Let $A\subset\mathbb{R}^p$ and $B\subset\mathbb{R}^q$, it’s not difficult to show that $$m^\*(A\times B)\leq m^\*(A)\cdot m^\*(B)$$, where $m^\*()$stands for the outter measure in Lebesgue meaning.


If A and B are measurable, then "=" holds. My question is whether "=" holds for all of/ none of/ some of the non-measurable A and B?


Construction of A and B in these different cases is needed.



Product of two non-measurable sets
For simplicity, assume $p = q = 1$ and $A, B \subseteq [0, 1]$. Let $\mu\_k, \mu\_k^{\star}$ denote the $k$-dimensional Lebesgue measure, outer measure respectively. Let $\{U(n) : n \geq 1\}$ be a sequence of decreasing open sets each containing $A \times B$ such that $\mu\_2(\bigcap\_n U(n)) = \mu\_2^{\star}(A \times B)$. For every $x \in A$, $B \subseteq U\_x(n) = \{y: (x, y) \in U(n)\}$ so $\mu\_1(U\_x(n)) \geq \mu\_1^{\star}(B)$. So the (measurable) set $\{x : \mu\_1(U\_x(n)) \geq \mu\_1^{\star}(B)\}$ contains $A$ and hence has measure $\geq \mu\_1^{\star}(A)$. By Fubini's theorem, $\mu\_2(U(n)) \geq \mu\_1^{\star}(A)\mu\_1^{\star}(B)$. It follows that $\mu\_2^{\star}(A \times B) \geq \mu\_1^{\star}(A)\mu\_1^{\star}(B)$.


