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 is of Lebesgue measure $\epsilon$ and a function $g$ continuous relative to $E$ such that $f=g$ on $E$.
My question is, what if something weaker is true? What if $f:[a,b]\rightarrow\mathbb{R}$ is a function such that for any $\epsilon>0$ there exists a compact subset $E$ of $[a,b]$ whose complement is of Lebesgue measure $\epsilon$ and a function $g$ Lebesgue measurable on $E$ such that $f=g$ on $E$? Then what properties does $f$ have to satisfy?
Does $f$ have to be Lebesgue measurable, or what?
Best Answer
Yes, $f$ does have to be Lebesgue measurable. Specifically, let $E_n, g_n$ be the corresponding compact set and measurable function for $\epsilon={1\over n}$, and let $F=[a,b]\setminus \bigcup_{n\in\mathbb{N}} E_n$ (note that $F$ is null). Now fix $c\in\mathbb{R}$; we want to show that $f^{-1}((-\infty,c))$ is measurable. Let $$A_n=E_n\cap g_n^{-1}((-\infty,c)).$$ Each $A_n$ is measurable, and we have $$\bigcup_{n\in\mathbb{N}} A_n\subseteq f^{-1}((-\infty,c))\subseteq F\cup\bigcup_{n\in\mathbb{N}} A_n.$$ This traps $f^{-1}((-\infty,c))$ between two measurable sets whose difference is null, and so $f^{-1}((-\infty,c))$ is measurable. So $f$ is a measurable function.
Note that we've actually used very little of the problem setup here (e.g. we've never used compactness at all). We've in fact shown something very general:
(Note that $(i)$ and $(ii)$ imply that $D$ is measurable.) In fact we can really go more general, but this seems like a good stopping point.