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Let {$f_n$} be a sequence of measurable functions on E that converges to the real-valued $f$ pointwise on E. Show that $E = E_0 \cup \bigcup_{k=1}^{\infty} E_k$, where all sets $E_0, E_1,...$ are measurable, and {$f_n$} converges uniformly to f on each $E_k$ if $k>0$, and $m(E_0) = 0$.

EDIT: I thought that Egoroff's Theorem would need to be used, but I don't have that E has finite measurable and I'm not sure how to get it. I know that since E is a countable union of measurable sets, E is measurable. Am I just supposed to take the cases of E having finite measure and m(E) = $\infty$ separately? Is there a better way to go about proving this?

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  • $\begingroup$ Countably union of measurable sets is measurable. So if $E$ is not measurable, you cannot get such decomposition. $\endgroup$ Commented Oct 19, 2017 at 1:48
  • $\begingroup$ @EclipseSun ok that makes sense. I actually made a typo. I meant how do I know E has finite measure? $\endgroup$ Commented Oct 19, 2017 at 1:56

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Consider $[-n,n]\cap E$ which is of finite measure. You can use Egoroff to get a set $E_n\subseteq [-n,n]\cap E$ such that $f_n$ converges uniformly on $E_n$, and $m([-n,n]\cap E)\setminus E_n)<1/n$. Then you have to show that $m(E\setminus \bigcup E_n)=0$.

To do this, note that $F_n:=(E\setminus\bigcup E_n)\cap [-n,n]\subseteq (E\cap[-n,n])\setminus E_n$. Hence, $m(F_n)<1/n$. But $F_n\uparrow (E\setminus \bigcup E_n)$. By continuity of measure $$m(E\setminus \bigcup E_n)=\lim_n m(F_n)=0.$$

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