TITLE: Measurability of an a.e. pointwise limit of measurable functions.
QUESTION [20 upvotes]: 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.

REPLY [4 votes]: 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.