Prove the extension of Lusin's Theorem to the case that $E$ has infinite measure.
This is my attempt:
Consider $E \subseteq \mathbb{R}$ measurable with $m(E)=\infty,$ and let $f:E \rightarrow \mathbb{R}$ be measurable. For each $n$, let $E_n=\{x \in E: n-1 \leq |x|<n\},$ so clearly $E_n$ is a bounded measurable set. Therefore, by the case when $m(E)<\infty$, we see that for each $n$ we can find a closed subset $F_n$ of $E_n$ such that $m(E_n \sim F_n)< \epsilon/2^n$ such that $f$ is continuous on $F_n.$ Let $F = \bigcup_{n \geq 1}F_n$.
I claim that $F$ is closed and $f$ is continuous on $F$. Since for each $n$, $F_n \subseteq E_n$ and $\{E_n\}_{n\geq 1}$ is a collection of disjoint measurable set, we see that $F_j \cap F_k=\emptyset$ for all $j \neq k$. Therefore, $F$ is the countable union of disjoint closed sets $F_n$,so $F$ is closed (I tried to prove that by Hein-Borel but I could not).
Now to prove the continuoity I am getting some problems, I am trying to use some thing like since $f$ is continuous on closed sets that are disjoint then it will be continuous on their union as well, but I am not sure if that would help.
Thanks for any hint or help with that.