# Product of two non-measurable sets

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.

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 $\mu_2^{\star}(A \times B) \geq \mu_1^{\star}(A)\mu_1^{\star}(B)$.