Let $A\subset \mathbb{R}^2$ be a closed set of positive Lebesgue measure. Can we find positive Lebesgue measure sets $A_1,A_2\subset \mathbb{R}$ such that $A_1\times A_2\subseteq A$?
Note that the above is not true if $A$ is not assumed to be closed. For example $$A=[0,1]\times [0,1]\setminus \{(x,y)\in [0,1]\times [0,1]:x-y\in \mathbb{Q}\}.$$
Best Answer
You already have an example of a set $A$ of positive measure such $A _1 \times A_2$ is not contained in $A$ for any $A_1$ and $A_2$ of positive measure. By regularity of Lebesgue measure on $\mathbb R^{2}$ $A$ contains a compact set $K$ of positive measure. It follows that $A _1 \times A_2$ is not contained in $K$ for any $A_1$ and $A_2$ of positive measure.