# Modes of Convergence, Real Analysis Folland, Ch 2 Problem 39

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.

• 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? – PhoemueX Dec 5 '15 at 22:34

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.
• 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. – math.h Sep 29 '19 at 14:15