Question

(ZFC)

Does there exist a function $f : \mathbb{R} \to \mathbb{R} \hspace{.1 in}$ such that for all $B$, if $B \subsetneq \mathbb{R}$ and $B$ is a nonempty Borel set, then $\lbrace x \in \mathbb{R} : f(x) \in B \rbrace$ is nonmeasurable?

Answer 1

There are continuum many pairwise disjoint subsets of [0,1] each having Lebesgue outer measure 1. (By transfinite recursion of length continuum.) Assume that they are ${A_x:0\leq x\leq 1}$ and their union is [0,1]. Then set $f:[0,1]\to[0,1]$ so that $f(y)=x$ is $y\in A_x$. Now the inverse image of any $X\subseteq [0,1]$, $X\neq\emptyset, [0,1]$ has innermeasure 0, outer measure 1, hence $f^{-1}[X]$ is nonmeasurable.