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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?

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    $\begingroup$ You need $\mu(E)< \infty.$ Please edit. $\endgroup$
    – zhw.
    Commented Nov 9, 2017 at 23:30
  • $\begingroup$ 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. $\endgroup$
    – user169852
    Commented Nov 9, 2017 at 23:34
  • $\begingroup$ @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. $\endgroup$
    – Hashimoto
    Commented Jul 24, 2018 at 22:18

1 Answer 1

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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 $$ \forall_m \exists_n E_\delta \subseteq E_n^m. $$

This way, the analytic part of the theorem (i.e. functions and convergence) is gone and we're left with a problem concerning only sets and measures.

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  • $\begingroup$ 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? $\endgroup$
    – Jan
    Commented Mar 6, 2019 at 13:05
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    $\begingroup$ 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. $\endgroup$ Commented Mar 6, 2019 at 21:14

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