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If $f_n\rightarrow f$ almost uniformly, then $f_n\rightarrow f$ a.e. and in measure.


Proof: Since $f_n\rightarrow f$ almost uniformly, then for every $\epsilon > 0$ there is a measurable set $F$ with $\mu(F) < \epsilon$ such that the sequence $\{f_n\}$ converges uniformly on $X\setminus E$.

I am not sure where to go from here, any suggestions is greatly appreciated.

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  • $\begingroup$ If $f_n \to f $ almost uniformly on each $E_n $, then $f_n \to f $ pointwise on $\bigcup_n E_n $ (why?). How does that help you? $\endgroup$ – PhoemueX Dec 5 '15 at 22:34
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Since $f_n$ converges almost uniformly, for any $\delta>0$, there exists a measurable set $E_\delta$ with measure less than $\delta$ such that $f_n$ converges uniformly on $E_\delta^c$. Thus there are $E_k$ with $\mu(E_k)<1/k$ such that $f_n$ converges uniformly on $E_k^c$.

Let $F_n=\bigcap_{k=1}^nE_k$. Then $$ \mu(F_n)\leqslant \mu(E_n)<1/n\quad\text{and so}\quad \lim_{n\to\infty}\mu(F_n)=0 $$ Since $F_1\supset\cdots\supset F_n\supset\cdots$, let $F=\bigcap_{n=1}^{\infty}E_n$ and by Monotone class theorem $$ \mu(F)=\mu\left(\bigcap_{n=1}^{\infty}E_n\right)=\lim_{n\to\infty}\mu(F_n)=0 $$ For any $x\in F^c=\bigcup_{n=1}^{\infty}E_n^c$, there is a $N$ that $x\in E_N^c$. Since $f_n$ converges uniformly on $E_N^c$, $f_n$ converges pointwise on $E_N^c$. This proves that $f_n$ converges pointwise a.e. Also $f_n$ converges in measure since $f_n$ converges pointwise a.e.

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  • $\begingroup$ Wouldn't it follow from the fact that $F = \cap E_k \subset E_k$ that $\mu(F) \leq \mu(E_k)<1/k$ for all $k$? Wouldn't it imply that $\mu(F) = 0$? Why did you first construct the sets $F_n$ above? Thank you. $\endgroup$ – math.h Sep 29 '19 at 14:15
  • $\begingroup$ It is more rigorous way to involve Monotone class theorem. $\endgroup$ – Math Wizard Sep 29 '19 at 14:28

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