# Borel Measurable Function but not Lebesgue Measurable

Can someone explain to me why this function $f\colon\mathbb R\to\mathbb R^2$ given by $f(x) = (x,0)$ is $\mathcal B$-$\mathcal B^2$-measurable but not $\mathcal L$-$\mathcal L^2$-measurable, where $\mathcal B$ and $\mathcal L$ denote the Borel and Lebesgue $\sigma$-algebra, respectively?

Regards

The map $f$ is continuous, so the preimage of a Borel subset of $\mathbb{R}^2$ will be a Borel subset of $\mathbb{R}$. Thus $f$ is Borel to Borel measurable.
Let $N\subseteq[0,1]$ be a Lebesgue nonmeasurable set. Then $N\times\{0\}$ is a Lebesgue measurable subset of $\mathbb{R}^2$ because the Lebesgue measure is complete, $N\times\{0\}\subseteq[0,1]\times\{0\}$, and $[0,1]\times\{0\}$ has Lebesgue measure zero. Then $f^{-1}(N\times\{0\})=N$ shows that $f$ is not Lebesgue to Lebesgue measurable.