2
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$

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.