# Is the cartesian product of a bounded set with a set of points, both in $\mathbb{R}^{n}$ Jordan-Measurable over $\mathbb{R}^{2n}$?

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 $$\mathbb{R}^{n}$$

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 $$\le n\epsilon^n\,m(D)$$.