# Understanding the proof to Egorov's Theorem

Theorem (Egorov).

Let $\{f_n\}$ be a sequence of measurable functions converging almost everywhere on a measurable set $E$ to a function $f$. Then, given any $\delta > 0$, there exists a measurable set $E_{\delta} \subset E$ such that

1. $\mu(E_{\delta}) > \mu(E) - \delta$
2. $\{f_{n}\}$ converges uniformly to f on $E_{\delta}$

proof (partial)

In the above link is a picture of (partially) the proof for the theorem in my book. They begin the proof by considering the following set

$$E_{n}^{m} = \bigcap_{i > n} \left \{ x \; : \; |f_{i}(x) - f(x)| < \frac{1}{m}\right \}$$

I do not understand the motivation in considering this set. To me, here is what I see. I know that this theorem is meant to show the relationship between convergence a.e. and uniform convergence. The definition of uniform convergence is

A sequence of function $\{f_n\}$ (with domain $D$) converges uniformly to $f$ if $$\forall \epsilon > 0 \; \exists N \in \mathbb{N}\; \forall n \geq N\; \forall x \in D, \; |f_n(x) - f(x)| < \epsilon$$

A sequence of functions converge to the function a.e. if the set of points for which the convergence fails to hold is of measure zero.

The set $E_{n}^{m}$ looks like the definition of uniform convergence I think. The $\frac{1}{m}$ is the $\epsilon"$ and the $i > n$ is the $n > N"$.

But I don't see the idea of what we want to do with this. Could someone explain the motivation to me please?

• You need $\mu(E)< \infty.$ Please edit.
– zhw.
Commented Nov 9, 2017 at 23:30
• You have the right idea. The set in the braces is the set on which $f_i$ agrees with $f$ to within $1/m$, uniformly. Using $1/m$ instead of arbitrary $\epsilon > 0$ is a standard technique in measure theory to allow us to express sets in terms of countable operations.
– user169852
Commented Nov 9, 2017 at 23:34
• @zhw. I think this is from Silverman's translation of Kolmogorov and Fomin, at a point in the book where sets with infinite measures are not introduced, and all sets have finite measures. Commented Jul 24, 2018 at 22:18

Your interpretation of $$1/m$$ as "$$\varepsilon$$" is correct. As already noted by Bungo, this is a standard technique. If we describe convergence as follows: $$a_i \to a \quad \iff \quad \forall_m \exists_n \forall_{i>n} |a_i-a| < \frac 1 m,$$ there is only countably many conditions to check. This is important in measure theory, since measures are by definition countably additive and $$\sigma$$-algebras are closed with respect to countable operations.
The idea behind introducing sets $$E_n^m$$ is to encode convergence in terms of sets, thus enabling us to use measures on them. You should be able to check that $$F := \bigcap_m \bigcup_n E_n^m$$ is exactly the set on which $$f_i(x) \to f(x)$$. Moreover, the statement $$f_i \rightrightarrows f$$ on $$E_\delta$$ is equivalent to $$\forall_m \exists_n E_\delta \subseteq E_n^m.$$
• Can you explain why we can describe convergence the way you do it here? I don't see why can we replace the $\varepsilon$ with the $\frac{1}{m}$ (uncountable to countable). Are the definitions equivalent?
• Sure. Given some $\varepsilon > 0$, you can always choose $m$ large enough so that $m \geqslant 1/\varepsilon$. If you can find $n$ such that $\forall_{i>n} |a_i-a| < \frac 1 m$, then the condition $\forall_{i>n} |a_i-a| < \varepsilon$ is also satisfied. Commented Mar 6, 2019 at 21:14