Prove that there must exist a measurable function $f$ on E and {$f_n$} converges to $f$ almost everywhere.

Suppose $$(X,R,\mu)$$ is a measure space,{$$f_n$$} is a sequence of measurable function on E.If {$$f_n$$} is convergence almost everywhere on E. Prove that there must exist a measurable function $$f$$ on E and {$$f_n$$} converges to $$f$$ almost everywhere.
If $$(X,R,\mu)$$ is a complete measure space, I can find a measurable set $$E_1$$,where $$\mu(E-E_1)=0$$ and {$$f_n$$} convergent to a measurable function $$f_1$$ on $$E_1$$.We set $$f = \begin{cases} f_1, & \text{if x\in E_1} \\ 0, & \text{if x\in E-E_1 } \end{cases}$$ So $$f$$ is the measurable function on E.However,if $$(X,R,\mu)$$ is only a measurable space,how to prove $$E_1$$ is a measurable set.

• If every $f_n$ is measurable then $\{x\mid (f_n(x))_n\text{ converges}\}$ is a measurable set, and you can take $E_1$ as that set. For a proof see here. – drhab Sep 29 '18 at 14:31
• Sorry, what is $(f_n(x))_n)$, I did not get it. – J.Guo Sep 29 '18 at 14:44
• It is the sequence $f_1(x),f_2(x),\dots$. – drhab Sep 29 '18 at 14:54
• @drhab For $$x\in E^{*}\iff\forall k\in\mathbb{N}\exists n\in\mathbb{N}\forall r,s\in\mathbb{N}\left[r,s\geq n\implies\left|F_{r}\left(x\right)-F_{s}\left(x\right)\right|<\frac{1}{k}\right]$$ It seems that on the left hand,$f_n(x)$ is convergence while on the right hand $f_n(x)$ is uniformly convergence. – J.Guo Sep 29 '18 at 15:20
• @drhab Sorry,I made a mistake minuates ago, I think I get it now. – J.Guo Sep 29 '18 at 15:29