2
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.