The version of Egorov's Theorem I am referring to is something like this: Let $\{f_k\}$ be a sequence of measurable functions that converges a.e. in a set $E$ of finite measure to a finite limit $f$. Then for all $\epsilon>0$, there is a closed subset $F\subseteq E$ such that $|E\setminus F|<\epsilon$ and $\{f_k\}$ converges uniformly to $f$ on $F$.

The question I have is:

Qn) Some sources say that without loss of generality, we can choose $F$ to be compact. However, I have not seen any proof of this in my notes /textbook. Why can we take $F$ compact? Is my below reasoning correct?

My attempt: Is the reason something like this:

Consider $F\cap [-n,n]\nearrow F$. By Monotone Convergence Theorem for measure, $|F\cap [-n,n]|$ tends to $|F|<\infty$.

So there exists a sufficiently large $N$ such that $|F|-|F\cap[-N,N]|<\epsilon/2$.

By the usual Egorov's Theorem, $|E\setminus F|<\epsilon/2$.

So, WLOG, we may take the closed and bounded set $F\cap[-N,N]$ instead and $$|E\setminus F\cap[-N,N]|=|E\setminus F|+|F\setminus (F\cap [-N,N])|<\epsilon/2+\epsilon/2$$.

Thanks for any help.

  • $\begingroup$ From your attempt, I assume your measure space is $\mathbb{R}$. And it looks like you're free to use the fact that the compact sets of $\mathbb{R}$ are exactly the closed and bounded sets. But if this is right, then the answer is trivial. If there exists a compact $F$ satisfying the conclusion of the theorem, then, since $F$ is closed, there exists a closed $F$ satisfying the conclusion of the theorem. $\endgroup$ – grndl Sep 13 '16 at 19:33
  • $\begingroup$ Thanks. What I mean is the other way around, the conclusion of Egorov's theorem gives a closed $F$, I wish to "strengthen" the theorem and have a compact $F$ instead. $\endgroup$ – yoyostein Sep 14 '16 at 0:06

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