If $X,Y\subset\mathbb{R}$ are not measure zero sets, how can I show that $X\times Y\subset \mathbb{R}^2$ is not a measure zero set too? If $X,Y\subset\mathbb{R}$ are not measure zero sets, how can I show  that $X\times Y\subset \mathbb{R}^2$ is not a measure zero set too?
or (the following is an easy case)
If $X\subset\mathbb{R}$ is not a measure zero set, how can I show that $X\times\mathbb{R} \subset \mathbb{R}^2$ is not a measure zero set too?
How can I show the assertion by using the definition of measure zero (as follows)?
(A subset $Z\subset \mathbb{R}$ is a mesuare zero set if $\forall\varepsilon>0$, $\exists$ countable open intervals $I_1, I_2, \cdots$ 
s. t. $Z\subset \cup_k I_k$ and $\sum_{k}length[I_k]<\varepsilon$). 
 A: We want to prove that if $X$ and $Y$ are Borel measurable sets and have nonzero measure, then $X \times Y$ has nonzero measure.
Take any set $X$ of nonzero measure and now ask yourself what sets $Y$ behave nicely. 
Of course if $Y$ is an interval your approach gives you the desired result. Suppose that $\lambda(X \times Y) = 0$ and choose intervals $I_k, J_k$ such that $$X \times Y \subset \bigcup_k I_k \times J_k$$ where the latter union is disjoint and it holds that: $\sum \lambda(I_k \times J_k) \le \epsilon.$ 
Without loss of generality you can take all $J_k = Y$ and you would get $\lambda(X)  \le \sum \lambda(I_k) \le \epsilon / \lambda(Y).$ So since $Y$ has nonzero measure you would get that $X$ has zero measure. This is a contradiction.
Now you can build the class of sets that behave nicely, i.e the class $\mathcal{M}$ of all the sets $Y$ such that if $X \times Y$ has measure zero than $Y$ must have measure zero: $$\mathcal{M} = \{ Y \subset \mathbb{R} \text{ measurable, s.t. either } \lambda(Y) = 0 \text{ or } \lambda(X \times Y) > 0\}$$
This class is closed under countable monotone unions and intersections and contains intervals. Hence it is a monotone class which contains intervals (which in turn form a $\pi-$system). Hence by the monotone class theorem it contains all Borel sets on the real line.
A: If $E\subset\mathbb{R}^{p+q}\quad$and$\quad mE=0$then $mE_x=0$a.e. in  $\mathbb{R}^p$
where$ E_x=\{y:y\in\mathbb{R}^p (x,y)\in E\}$
