Lemma 18.6.3. Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_{n} \to f$ almost everywhere, then $f_{n} \to f$ in measure.
Proof. Let $\eps > 0$, then for almost every $x \in X$, there exists $N \in \natp$ such that $d(f_{n}(x), f(x)) < \eps$ for all $n \ge N$, so
\[\mu\paren{\bigcup_{N \in \natp}\bigcap_{n \ge N}\bracs{d(f_n, f) < \eps}}= \mu(X)\]
By continuity from above (Proposition 15.1.5),
\[\limv{N}\mu{\bracs{d(f_N, f) \ge \eps}}\le \limv{N}\mu\paren{\bigcup_{n \ge N}\bracs{d(f_n, f) \ge \eps}}= 0\]
$\square$