outer measure and Cartesian product

Question

Let $A$ be a subset of $\mathbb{R}^n$, and let $B$ be a subset of $\mathbb{R}^m$. Note that the Cartesian product $\{(a, b) : a \in A, b\in B\}$ is then a subset of $\mathbb{R}^{n+m}$. Show that $m^{*}_{n+m} (A \times B) \le m^{*}_n(A)m^{*}_m(B)$.

$m^*_n$ denotes an n-dimensional outer measure.

The definition of an outer measure is that $m^*_{n+m}(A \times B) = \inf \{\sum_{j \in J} \text{vol}(C_j) : A \times B \subseteq \bigcup_{j \in J} C_j\}$ and similarly $m^{*}_n = \inf \{\sum_{i \in I} \text{vol}(A_i) : A \subseteq \bigcup_{i \in I} A_i\}$ and $m^{*}_m(B) = \inf \{\sum_{k \in K} \text{vol}(B_k) : B \subseteq \bigcup_{k \in K} B_k\}$.

I can also observe that the Cartesian product of a cover of $A$ and a cover of $B$ covers $A \times B$, but I am not sure how to proceed from here. I appreciate if you give some help.

Does this answer your question? Outer measure of product of sets — qqo, Apr 05 '20 at 19:55
Yes it answers my question, but I don't understand one thing there. How do we know that $\sum_{i,j} \mu^(A_i \times B_j) = \sum_{i,j} \mu^(A_i) \mu^*(B_j) $ ? — shk910, Apr 05 '20 at 22:37
@shk910 You should probably ask that as a separate question. — Chill2Macht, Apr 06 '20 at 02:15
Remember that $A_i$ and $B_j$ are rectangles. What is their area? — qqo, Apr 06 '20 at 06:26
$\text{vol}(A_i) = \prod_{k=1}^n(b_{i,k} - a_{i,k})$ given that $A_i = {(x_1, ..., x_n): x_k \in (a_{i,k}, b_{i,k}) }$, and similarly, $\text{vol}(B_j) = \prod_{k=1}^n(b_{j,k} - a_{j,k})$, but I am not sure about $\text{vol}(A_i \times B_j)$. — shk910, Apr 06 '20 at 06:40

outer measure and Cartesian product

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