Suppose $f$ is a function defined on an open set of $\mathbb{R}^{n}$ ($n\geq1$). In order to apply Lusin's theorem to $f$ (for all $\epsilon>0$ there is an open set of Lebesgue measure $<\epsilon$ on the complement of which $f$ is continuous) does $f$ need to be BOREL measurable or Lebesgue measurability is sufficient?
Does Lusin’s theorem require Borel measurability
measure-theoryreal-analysis
Best Answer
Hint: If $f$ is merely Lebesgue measurable on an open set $U,$ then there is a Borel measurable function $g$ on $U$ and a set $E\subset U$ of measure $0$ such that $f=g$ on $U\setminus E.$