Question regarding generalized Egoroff's Theorem I'm looking at this problem, which asks one to prove a generalized version of Egoroff's Theorem. While I understand the OP's approach to divide $X$ into a countable number of subsets (and apply Egoroff's Theorem to those with finite measure), I don't understand his final claim, that $f_n \to f$ uniformly on $E^{c}$. Roughly speaking, the OP's argument is "if $f_n \to f$ uniformly on a countable  collection of sets $\{A_k\}_{k_1}^{\infty}$, then $f_n \to f$ uniformly on $\bigcup_{k=1}^{\infty} A_k$". However, this is not necessarily true. I can't find a good justification for his last step.
 A: David Giraudo in this posting gives a nice solution to the problem. I present the outline and some details
Define $g_n=\sup_{k\geq n}|f_k-f|$.
Notice that

*

*$g_n$ is monotone nonincreasing,

*$0\leq g_n\leq 2g$,

*and $g_n\xrightarrow{n\rightarrow\infty}0$ $\mu$-a.s. since $f_n\xrightarrow{n\rightarrow\infty}f$ $\mu$-a.s.

*By Markov-Chebyshev's inequality, for all $n,m\in\mathbb{N}$
$$\mu(g_n> 1/m)\leq m\int_Xg_n\,d\mu\leq 2\int_X g\,d\mu<\infty$$

*Dominated converges implies that
$$\lim_n\int_X|g_n|\,d\mu=0$$
Hence, given $\varepsilon>0$, one can choose a subsequence $n_m$ along which
$$\mu(g_{n_m}>1/m)\leq \varepsilon 2^{-m}$$
Define $A_m=\{g_{n_m}>1/m\}$ and $B:=\bigcap_m(X\setminus A_m)$.


*For all $x\in B$,  $0\leq g_{n_m}(x)\leq \tfrac1m$ for all $m$. This means that $f_n$ converges uniformly on $B$. Indeed, for any $\varepsilon'>0$ choose $M$ large enough so that $1/M <\varepsilon'$. Then for all $n\geq n_M$,
$$\sup_{x\in B}|f_n(x)-f(x)|\leq\sup_{x\in B}g_{n_M}(x)\leq \tfrac1M<\varepsilon'$$


*$\mu(X\setminus B)=\mu\Big(\bigcup_mA_m\Big)\leq\sum_m\varepsilon 2^{-m}=\varepsilon$.
