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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?

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    $\begingroup$ The ideas in my answer here (math.stackexchange.com/questions/1068043/…) should be helpful. $\endgroup$ – PhoemueX Jul 31 '15 at 10:10
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    $\begingroup$ @PhoemueX thanks tried your other answer but I still cannot understand it $\endgroup$ – Don John Prep Jul 31 '15 at 11:09
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    $\begingroup$ If you start with the stronger assumption that $\mu$ is a finite measure, can you see how the assertion is obtained from Egorov's theorem? $\endgroup$ – Daniel Fischer Jul 31 '15 at 13:21
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    $\begingroup$ Let $F = \bigcup E_n$. Then you have $X\setminus F \subset X \setminus E_k$, and hence $\mu(X\setminus F) < 1/k$. Since that holds for all $k$, $\mu(X\setminus F) = 0$. Okay, the case of finite measure done. Now, for $\sigma$-finite measure, write $X = \bigcup A_m$, where each $A_m$ has finite measure. Perhaps it makes things easier to assume the $A_m$ disjoint, one can do that. So by the preceding, for each $m$, we have a sequence $E^{(m)}_n$ of subsets of $A_m$. How can you use that to get the desired sequence of subsets of $X$? $\endgroup$ – Daniel Fischer Jul 31 '15 at 13:48
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    $\begingroup$ @AbeerAbassi: When you have got it worked out, you can post your own answer to this question, with your solution. $\endgroup$ – Nate Eldredge Jul 31 '15 at 14:11
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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.

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    $\begingroup$ Hi, when you write $\mu(E_n^c) < \frac{1}{n}$ you actually mean $\mu(A_n - E_n)$. Thus your argument doesn't work as stated. But you can actually prove that Egoroff works for $\sigma- $ finite case by using your construction and using a standard $\frac{\epsilon}{2^n}$ argument. $\endgroup$ – Soham Dec 11 '18 at 5:36
  • $\begingroup$ @LucyferZedd I did not realize that! I fixed it :) Thank you! $\endgroup$ – Lev Bahn Dec 14 '18 at 16:06
  • $\begingroup$ Are the $A_n$ should be pairwise disjoint? $\endgroup$ – jackson May 14 '19 at 3:15
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    $\begingroup$ @jackson Thanks for commenting on my proof. My proof is still valid. Firstly, the definition of sigma finite measure does not require $A_n$ to be pairwise disjoint. And $\epsilon$ does not depend on $A_n$. Even we choose $A_n$ before saying $\epsilon$. $\endgroup$ – Lev Bahn May 14 '19 at 4:12
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    $\begingroup$ @jackson Thanks! you are right. I have corrected the answer. Sorry for late reply. $\endgroup$ – Lev Bahn May 14 '19 at 12:58

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