Egorov's theorem

Remarks

For $k, n \ge 1$ let $E_{n,k} = \bigcup_{m \ge n} \set{x : \abs{f_m(x) - f(x)} \ge 1/k}$. For fixed $k$ these sets decrease in $n$ to a null set, so continuity from above, which needs $\mu(X) < \infty$, gives $n_k$ with $\mu(E_{n_k, k}) < \varepsilon / 2^k$. The set $E = \bigcup_k E_{n_k, k}$ works. Example: $f_n(x) = x^n$ on $[0, 1]$ with Lebesgue measure tends to $0$ almost everywhere, not uniformly; it does converge uniformly on $[0, 1 - \delta]$ for each $\delta > 0$. The finite measure hypothesis is necessary: on $\R$ with Lebesgue measure, $f_n = \mathbf{1}_{[n, \infty)}$ tends to $0$ everywhere. Every set $E$ of finite measure misses points of $[n, \infty)$ for each $n$, so $\sup_{X \setminus E} f_n = 1$ and the convergence is never uniform there.

Going deeper

For a student

This is the third of Littlewood's three principles: pointwise convergence is nearly uniform convergence. It is used to prove convergence theorems for integrals, such as the bounded convergence theorem and the Vitali convergence theorem. The convergence it gives, uniform outside sets of small measure, is called almost uniform convergence. Almost uniform convergence implies convergence in measure, so on finite measure spaces almost everywhere convergence implies convergence in measure.