TITLE: Convergence in measure implies convergence almost everywhere of a subsequence
QUESTION [39 upvotes]: How can I prove that if a sequence of functions $\{f_n\}$ that converges to $f$ in measure on a space of finite measure, then there exists a subsequence of $\{f_n\}$ that converges to $f$ almost everywhere?

REPLY [4 votes]: We can enhance the result to almost uniform convergence by going to subsequences. Fix $e$, let $E_{n,k} = {x: |f_n - f| > 1/k}$, then $\lim \mu E_{n,k} = 0$, we can pick $n_j$, such that $\mu E_{n_j,k} < e/2^{k+j}$, then we have $\mu \bigcup_j E_{n_j,k} < e/2^k$ and $\mu \bigcup_{k} \bigcup_j E_{n_j,k} < e$, let the last union of sets be $T$, then on $T^c$ we have uniform convergence of $f_n$. cf. Egoroff's theorem.