f is measurable function iff $f ^{-1} (A) $ is measurable for every finite set?

Question

$f : \mathbb{R} \to \mathbb{R} $ is measurable function iff $f ^{-1} (A) $ is measurable for every finite set $ A \subset \mathbb{R} $ ?. ($\mathbb{R} $ with Lebesgue measure).

since every finite set is measurable then if $f ^{-1} (A) $ is measurable for every finite set $ A \subset \mathbb{R} $ then $f$ is measurable function . but I think conversely is not true and we must construct a function $f : \mathbb{R} \to \mathbb{R} $ that is not one to one because $f ^{-1} (A) $ is finite set for every finite set $ A \subset \mathbb{R} $.

Answer 1

Indeed, the converse is not true.

Consider $E$ a non measurable subset of $]0,+\infty[$. Then define $$f : x \mapsto \left\{ \begin{array}{ll} |x| \mbox{ if $|x| \in E$} \\ -|x| \mbox{ if $|x| \notin E$} \end{array} \right.$$

It is easy to see that for all $y \in \mathbb{R}$, $f^{-1}(\{y\})$ has at most $2$ elements, so for all $A$ finite (and even $A$ countable), $f^{-1}(A)$ is measurable.

Yet $]0,+\infty[ \cap f^{-1}(]0,+\infty[) = E $ is not measurable, so $f^{-1}(]0,+\infty[)$ is not measurable.

Hence $f$ is not measurable.