# If $X,Y \subset \mathbb{R}$ are measure zero sets, how can I show that $X \times Y \subset \mathbb{R^2}$ is a measure zero set too?

If $X,Y \subset \mathbb{R}$ are measure zero sets, how can I show that $X \times Y \subset \mathbb{R^2}$ is a measure zero set too?

My outline is the following:

Since $X,Y$ is a measure zero set, we can find open intervals $U_i$ and $V_j$ for $X$ and $Y$, respectively. Then, the union of the open intervals is a cover for $X$ and $Y$. That part I have been able to do. However, I need to show now that the sum of the measures of $(U_i \times V_j)$ can be written less than any arbitrary $\epsilon >0$. That is:

$$\sum_{i,j}|(U_i \times V_j)| < \epsilon$$

Since by definition of $X,Y$ being a measure zero set, we have

$$\sum_{i}|U_i| < \sqrt{\epsilon} \\ \text{and} \sum_{j}|V_j| < \sqrt{\epsilon}$$

It then makes logical sense that we want

$$\sum_{i,j}|(U_i \times V_j)| < \sum_{i}|U_i| \cdot \sum_{j}|V_j| < \sqrt{\epsilon} \cdot \sqrt{\epsilon} < \epsilon$$

Does anyone know how I can get the relation:

$$\sum_{i,j}|(U_i \times V_j)| < \sum_{i}|U_i| \cdot \sum_{j}|V_j|$$

by something simple like the triangle inequality without having to resort to heavy measure theory? Thanks!

$U_i \times V_i$ is an open rectangle. Use the formula for the measure of a rectangle, and apply the distributive law:
$$\sum_{i=1}^m\sum_{j=1}^na_ib_j = \left(\sum_{i=1}^ma_i\right)\left(\sum_{j=1}^jb_j\right)$$
If $X\subseteq\mathbb{R}$ has measure zero, the product will have measure zero no matter how large $Y$ is. So morally, the proof should not depend on $Y$ having measure zero.
It clearly suffices to show that $X\times\mathbb{R}$ has measure zero. Since $$X\times\mathbb{R}=\bigcup_{n\in\mathbb{N}}\big(X\times [-n,n]\big),$$ and the countable union of measure zero sets is a set of measure zero, it suffices to show that $X\times [-n,n]$ has measure zero for all $n$.
So take any $\epsilon>0$. Since $X$ has measure zero, there is a sequence of open intervals $(O_k)$ such that $X\subseteq\bigcup_k O_k$ and the sum of the length of the intervals is less than $(2n+1)\epsilon$. But then the sum of the areas of the products of open intervals $O_k\times (-n-1/2,n+1/2)$ will be less than $\epsilon$ and $$X\times[-n,n]\subseteq\bigcup_kO_k\times (-n-1/2,n+1/2).$$