Egorov's Theorem

Theorem
Suppose we are given a measure space $$(X, \Sigma, \mu)\ $$ and a sequence of $\Sigma$-measurable functions $$f_n:D\to\R$$ for $$D\in\Sigma$$.

Then if $$f_n\ $$ converges a.e. to a function $$f\ $$ on $$D\ $$ and $$\mu(D) < \infty\ $$, then $$f_n\ $$ almost uniformly converges to $$f\ $$.

Proof
Pick $$\epsilon > 0\ $$. By definition of convergence a.e., there is a set $$E\in\Sigma$$ such that
 * $$E\subseteq D$$,
 * $$\mu(E) = 0\ $$, and
 * $$f_n(x)\to f(x)$$ for each $$x\in A \equiv D - E$$.

For each $$n, k\in\N$$, define $$A_{n, k} \equiv \left\{{x\in A : |f_n(x) - f(x)| \geq \frac{1}{k}}\right\}$$. Also define $$B_{n, k} \equiv \bigcup_{i=n}^\infty A_{i, k}$$.

Since $$f_n\to f$$ on $$A\ $$, it follows by definition of convergence that for each $$x\in A$$, $$|f_i(x) - f(x)| < \frac{1}{k}$$ for all $$i$$ sufficiently large.

Thus, when $$k$$ is fixed, no element of $$A\ $$ belongs to $$A_{n, k}\ $$ infinitely often. Hence by Characterization of Limit Superior of Sets, $$\limsup_{n\to\infty}A_{n, k} = \varnothing$$.

So we have

$$ $$ $$ $$ $$

Thus, since $$k$$ was arbitrary, we can associate some $$n_k\in N$$ with each $$k$$ such that $$\mu(B_{n_k, k}) < \frac{\epsilon}{2^{k+1}}$$.

Setting $$B \equiv \bigcup_{k \in \N} B_{n_k, k}$$, we have $$\mu (B) \leq \sum_{k \in N} \mu (B_{n_k, k}) \leq \sum_{k \in \N} \frac{\epsilon}{2^{k+1}} = \epsilon$$ by the countable additivity of $\mu\ $ and by Sum of Infinite Geometric Progression.

Also, given any $$\frac{1}{k}$$, we have $$x \in A - B$$ implies $$x \notin B_{n_k, k}$$, which means $$|f_i(x) - f(x)| < \frac{1}{k}$$ for all $$i \geq n_k$$. Since this is true for all $$x\in A - B$$, it follows that $$f_n\ $$ converges to $$f\ $$ uniformly on $$A - B\ $$.

Finally, note that $$A - B = D - (E\cup B)$$, and $$\mu (E\cup B) \leq \mu (B) + \mu (E) = \mu (B) + 0 < \epsilon$$. Since $$\epsilon\ $$ was arbitrary, the convergence is uniform a.e.