Analysis Problem of the Day 51

Today’s problem appeared as Problem 3 on the UCLA Fall 2021 Analysis Qual:

Problem 3: Let $\phi: [0,1] \to [0,1]$ be Borel measurable. Show that there exists a Borel subset $B \subseteq \phi([0,1])$ such that $\mu(\phi^{-1}(B))=1$ (Hint: $\phi([0,1])$ need not be Borel).

Solution: This is quite a tricky problem if one does not go off of anything. However, we do have a general result for measurable functions known as Lusin’s theorem, which states that one may restrict a measurable function to a set of arbitrarily large measure and get a continuous function. Thus, let $E_n$ be a sequence of such sets s.t. $\mu(E_n)\to 1, \phi|_{E_n}$ is continuous. By the regularity of the Lebesgue measure, one may approximate from below by compact sets, so we may assume that the $E_n$ are compact. Then, $\phi(E_n) \subseteq \phi([0,1])$ are compact, so $B:=\bigcup_n \phi(E_n)$ is Borel and $\phi^{-1}(B) \supseteq \bigcup_n E_n,$ so since $\mu(\bigcup_n E_n) =1,$ $\mu(\phi^{-1}(B))=1$ as desired.

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