B\bigcap A)=(B\setminus A)\bigcup (A\setminus (A\setminus B))$ is measurable. Hence, $g$ is measurable.
Therefore, every measurable function is the limit a.e. of a sequence of continuous functions.
The grader took off half the points and it is too late to ask my professor(exam tomorrow!). I would appreciate it anyone helps me understand this and maybe fix my proof. Also note that various sources helped me with this proof. Thanks.