Theorem 18.1.7 (Egoroff). 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 for any $\eps > 0$, there exists $E \in \cm$ such that:
$f_{n} \to f$ uniformly on $E$.
$\mu(X \setminus E) < \eps$
Proof. Since $f_{n} \to f$ almost everywhere, for any $\eps > 0$,
\[\mu\paren{\limsup_{n \to \infty}\bracs{d(f_n, f) > \eps}}= 0\]
By continuity from above (Proposition 15.1.5),
\[\lim_{N \to \infty}\paren{\bigcup_{n \ge N}\bracs{d(f_n, f) > \eps}}= 0\]
For each $k \in \natp$, let $N_{k} \in \natp$ such that
\[\paren{\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}}< 2^{-k}\eps\]
then by subadditivity,
\[\mu\paren{\bigcup_{k \in \natp}\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}}\le \eps\]
and $f_{n} \to f$ uniformly on
\[\braks{\bigcup_{k \in \natp}\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}}^{c}\]
$\square$