Sequence of Measurable Functions Converging Pointwise 
Let $\{f_n\}$ be a sequence of measurable functions on $E$ that converges pointwise almost every on $E$ to the function $f$. Then $f$ is measurable. 

Proof:
Let $g : E \to \Bbb{R}$ be defined by $g(x) = \lim_{n \to \infty} f_n(x)$ for every $x \in E$. Then $f = g$ almost every on $E$, and therefore it suffices to prove that $g$ is measurable. Let $C$ be some closed set in $E$. I trying to show that $ g^{-1}(C) = \bigcap_{n=1}^\infty f_n^{-1}(C)$, but I don't think this is exactly right. I was able to prove the the RHS side is contained in the LHS, as I will now show:

Let $x$ be in the set on the LHS. Then $x \in f_n^{-1}(C)$ for every $n \in \Bbb{N}$ or $f_n(x) \in C$ for every $n \in \Bbb{N}$. Since $C$ is closed, $g(x) = \lim_{n \to \infty} f_n(x) \in C$ or $x \in g^{-1}(C)$.

As for proving the other set inclusion, I have been quite unsuccessful. Is there any way of getting this strategy to work? 

*

*Does $\lim_{n\ge 1} f_n$ exist everywhere? (i.e. is $g$ well-defined?)

*The implication holds iff $\mu$ (the underlying measure) is complete (see Folland, Proposition 2.11). Specifically, $f_n$ converges everywhere on $A$ s.t. $\mu(E\setminus A)=0$, which means that $f\times 1_A$ is measurable. However, since $f\times 1_A=f$ $\mu$-a.e., $f$ is also measurable.
