Must $f$ be measurable if each $f^{-1}(c)$ is?

Question

Suppose $f$ is a real-valued function on $\mathbb R$ such that $f^{−1}(c)$ is measurable for each number $c$. Is $f$ necessarily measurable?

Answer 1

Hint: Note that the Vitali set has size $2^{\aleph_0}$ and therefore there is a bijection between $\mathbb R$ and the Vitali set.

Answer 2

Hint: Every injective function satisfies the hypothesis. You can take any nonmeasurable set, map it injectively into $(0,\infty)$, and map its complement injectively into $(-\infty,0)$ (for example using $e^x$ for a simple formula).

Answer 3

Consider for example a set $E \subset [0,1]$; then the function $$ f(x)=\begin{cases} x & x \in E\\ -x & x \in [0,1]\setminus E \end{cases} $$ is measurable if and only if $E$ is measurable. On the other hand, $f$ is injective, hence $f^{-1}(c)$ is either empty, either a singleton (in both cases, anyway, $f^{-1}(c)$ is measurable).