Let $X\subset \mathbb R^m$ be a set with measure zero, i. e., for every $\epsilon>0$ there is a covering of blocks $\{C_i\}_{i=1}^{\infty}$ such that $\sum_{i=1}^{\infty} \text{vol}\ C_i<\epsilon$. I'm trying to prove for every $Y\subset\mathbb R^n$ the $X\times Y$ has measure zero in $\mathbb R^{m+n}$
For every $\epsilon>0$, I'm trying to find a covering for $X\times Y$ without success. I need a hint to begin to tackle this question
first prove that a countable union of sets of measure zero has measure zero:
Suppose that $A_1,A_2,\dots$ all have measure zero, pick boxes $B_{n,1}, B_{n,2},\dots$ that cover $A_n$ and have sum of volumes less than $\epsilon/2^n$. Then the boxes of the form $B_{n,m}$ cover $\bigcup\limits_{n=1}^\infty A_n$ and the sum of their volumes is at most $\epsilon/2+\epsilon/4+\dots=\epsilon$.
Using this we prove that $X\times \mathbb R^n$ has measure zero.
First cut $\mathbb R^n$ into unit cubes $C_1,C_2,\dots $. It is easy to prove that $X\times C_k$ has measure zero because if $B_1,B_2,\dots $ are cubes in $\mathbb R^m$ that cover $X$ and have sum of volumes less than $\epsilon$ then the cubes $B_1\times C_k,B_2\times C_k\dots$ have the same sum of volumes and cover $X\times C_k$.
Since $X\times C_k$ has measure zero for all $X$ and $X\times \mathbb R^n=\bigcup\limits_{n=1}^\infty X\times C_n$ we conclude $X\times \mathbb R^n$ has measure zero. It follows directly that $X\times Y$ also has measure zero for all $Y\subseteq \mathbb R^n$
Let $\epsilon > 0$. Then, there is a covering $\{C_k\}_{k=1}^{\infty}$ of $X$ such that $$V(C_k)<\frac{\epsilon}{2^{k+1}(2k)^n}$$ for all $k\in \mathbb{N}$, so that $$\sum_{k=1}^{\infty}V(C_k)<\sum_{k=1}^{\infty}\frac{\epsilon}{2^{k+1}(2k)^n}=\frac{\epsilon}{(2k)^n}<\epsilon.$$
Now, let $Y \subset \mathbb{R}^n$. Cover $\mathbb{R}^n$ by the rectangles $[-k,k]\times \cdots \times [-k,k]$, $k \in \mathbb{N}$. In particular, this is a covering of $Y$. Now let $R_k = C_k \times [-k,k]\times \cdots \times [-k,k]$, $k \in \mathbb{N}$. We have $$ X \times Y \subset \bigcup_{k=1}^{\infty} R_k.$$ Furthermore, $V(R_k)=V(C_k) \cdot (2k)^{n}$ for all $k$. Thus, $$V(R_k)=V(C_k) \cdot (2k)^{n}<\frac{\epsilon}{2^{k+1}(2k)^n}\cdot (2k)^{n}=\frac{\epsilon}{2^{k+1}}.$$ Therefore $$\sum_{k=1}^{\infty}V(R_k) = \sum_{k=1}^{\infty}V(C_k) \cdot (2k)^{n}<\sum_{k=1}^{\infty}\frac{\epsilon}{2^{k+1}} = \epsilon.$$