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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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I don't see any mention of continuous functions, except for the last sentence. So you must have forgotten something. – azarel Mar 27 '12 at 23:19
See also this thread – t.b. Mar 27 '12 at 23:31
@Sri Pot : yes, I agree with azarel's comment ; I understand that if $f=g$ a.e. then $g$ is measurable, and there is an easier way to do this : $f=g$ almost everywhere means $f-g = 0$ almost everywhere, but then the pre-image of a measurable set is always of the form $(f-g)^{-1}(\{0\}) \cup N$, where $N \subseteq (f-g)^{-1}(\mathbb R \backslash \{0\})$, which is of zero measure by assumption. Therefore $f-g$ is measurable, hence $f - (f-g) = g$ also is. – Patrick Da Silva Mar 27 '12 at 23:31
But either way, you have not said anywhere WHY measurable functions are limits of continuous functions. – Patrick Da Silva Mar 27 '12 at 23:31
You prove the converse of what was asked - that is $f$ is an a.e. limit of continuous functions, then it is measurable. – D. Thomine Mar 27 '12 at 23:36

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. – Greg Martin Mar 28 '12 at 0:48

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!

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