Lebesgue outer measure of product of sets

Question

Let $\lambda^\ast$ be the Lebesgue outer measure on $\mathbb{R}$, and $\lambda^\ast_2$ the Lebesgue outer measure on $\mathbb{R}^2$, defined as the infimum of the sum of the rectangles covering a set. Let $A,B\subseteq\mathbb{R}$ be subsets such that $\lambda^\ast(A)=\infty$ and $\lambda^\ast(B)>0$. Then is it true that $\lambda^\ast_2(A\times B)=\infty$ ? It seems intuitively obvious, but I can't seem to prove it. I would prefer a proof which is as elementary as possible.

Answer 1

I don't know if it is true for any sets $A,B\subseteq\mathbb{R}$, but I think I can prove it when $A$ and $B$ are Lebesgue-measurable. The idea is that we first prove \begin{equation} \lambda(A\times B)=\lambda(A)\lambda(B) \tag{$\ast$}\label{ab} \end{equation} when $\lambda(A)<\infty$ and $\lambda(B)<\infty$. Then with $A_n=A\cap[-n,n]$ and $B_n=B\cap[-n,n]$, $$\lambda(A\times B)=\lambda\left(\bigcup_n(A_n\times B_n)\right)=\lim_{n\to\infty}\lambda(A_n\times B_n)=\lim_{n\to\infty}\lambda(A_n)\lambda(B_n)=\infty.$$ To prove \eqref{ab} when $\lambda(A)<\infty$ and $\lambda(B)<\infty$, we first prove $$\lambda(A\times B)\leq\lambda(A)\lambda(B),$$ which follows easily from the definition of outer measure. (In fact, it holds even when $A$ or $B$ is not Lebesgue-measurable.) Here is a link for the proof. For the opposite inequality, we first prove it when $A$ and $B$ are open sets, then use the regularity of Lebesgue-measurable sets. The same link shows a proof, although it contains some errors, which can be easily fixed.