Measurability of an a.e. pointwise limit of measurable functions. Suppose that $(f_n)_n$ is a sequence of measurable functions on a set $E$ and that $f_n \to f$ a.e.on $E$. Does this imply that $f$ is  measurable?
I know that pointwise limit of measurable function is measurable. But here we only have convergence a.e. So I got confused. 
 A: I agree with the answer by @PhoemueX, but maybe not with the question ! Consider an alternative .......
Suppose that $\{f_n\}$ is a $\mu$-.a.e. convergent sequence and define $f = \lim_{n \rightarrow \infty} f_n$ where the limit exists. Then $f$ is a $\mu$-.a.e. defined function and $f$ is measurable. 
Clearly $f$ is $\mu$-.a.e. defined, to prove measurability.....
... let A be the null-set on which $\{f_n\}$ fails to converge.
Then $f_n |_{X \setminus A}$ is a measurable function on $X \setminus A$ and converges to a limit $f$ on $X \setminus A$ which is measurable by the normal convergence theorem. Then $f$ is a  $\mu$-.a.e. defined measurable function on X.
In terms of the counterexample in the previous answer, only constant functions are measurable and therefore a sequence of constants converge everywhere or nowhere. In the latter case, $X$ is the "null-set" where the sequence fails to converge and $f$ is defined "almost everywhere" except on $X$ - i.e. it is nowhere defined, but this still fits the definition of $\mu$-.a.e. defined in this case.
A: That depends on the exact context.
In general, $f$ will not be measurable. To see this, simply take $(X, \{\emptyset, X\}, \mu)$ as your measure space, with $\mu(\emptyset) = 0 = \mu(X)$.
Then $f_n \to f$ almost everywhere holds for every sequence $(f_n)_n$ and every map $f$, but only constant maps are measurable.

Now assume that your measure space $(X, \Sigma, \mu)$ is complete. This means that if $A \in \Sigma$ with $B \subset A$ and $\mu(A) = 0$, then also $B \in \Sigma$.
Then $f$ is measurable. To see this, first note that if $f = g$ almost everywhere (i.e. on $N^c$ with $\mu(N) = 0$) and $g$ is measurable, then so is $f$, because for every interval $I \subset \Bbb{R}$, we have
\begin{eqnarray*}
f^{-1}(I) &=& [f^{-1}(I) \cap N^c]  \cup [f^{-1}(I) \cap N] \\
&=& [g^{-1}(I) \cap N^c] \cup [f^{-1}(I) \cap N].
\end{eqnarray*}
But $N,N^c$ are measurable and $g$ is measurable. Hence, $g^{-1}(I) \cap N^c$ is measurable.
Finally, since your measure space is complete, $f^{-1}(I) \cap N$ is measurable (because it is a subset of the null-set $N$).
Hence, $f^{-1}(I)$ is measurable.
Now let $g := \liminf_n f_n$. Then $g$ is measurable and $f = g$ almost everywhere because of $f_n \to f$ almost everywhere. By the above, this implies that $f$ is measurable.

Finally note that the Lebesgue measure equipped with the $\sigma$-algebra of Lebesgue measurable sets is complete, but equipped with the $\sigma$-algebra of Borel measurable sets, it is not complete.
