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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.

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    $\begingroup$ 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. $\endgroup$ – drhab Sep 29 '18 at 14:31
  • $\begingroup$ Sorry, what is $(f_n(x))_n)$, I did not get it. $\endgroup$ – J.Guo Sep 29 '18 at 14:44
  • $\begingroup$ It is the sequence $f_1(x),f_2(x),\dots$. $\endgroup$ – drhab Sep 29 '18 at 14:54
  • $\begingroup$ @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. $\endgroup$ – J.Guo Sep 29 '18 at 15:20
  • $\begingroup$ @drhab Sorry,I made a mistake minuates ago, I think I get it now. $\endgroup$ – J.Guo Sep 29 '18 at 15:29

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