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.