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