# Cartesian product involving non-measurable Lebesgue sets

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?

• 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. Nov 30, 2022 at 7:11

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.