The following is Folland's proof of Egoroff's theorem.
Without loss of generality we may assume that $f_n \rightarrow f$ everywhere on $X$. For $k, n \in \mathbb{N}$ let$$E_n(k) = \bigcup_{m=n}^\infty \{x : |f_m(x) - f(x)| \geq k^{-1}\}.$$Then, for fixed $k$, $E_n(k)$ decreases as $n$ increases, and $\bigcap_{n=1}^\infty E_n(k) = \emptyset$, so since $\mu(X) < \infty$ we conclude that $\mu(E_n(k)) \rightarrow 0$ as $n\rightarrow \infty$. Given $\epsilon > 0$ and $k \in \mathbb{N}$, choose $n_k$ so large that $\mu(E_{n_k}(k)) < \epsilon 2^{-k}$ and let $E = \bigcup_{k=1}^\infty E_{n_k}(k)$. Then $\mu(E) < \epsilon$, and we have $|f_n(x) - f(x)| < k^{-1}$ for $n > n_k$ and $x \not\in E$. Thus $f_n \rightarrow f$ uniformly on $E^c$.
Why is it important that $X$ have finite measure and how does it imply $\mu(E_n(k)) \rightarrow \infty$?
