Here is the problem (note: all of the talk of measure and measurable refers to Lebesgue measure here):
Let $f$ be measurable and defined on a (measurable) set $E\subset \mathbb{R}^n$. Show that there is a Borel set $H\subset E$ with $\rvert H\rvert = \rvert E\rvert$, where $f\mid_H$ is Borel measurable.
There is a hint, which is to use Lusin's theorem, recall this below:
Lusin's Theorem. Let $f$ be defined and finite on a measurable set $E$. Then $f$ is measurable iff $f$ has property $\mathscr{C}$ on $E$ (that is, $\forall \epsilon > 0$, there is a closed set $F\subset E$ s.t. $\rvert E\setminus F\rvert < \epsilon$ and $f$ is continuous relative to $F$).
What I have tried so far:
So immediatly, $f$ has property $\mathscr{C}$ on $E$. The issue I am having is that I just am having a lot of trouble seeing how this is actually helpful here (other than maybe somehow using the fact that a countable union of closed sets is Borel). So for instance, one could say that $\forall k\in \mathbb{Z}_+$ there is a closed set $F_k\subset E$ s.t. $\rvert E\setminus F_k\rvert < 1/k$, and then I could take the union of these $F_k$, and then I think that $\rvert E\setminus (\cup F_k)\rvert<1/k$ $\forall k$ by monotonicity, so I guess $\rvert E\rvert = \rvert \cup F_k\rvert$, where $ \cup F_k$ is $F_{\sigma}$ and thus Borel, but I am not really sure if this is helpful since we get these functions we get from Lusin's theorem: $h_k:=f\mid_{F_k}$, but I am really not sure what to do with these to get some Borel measurable function $h$ on $F:= \cup F_k$.
(yes I am aware there is another MSE post which constructs a Borel susbset of $E$ with measure zero compliment in $E$, but I need this subset to be s.t. when I restrict $f$ to it, I get something Borel measurable, so it seems like a proof of the above theorem can't really be taken partially from that other post).
Best Answer
Choose closed sets $F_n \subset E$ such that $|E\setminus F_n| <\frac 1 n$ and let $H=\bigcup_n F_n$. Continuity of $f|_{F_n}$ shows that for any Borel set $A$ in $\mathbb R$ the set $F_n \cap f^{-1}(A)$ is a Borel set: This is because the set is open in $F_n$ when $A$ is open. The collection of all Borel sets $A$ for which $F_n \cap f^{-1}(A)$ is a Borel set is a sigma algerba which contains all open sets so it contains all Borel sets.
It follows that $f|_H^{-1}(A)=\bigcup_n [f^{-1}(A) \cap F_n]$ is a Borel set in $H$.