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.
Best Answer
When you write "Therefore, every measurable function is the limit a.e. of a sequence of continuous functions.": until that point, you have not mentioned continuous functions at all. In other words, you did some computations and then, out of the blue, claim that something follows of it.
I'm surprised you only lost half the points.