If $A\subseteq R^n$ and $B\subseteq R^m$, such that $A \times B\subseteq R^{n+m}$ Prove that $μ^{*}_{n+m}(A\times B)\leq μ^*_n(A)μ^*_m(B)$, where $μ^*_q$ is the outer measure of $ \mathbb{R}^q $.
My attempt $A⊆⋃_iA_i,B⊆⋃_jB_j $, and since $A \times B \subseteq ⋃_{i,j}A_iB_j$ I have the inequality (because of the outer measure monotocity) that states
$m^*_{n+m}(A \times B) \leq m^*_{n+m}(⋃_{i,j}A_iB_j) $ But I don't think that is going to take me somewhere.
Thanks!
I think that the critical point you have to show is that the inequality holds for $A,B$ bounded. After that, one uses some standard argument on $\sigma$-finiteness.
Clearly, if $A,B$ are rectangles, the inequality holds trivially (in fact equality holds).
Now, for all $\varepsilon >0$, take some rectangles $A_i$, $B_j$ covering $A,B$ such that $$\mu^*(A) \geq \sum_i \mu^*(A_i) - \varepsilon \qquad \mbox{ and } \qquad \mu^*(B) \geq \sum_j \mu^*(B_j) - \varepsilon$$
Then $$A \times B \subset \bigcup_{i,j} A_i \times B_j$$ moreover $$\mu^* \left( \bigcup_{i,j} A_i \times B_j \right) \leq \sum_{i,j} \mu^*(A_i \times B_j) = \sum_{i,j} \mu^*(A_i) \mu^*(B_j) =$$ $$= \left( \sum_i \mu^*(A_i) \right) \left( \sum_j \mu^*(B_j) \right) \leq (\mu^*(A)+ \varepsilon )(\mu^*(B)+ \varepsilon )$$
Since $A,B$ have finite measure, you can send $\varepsilon \to 0$, and get the inequality.
