Convergence in Measure Implies Convergence a.e. of Subsequence

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f_n: D \to \R$ be a sequence of $\Sigma$-measurable functions for $D \in \Sigma$.

Let $f_n$ converge in measure to a function $f$ on $D$.

Then there is a subsequence $f_{n_k}$ of $f_n$ that converges a.e. to $f$.

Proof
For each $n, k\geq 1$, define $B_{n,k} = \{ x \in X : |f(x) - f_n(x)| > \frac{1}{k} \}$.

For fixed $n$, the sequence $\mu(B_{n,k})$ converges to zero by our assumption that $f_n$ converges in measure to $f$. Hence for each $n$ we can select a $k_n$ such that $\mu(B_{n,k_n}) < \frac{1}{2^n}$.

It follows that $\sum_{n=1}^\infty \mu(B_{n,k_n}) < \sum_{n=1}^\infty \frac{1}{2^n} = 2 < \infty$.

Therefore, by the Borel-Cantelli Lemma, the set $\limsup_n B_{n,k_n}$ has measure zero. By Equivalence_of_Definitions_of_Limit_Superior_of_Sequence_of_Sets, this means the set of $x\in X$ such that $|f(x) - f_{n_k}(x)| > \frac{1}{n_k}$ infinitely often has measure zero. Since $\frac{1}{n_k} \to 0$, it follows that almost every $x\in X$ is such that $|f(x) - f_{n_k}(x)| \to 0$, which means $f_{n_k}$ converges to $f$ almost everywhere.