Egorov's Theorem

Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $D \in \Sigma$ be such that $\mu \left({D}\right) < +\infty$.

Let $\left({f_n}\right)_{n \in \N}, f_n: D \to \R$ be a sequence of $\Sigma$-measurable functions.

Suppose that $f_n$ converges a.e. to $f$, for some $\Sigma$-measurable function $f: D \to \R$.

Then $f_n$ converges uniformly a.e. to $f$.

Proof
Pick $\epsilon > 0$.

By definition of convergence a.e., there is a set $E \in \Sigma$ such that
 * $E \subseteq D$
 * $\mu \left({E}\right) = 0$
 * $f_n \left({x}\right) \to f \left({x}\right)$ for each $x \in A \equiv D - E$.

For each $n, k \in \N$, define:
 * $\displaystyle A_{n, k} \equiv \left\{ {x \in A : \left| {f_n \left({x}\right) - f \left({x}\right)} \right| \geq \frac 1 k}\right\}$

Also define:
 * $\displaystyle B_{n, k} \equiv \bigcup_{i=n}^\infty A_{i, k}$

Since $f_n \to f$ on $A$, it follows by definition of convergence that for each $x \in A$, $\displaystyle \left| {f_i \left({x}\right) - f \left({x}\right)}\right| < \frac 1 k$ 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, $\displaystyle \limsup_{n \to \infty} A_{n, k} = \varnothing$.

So we have:

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

Setting $\displaystyle B \equiv \bigcup_{k \in \N} B_{n_k, k}$, we have:
 * $\displaystyle \mu (B) \leq \sum_{k \in N} \mu (B_{n_k, k}) \leq \sum_{k \in \N} \frac{\epsilon}{2^{k+1}} = \epsilon$

by the countable additivity of $\mu$ and by Sum of Infinite Geometric Progression.

Also, given any $\displaystyle \frac 1 k$, 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.