Product of outer measure equals outer measure of product

Question

Define the outer measure $M^*$ on $\mathbb R$ by $ M^*(A) =inf\{\sum|I_k|: {I_k} $is an covering of A by open intervals $ \} $, Similarly the outer measure $M^*$ on $\mathbb R^2$ is defined by $ M^*(A) =inf\{\sum|R_k|: {R_k} $is an covering of A by open rectangles $ \} $. Suppose $A \subset \mathbb R$, $B \subset \mathbb R$, is it true that $ M^*(A \times B) = M^*(A)M^*(B)$? It is trivial that $ M^*(A \times B) \leq M^*(A)M^*(B)$, but I am not sure if $ M^*(A \times B) \ge M^*(A)M^*(B)$ is ture.

Answer 1

$\mu_n$ (resp. $\mu_n^{\star}$) denotes the $n$-dimensional Leb. measure (resp. outer measure). WLOG $A, B$ are bounded sets of reals. Let $G$ be a $G_{\delta}$ set containing $A \times B$ with $\mu_2^{\star}(A \times B) = \mu_2(G)$. Then for every $x \in A$, $G_x = \{y : (x, y) \in G\} \supseteq B$. Hence $\mu_1(G_x) \geq \mu_1^{\star}(B)$. Now use Fubini's theorem.