Is $\partial (A\times B)$ jordan measurable when both of $A$ and $B$ are jordan measurable?

Question

If $A\subseteq \mathbb{R}^{n} $ is Jordan measurable, $B\subseteq \mathbb{R}^{m} $ is Jordan measurable, then $A \times B \subseteq \mathbb{R}^{n+m}$ is Jordan measurable?

We have $$\partial (A\times B)=\left( \partial A \times \overline{B}\right)\cup\left( \partial B \times \overline{A}\right)$$

If $A\subseteq \mathbb{R}^{n}, B\subseteq \mathbb{R}^{m} $ are Jordan measurable, hence the Lebesgue measure of $\partial A $ and $\partial B $ is zero .(i.e. $m(\partial A )=0，m(\partial B )=0$.)

For every $\epsilon >0$, there is a finite mumber of rectangels $ R_{1},\cdots,R_{K} $ such that $\partial A\subset \bigcup_{i=1}^{K}R_{i}$ with $\sum_{i=1}^{K}\mathcal{v}(R_{i})<\epsilon .$ $\mathcal{v}(R_{i})$ is the volume of $R_{i}$. But we can not say $\sum_{i=1}^{k}\mathcal{v}(R_{i}\times \overline{B})<\epsilon.$

Similarly,

For every $\epsilon >0$, there is a finite mumber of rectangles $ S_{1},\cdots,S_{N} $ such that $\partial B\subset \bigcup_{i=1}^{N}S_{i}$ with $\sum_{i=1}^{N}\mathcal{v}(S_{i})<\epsilon .$$\mathcal{v}(S_{i})$ is the volume of $S_{i}$. But we can not say $\sum_{i=1}^{N}\mathcal{v}(S_{i}\times \overline{A})<\epsilon.$

So we must find other ways to clarify this question. I need some hints. Thanks for any help in advance.

Answer 1

Since $$ \partial (A\times B)=\left( \partial A \times \overline{B}\right)\cup\left( \partial B \times \overline{A}\right), $$ then $$ \mu(\partial (A\times B))\le\mu\left( \partial A \times \overline{B}\right)+\mu\left( \partial B \times \overline{A}\right)=\mu(A)\cdot\mu(\partial B)+\mu(B)\cdot\mu(\partial A)=0. $$