$\{f_n\}_1^\infty$ are $(X,\mathcal{F})\rightarrow(\mathbb{R},\mathcal{B}_{})$ measurable, and $\lim_{n\rightarrow \infty} \{f_n\}= f$ $\mu.a.s$. Does $f$ is measurable in case $(X,\mathcal{F})$ is not a complete measurable space? If not, are there any additional conditions in which this might be true in incomplete measurable space?

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    $\begingroup$ If the sequence converges almost everywhere to $f$ and $f$ equals $g$ almost everywhere, then the sequence also converges almost everywhere to $g$. Whenever the underlying measure space is not complete, you can find functions $f$ and $g$ that are equal almost everywhere but only $f$ is measurable. The constant sequence consisting of repeating $f$ gives you then a counterexample. There is no problem if you use convergence everywhere instead. $\endgroup$ – Michael Greinecker Mar 5 '18 at 21:30
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    $\begingroup$ Reference for what Greinecker said: math.stackexchange.com/questions/344791/… $\endgroup$ – Martin Mar 5 '18 at 21:39

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