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Let $A=\{\vec{x_{1}},\dots,\vec{x_{k}}\}\subset\mathbb{R}^{n}$. And let $B\subset\mathbb{R}^{n}$ a bounded set. Then $A\times B$ is Jordan-Measurable on $\mathbb{R}^{2n}$?

In the beginning, I thought that the counterexample could be $\{\frac{1}{2}\}\times ((\mathbb{Q}\times\mathbb{Q})\bigcap([0,1]\times [0,1]))$ But then I realized this doesn't works. I started to think that maybe this is true. I will appreciate any hint.

$\vec{x_{i}}$ is a vector on $\mathbb{R}^{n}$

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Since $B$ is bounded, it fits inside some $n$-rectangle $D$. Each $\vec x_k$ can be placed in an $n$-cube $C_k$ of sidelength $\epsilon$ for any $\epsilon > 0$. Which means that $$A \times B \subseteq \bigcup_k C_k \times D$$ whose Jordan measure is $\le n\epsilon^n\,m(D)$.

Can you take it from there?

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