# Convergence in measure implies convergence almost everywhere of a subsequence

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?

Let $$(X,\mathcal{A},\mu)$$ be a measure space and $$(f_n)_{n \in \mathbb{N}}$$ such that $$f_n \to f$$ in measure, i.e.

$$\mu(|f_n-f|>\varepsilon) \stackrel{n \to \infty}{\to} 0$$

for any $$\varepsilon >0$$. Setting $$\varepsilon=2^{-k}$$, $$k \in \mathbb{N}$$, we can choose $$n_k$$ such that

$$\mu(|f_n-f|> 2^{-k}) \leq 2^{-k}$$

for all $$n \geq n_k$$. Without loss of generality, $$n_{k+1} \geq n_k$$ for all $$k \in \mathbb{N}$$. Set

$$A_k := \{x \in X; |f_{n_k}(x)-f(x)| > 2^{-k}\}.$$

As $$\sum_{k \geq 1} \mu(A_k) \leq \sum_{k=1}^{\infty} 2^{-k} < \infty,$$ the Borel-Cantelli lemma yields

$$\mu \left( \limsup_{k \to \infty} A_k \right) =0.$$

It is not difficult to see that this implies

$$\lim_{k \to \infty} f_{n_k}(x) =f(x)$$

$$\mu$$-almost everywhere.

• I just came across this answer and I'm struggling to understand the last step. Why is $\limsup\limits_{k \rightarrow \infty} A_k$ the opposite of $\{ f_{n_k}(x) \rightarrow f(x), k \rightarrow \infty \}$? Jun 6 at 19:19

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.