Let's call $\mu_{n}$ the Lebesgue measure on $\mathbb R^n$. Prove that if $A \subset \mathbb R^n$ and $B\subset \mathbb R^m$ are open subsets then $ \mu_{n+m}(A\times B) = \mu_n(A)\mu_m(B). $
In other words, this tells us that the Lebesgue $(n+m)$-dimensional measure and the product measure of $\mu_n$ and $\mu_m$ agree on open sets. How can I do? Suppose that $(A_i)_{i \in \mathbb N}$ and $(B_i)_{i \in \mathbb N}$ are sequences s.t. $ A \subseteq \bigcup_{i} A_i, \qquad B \subseteq \bigcup_{j} B_j $ Then $ A \times B \subseteq \bigcup_{i} A_i \times\bigcup_{j} B_j $ Now may I write - by $\sigma$-subadditivity $ \mu_{n+m}(A \times B) \le \sum_{i,j} \mu_n(A_i)\mu_m(B_j) $ Is it correct? Now I would like to conclude - by taking $\inf$ on RHS - that $\mu_{n+m}(A \times B) \le \mu_n(A)\mu_m(B)$: how can I do? And what about the opposite inequality?
Bonus (self-posed) question: what happens if we remove the hypothesis $A,B$ are open? For example, is it true that if $\mu_n(A)<\infty$ and $\mu_m(B)<\infty$ then $ \mu_{n+m}^{\star}(A \times B) = \mu_n(A)\mu_m(B) $ where $\mu^{\star}_{n+m}$ is the outer $(n+m)$-dimensional Lebesgue measure? Thanks in advance.