I've recently been studying the functional analysis and I think I need some help with one exercise in the book.
We have the Lusin's theorem, which is stated the following way:
Given some measurable $E$ where $\mu(E)<\infty$ and some function $f$, which is measurable and finite almost everywhere in $E$, the following statement is true:
$\forall \epsilon>0 : \space \exists F_\epsilon \subset E$, that ($F_\epsilon$ has to be closed):
1) $\mu(E \setminus F_\epsilon) < \epsilon$
2) $f$ is continuous on $F_\epsilon$
Now - the exercise is to tell if the inverse theorem is correct (and give the proof if it is) - so that if there exists that $F_\epsilon$, on which our function is continuous (and I assume finite almost everywhere), then it's also measurable on the corresponding set $E$.
I assume this statement is correct, but unfortunately I can't come up with a proper proof. Proof of the Lusin's theorem doesn't help here, because we have to prove the continuity -> measurableness on the corresponding sets and the Egorov's theorem is no help in that case.
Could someone share the proof please or give me some clues on accomplishing it.