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

Question

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.

Answer 1

I figured it out:

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.