Question on Egoroff-like theorem Hi all I was tackled by this question from Folland's real analysis second edition in the second chapter, it looks like a modified Egoroff theorem but I cannot really tackle it, it is question 41 of chapter 2 which reads as follows:

Let $\mu$ be a $\sigma$-finite  measure and $ f_n \rightarrow f $ a.e. Then there exist measurable sets $E_1,E_2,\ldots \subset X $ such that 
  $$\mu\left(\left(\bigcup_{i=1}^\infty E_i\right)^C\right)=0$$ and $f_n \rightarrow f $ uniformly on each $ E_i. $

I think this might have something to do with Egoroff's theorem but that theorem mentions nothing about complement having measure zero, only as small as you would like, which is what confuses me. Can anyone point out a proof of this with an explanation?
 A: Since $\mu$ is a $\sigma-$finite measure, there exists $\{A_n\}_{n\in \mathbb{N}}$ such that $\bigcup_{n}A_n=X$ and $\mu(A_n)<\infty$ $\forall n$.
Now, let $n\in \mathbb{Z}^+$ be given. Since $f_n\rightarrow f$ a.e. on $A_n$, by Egoroff's theorem, for any $k\in \mathbb{Z}^+$ there is $E_{n,k}\subset A_n$ such that $\mu(A_n\setminus E_{n,k})<\frac{1}{2^nk}$ and $f_n\rightarrow f$ uniformly on $E_{n,k}$.
Then observe that, for given $j\in \mathbb{Z}^+$,
\begin{align*}
\mu(\left(\bigcup_{n,k}E_{n,k}  \right)^c)&= \mu\left( \bigcup_{n} A_n \setminus \bigcup_{n,k} E_n  \right)\\ &= \mu\left(\bigcup_{n}\left(A_n\setminus \bigcup_{k} E_{n,k}\right)\right)\\&\leq \sum_n \mu\left(A_n\setminus \bigcup_k  E_n\right)\\&=\sum_n \mu\left(\bigcap_k\left( A_n\setminus E_{n,k} \right)  \right)\\& \leq \sum_n \mu(A_n\setminus E_{n,j})  \\&<\sum_n\frac{1}{2^nj}\\&<\frac{1}{j}.
\end{align*}
Since $j\in \mathbb{Z}^+$ is arbitrary, $\mu(\left(\bigcup_{n,k}E_{n,k}  \right)^c)=0$. 
Since $f_n\rightarrow f$ uniformly on each $E_{n,k}$ by construction, $\{E_{n,k}\}_{n,k}$ is the countable  collection of the measurable subsets of $X$. Thus,  we are done. 
