Lusin's theorem states that if $f:[a,b]\rightarrow\mathbb{R}$ is a Lebesgue measurable function, then for any $\epsilon>0$ there exists a compact subset $E$ of $[a,b]$ whose complement has Lebesgue measure less than $\epsilon$ and a function $g$ continuous relative to $E$ such that $f=g$ on $E$. My question is, what is an example of a measure on $\mathbb{R}$ for which this is not true?
I know that it's true for all Borel measures on $\mathbb{R}$ which assign finite measure to all bounded intervals, so a counterexample would either have to be a non-Borel measure or a Borel measure which assigns infinite measure to some bounded intervals.
Best Answer
Try with the counting measure and the indicator function of the rationals.