Egorov's theorem

Let $(X, \mathcal{A}, \mu)$ be a measure space with $\mu(X) < \infty$, and let $f_n, f : X \to \mathbb{R}$ be measurable with $f_n \to f$ pointwise almost everywhere. Then for every $\varepsilon > 0$ there is a measurable $E \subseteq X$ with

$\mu(X \setminus E) < \varepsilon \qquad \text{and} \qquad f_n \to f \text{ uniformly on } E.$

The finiteness of $\mu(X)$ is essential: on $\mathbb{R}$ with Lebesgue measure, $f_n = \mathbf{1}_{[n, n+1]} \to 0$ pointwise, but on any set whose complement has finite measure the convergence is not uniform. The conclusion cannot be improved to $\varepsilon = 0$: convergence is in general not uniform off a null set, as $f_n = x^n$ on $[0,1]$ shows.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. Countable subadditivity and continuity from above of a finite measure, applied to the sets $E_{n,k} = \bigcup_{m \ge n} \{ |f_m - f| \ge 1/k \}$: each decreases in $n$ to a null set, so some $E_{n_k, k}$ has measure below $\varepsilon 2^{-k}$, and the complement of their union works. The argument is short but is entirely about the measure, which this library does not have (Lebesgue measure and the Lebesgue integral ‡). Continuity from above is where $\mu(X) < \infty$ is spent.

Which page it serves. The uniform convergence page. That page proves what uniform convergence buys, and it can exhibit sequences that converge pointwise but not uniformly; Egorov is the theorem that says the failure is always confined to a small set, and it is the standard bridge from pointwise hypotheses to uniform conclusions. It is also the usual route to Lusin's theorem (Lusin's theorem ‡).