# Existence of Lebesgue measurable set whose preimage is not Lebesgue measurable

Let $f:\mathbb{R} \to \mathbb{R}^2, f(x)=(x,0)$. Then there exists a set $A \in \mathscr{L}_2$ with $f^{-1}(A) \notin \mathscr{L}_1$, where $\mathscr{L}_n$ denotes the $n$-dimensional Lebesgue $\sigma$-Algebra.

Can somebody help me with this exercise? I already figured out that the statement does not hold if I replace the Lebesgue $\sigma$-Algebra by the Borel $\sigma$-Algebra but in the Lebesgue case I don't know how to show the existence of such a set $A$. I know that $\mathscr{L}_d \neq \mathscr{P}(\mathbb{R}^d)$, so there are non Lebesgue measurable sets.

Let $N$ be a subset of $[0,1]$ that is not Lebesgue measurable, and consider the set $A:=N\times\{0\}$. Now $A$ is Lebesgue measurable with measure zero since $A\subseteq [0,1]\times\{0\}$ and $[0,1]\times\{0\}$ has Lebesgue measure zero. However $f^{-1}(A)=N$ is Lebesgue nonmeasurable.