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? 
 A: 
Proposition
Let $f_n:E \to \Bbb{R}$ a sequence of measurable functions such that $f_n(x) \to f(x)$ pointwise $\forall x \in E$.Then $f$ is measurable.
Proof
If $f_n$ are measurable ,then $\limsup_n f_n,\liminf_n f_n$ are also measurable.
But $f_n$ converges to $f,\forall x \in E\Rightarrow f(x)=\limsup_nf_n(x)=\liminf_nf(x)$
Thus $f(x)$ is measurable.

Let $N$ be the set of elements $x$,where $f_n(x)$ does not converge to $f(x)$.
Then from hypothesis $m(N)=0$
Define the function: $g:E \to \Bbb{R}$ such that $$g(x)=\begin{cases}\
                                                        f(x) & E\setminus N\\
                                                         0   & x \in N\\
                                                         \end{cases}$$
Then from the proposition, $g$ is measurable because $g=(f)1_{E \setminus N}$.
Now  we have that $g=f$ almost everywhere  and we know that if a function $h$ is almost everywhere equal to a measurable function,then $h$ is measurable.
A: *

*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.
