Sequence of measurable functions converging to function $f$ a.e ,then $f$ is measurable

Question

if $f_n(x)$ is sequence of measurable functions,all defined on some measurable set $D$. and we have$$ f(x)=\lim_{n\to \infty} f_n(x)$$ almost every where. then show $f$ is measurable function.

now this mean there is set say $E$ such that $f(x)=\lim_{n\to \infty} f_n(x)$ for all $x\in D/E$ with $m(E)=0$

now what is domain of $f$ ??and how do we show $f$ is measurable ? please help

Answer 1

First show that $g_1(x)=\sup_j f_j(x)$, $g_2(x)=\inf_j f_j(x)$, $g_3(x)=\limsup_{j\rightarrow \infty} f_j(x)$, $g_4(x)=\liminf_{j\rightarrow \infty} f_j(x)$ are all measurable. Then $f=g_3=g_4$.

Answer 2

assuming domain of $f$ to be $D$.

then let $$A=\{x\in D:f(x)> \alpha\}$$ then$$A=\{x\in D/E:f(x)> \alpha\}\ \cup \{x\in E:f(x)> \alpha\}$$. now $$m(\{x\in E:f(x)> \alpha\})=0$$ as being a subset of set of measure zero. now $$\{x\in D/E:f(x)> \alpha\}=\cup_{n=1}^\infty \{x\in D/E:f_n(x)> \alpha\}$$

now since $f_n$ are measurable function on $D$ hence they are measurable on $D/E$ and hence right hand side is measurable set. which prove the result.

is it correct ?? or any better method? what about the domain of $f$??