UPDATE: I added an answer based off the hints provided by Robert Israel. It may, however, need some adjustment.
I'm trying to solve #18 (pg. 42) from Stein and Sharkarchi's analysis text.
(Before reading, I added the if part (<=) to the question since it didn't seem like much extra work to prove the statements were equivalent).
Problem Statement -- A function is measurable if and only if it is the a.e. limit of continuous functions.
My proof: Assume $\{f_{n}\}\to f\;a.e.\;x$ is a sequence of continuos functions. Then by property 2, each $f_{n}$ is measurable, and by property 4, their $a.e.$ limit is measurable. Thus, $f$ is measurable. Now suppose $f$ is measurable, and further assume $f$ is defined on a set $A\subset\mathbb{R}^{d}$ of finite measure and is infinite only on a set of measure zero. Let $B\subset A$ be the set where $f$ is finite so that $m(A-B)\leq\epsilon\;\forall\epsilon>0$. Then Lusin's Theorem guarantees the existence of a compact set $C\subset B$ where $m(B-C)\leq\epsilon\;\forall\epsilon>0$ and $f_{C}$ uniformly continuous. An application of the Tietze Extension Theorem allows us to construct a new function $F$ such that $F = f_{C}$ on $C$ and is continuous on all of $B$. Define a sequence of functions on $B$ by $f_{n} = F\;\forall\;n=1,2\ldots$ Then $\{f_{n}\}$ is (trivially) a sequence of continuous functions which uniformly converges to $F$ on $B$ (to obtain a non-trivial sequence of continuous functions, one could apply the generalized Stone-Weirstrass Approximation Theorem to $F$). As consequence, $f_{n}\to f\;a.e.\; x\in B$, since $m(B-C)\leq\epsilon$. But the union of two sets of measure zero is again a set of measure zero, so in fact $f_{n}\to\; f\; a.e.\; x\in A$
I'm concerned that my proof strategy is off the mark, particularly since I construct a rather trivial sequence of continuous functions. Secondly, how can I strengthen the theorem by getting rid of the hypotheses of $A$ having finite measure and $f$ being finite? (I managed to make $f$ finite a.e., but I don't think that's strong enough?).
Thanks!