2
$\begingroup$

This has been asked before here and here but has not been answered correctly/completely.

Here's the problem:

Given, $A, B \subset \mathbb{R}$, where $A$ is not Lebesgue-measurable while $B$ has positive Lebesgue measure. Show that $A \times B$ is $\mathscr{L}^2$ non-measurable. However, if $B$ has zero measure, then $A \times B$ is $\mathscr{L}^2$ measurable.

Here $\mathscr{L}^2$ is the space of Lebesgue measurable sets in $\mathbb{R}^2$ which is obtained by completing the product sigma algebra $\mathscr{L}\otimes \mathscr{L}$, where $\mathscr{L}$ is the sigma algebra of Lebesgue measurable sets in $\mathbb{R}$

Attempt: I could prove that $A\times B$ is $\mathscr{L}\otimes \mathscr{L}$ measurable and hence $\mathscr{L}^2$ measurable when $m(B)=0$, since $A\times B \subset \mathbb{R}\times B$ and $m\times m( \mathbb{R}\times B)=m(\mathbb{R}) \times m(B)=0$, i.e. $\mathbb{R}\times B$ is a null set.

Since $\mathscr{L}^2$ is complete, we get $A \times B$ is $\mathscr{L}^2$ measurable.

How do I show the general statement?

$\endgroup$
1
  • $\begingroup$ If we replace Lebesgue measurability with Borel measurability everywhere in the problem statement, then $A \times B$ being measurable, and $B$ being nonempty implies that $A$ is measurable. When we deal with Lebesgue measurability, null sets get involved, but the same idea should work. $\endgroup$
    – Mason
    Nov 30, 2022 at 7:11

1 Answer 1

1
$\begingroup$

Let $\nu$ denote the outer measure on $\mathbb{R}$, and $\nu^2$ the outer measure on $\mathbb{R}^2$. Since $A$ is non-measurable, there is some $E\subseteq\mathbb{R}$ with $\nu(E)<\nu(E\cap A)+\nu(E\cap A^c)$. Now $E\times B$ is a subset of $\mathbb{R}^2$. We have $$(E\times B)\cap (A\times B)=(E\cap A)\times B$$$$(E\times B)\cap (A\times B)^c=(E\cap A^c)\times B.$$ Moreover, by Fubini's theorem, eg as argued here, we have $$\nu^2(A\times B)=\nu(A)\nu(B)\text{, and}$$$$\nu^2((E\cap A)\times B)=\nu(E\cap A)\nu(B)$$ $$\nu^2((E\cap A^c)\times B)=\nu(E\cap A^c)\nu(B)$$ Since $\nu(B)=\mu(B)>0$, it follows that $$\nu^2(A\times B)<\nu^2\big((E\times B)\cap (A\times B)\big)+\nu^2\big((E\times B)\cap (A\times B)^c\big),$$ whence indeed $A\times B$ is not measurable.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .