# 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

• Since I'm making a substantial edit to the title over 4 years after the question was asked and answered, I'm putting in a comment about it here (anticipating that the edit will be approved): The usual meaning of Lebesgue measurability for functions is Lebesgue–Borel, but here we want Lebesgue–Lebesgue. Indeed, the question title confused me at first, since every Borel–Borel measurable function is Lebesgue–Borel measurable, but not Lebesgue–Lebesgue measurable. Having put in the missing second half in the Lebesgue case, I put it in the Borel case too, for symmetry. Commented Oct 1, 2020 at 18:12

## 1 Answer

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.