Measurable Functions - Main Theorem

TheoremEgorov's Theorem

Let $(X, \mathcal{F}, \mu)$ be a finite measure space (i.e., $\mu(X) < \infty$ ). Let $\{f_n\}$ be a sequence of measurable functions converging pointwise almost everywhere to a measurable function $f$ .

Then for every $\epsilon > 0$ , there exists a measurable set $E \subseteq X$ such that:

$\mu(E) < \epsilon$
$f_n \to f$ uniformly on $E^c$

In other words, pointwise convergence almost everywhere implies uniform convergence except on a set of arbitrarily small measure.

Egorov's Theorem bridges pointwise and uniform convergence. It shows that on finite measure spaces, pointwise convergence a.e. is "almost" uniform convergence - we only need to remove a small exceptional set.

ExampleApplication to Function Sequences

Consider $f_n(x) = x^n$ on $[0, 1]$ with Lebesgue measure. We have $f_n \to 0$ pointwise on $[0, 1)$ and $f_n(1) = 1$ .

Given $\epsilon > 0$ , Egorov's theorem guarantees a set $E$ with $\lambda(E) < \epsilon$ such that convergence is uniform on $[0, 1] \setminus E$ .

We can take $E = [1 - \delta, 1]$ for appropriate $\delta$ . On $[0, 1 - \delta]$ , we have $x^n \leq (1-\delta)^n \to 0$ uniformly, confirming the theorem.

Remark

Important observations about Egorov's Theorem:

Finiteness is essential: The theorem fails for infinite measure spaces. On $\mathbb{R}$ with Lebesgue measure, consider $f_n = \chi_{[n, n+1]}$ . Then $f_n \to 0$ pointwise everywhere, but for uniform convergence on a set $F$ , we need $F$ to miss infinitely many intervals, so $\lambda(F^c) = \infty$ .
Relationship to integration: Egorov's theorem helps prove that convergence in measure follows from pointwise convergence on finite measure spaces.
Lusin's theorem complement: While Lusin's theorem makes measurable functions "nearly continuous," Egorov's theorem makes convergent sequences "nearly uniformly convergent."

The proof uses the fact that if $f_n \to f$ pointwise, then for each $k$ , the sets $E_k^N = \bigcup_{n=N}^{\infty} \{x : |f_n(x) - f(x)| \geq 1/k\}$ decrease to measure zero as $N \to \infty$ . We can choose $N_k$ so that $\mu(E_k^{N_k}) < \epsilon/2^k$ , and take $E = \bigcup_k E_k^{N_k}$ .