# Egorov Like Theorem, Finding a Sequence of Sets where Uniform Convergence Holds

Exercise: Under the assumptions in Egorov’s theorem, prove that there exists a sequence of measurable sets $$E_n$$ such that $$m(E\setminus \bigcup_{n=1}^\infty E_n) = 0$$ and the sequence $$\{f_n\}_{n\geq1}$$ converges uniformly on each $$E_n$$.

Here's my attempt:

Let $$\epsilon > 0$$ be arbitrary. Then let us define $$\begin{equation*} G_m := \{x \in E: |f_m(x) - f(x)| \geq \epsilon\}. \end{equation*}$$ and concurrently $$\begin{equation*} E_n^c := \bigcup_{m = n}^\infty G_m = \{x \in E: |f_m(x) - f(x)| \geq \epsilon \text{ for some }m \geq n\}. \end{equation*}$$ Now, first, notice that $$E_{n+1}^c \subseteq E_n^c$$. Moreover since we assume that $$f_n \rightarrow f$$ pointwise for $$x \in E$$ almost everywhere, it follows that there should be some $$E_n^c$$ for which $$x$$ does not belong for all $$x \in E$$ almost everywhere. Thus, $$\bigcap_{n=1}^\infty E_n^c = Z$$, where $$Z$$ is a measure zero set. Thus $$m(\bigcap_{n=1}^\infty E_n^c) = 0.$$ Now, since $$E_n^c$$ is a sequence of sets where $$f_n$$ does not uniformly converge to $$f$$, it follows that $$E_n$$ is a sequence of sets where $$f_n$$ converges to $$f$$ uniformly. Now, since $$\begin{equation*} E \setminus \bigcup_{n=1}^\infty E_n = E \cap \biggr(\bigcup_{n=1}^\infty E_n\biggr)^c = E \cap \bigcap_{n=1}^\infty E_n^c = \bigcap_{n=1}^\infty E_n^c, \end{equation*}$$ it follows that we have created the desired sequence of sets by taking the complement of each $$E_n^c$$.

Where I'm unsure: I don't use the finite condition, $$m(E) < \infty$$, and I'm concerned that I'm missing some counterexample (something along the lines of $$\chi_{[n,n+1]}$$) but I don't know where the finite condition fits into the puzzle.

By Egorov's Theorem, there sequence of measurable sets $$\{E_n\}_{n=1}^\infty$$ such that for each $$n \in \mathbb{N}$$, $$f_m \rightarrow f$$ uniformly and $$m(E \setminus E_n) \leq \frac{1}{n}$$. Now first notice that $$E \setminus E_{n+1} \subset E \setminus E_{n}$$ for all $$n \in \mathbb{N}$$, and since $$E \setminus E_n \subset E$$ for all $$n \in \mathbb{N}$$, we have $$m(E \setminus E_n) < \infty$$ for all $$n \in \mathbb{N}$$. Therefore, it follows that $$\begin{equation*} m\biggr(\bigcap_{n=1}^\infty E \setminus E_n\biggr) = \lim_{n\rightarrow\infty} m(E \setminus E_n) = \lim_{n\rightarrow \infty} \frac{1}{n} = 0. \end{equation*}$$ Lastly, notice that $$\begin{equation*} E \setminus \bigcup_{n=1}^\infty E_n = E \cap (\bigcup_{n=1}^\infty E_n\biggr)^c = E \cap \biggr(\bigcap_{n=1}^\infty E_n^c \biggr) = \bigcap_{n=1}^\infty E\setminus A_n. \end{equation*}$$ Therefore, it follows that the sequence $$\{E_n\}_{n=1}^\infty$$ is exactly the sequence desired.