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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.

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    Lusin + Tietze (some chars)2012-03-29

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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.

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    No, you didn't prove the converse either. Nowhere in your argument does a sequence of continuous functions appear (not even a single continuous function). So you can't have proved either the statement itself or its converse.2012-03-28
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You have to use that every measurable function is the limit pointwise of a simple functions and then that every simple function is approximated by step function a.e and (you need to show only for the characteristic) and then every step function is the limit a.e. of a continuous function!