---
tag: 762Y2NYA7W
kind: statement
role: theorem
name: Egorov's theorem
msc: [28A20]
uses: [P15VQS5DJ9, 63Z1Q4FS23, YYNCJSR0PJ, RARTMA0ZEC, 08VZ57PT26]
about: [YYNCJSR0PJ]
sources:
  - url: https://en.wikipedia.org/wiki/Egorov%27s_theorem
same_as:
  wikidata: Q1191709
  wikipedia: Egorov's theorem
---

## Statement

Let $(X, \Sigma, \mu)$ be a [[measure space|P15VQS5DJ9]] with $\mu(X) < \infty$.
Let $f_1, f_2, \ldots$ and $f$ be [[measurable functions|YYNCJSR0PJ]] from $X$ to $\R$ such that $f_n(x) \to f(x)$ for [[almost every|RARTMA0ZEC]] $x \in X$.
Then for every $\varepsilon > 0$ there is a set $E \in \Sigma$ with $\mu(E) < \varepsilon$ such that $f_n \to f$ [[uniformly|08VZ57PT26]] on $X \setminus E$.

## 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.

## Explanations

### L1 human

A sequence of functions may converge at every point but at very different speeds, so that it is not uniformly convergent.
Egorov's theorem says that on a space of finite measure this is a small problem.
After removing a set as small as one likes, the convergence becomes uniform on what is left.

### L2 human

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|G5WT6QN8WR]], so on finite measure spaces almost everywhere convergence implies convergence in measure.