The measure of $([0,1]\cap \mathbb{Q})×([0,1]\cap\mathbb{Q})$ We know that $[0,1]\cap \mathbb{Q}$ is a dense subset of $[0,1]$ and has measure zero, but what about $([0,1]\cap \mathbb{Q})\times([0,1]\cap \mathbb{Q})$? Is it also a dense subset of $[0,1]\times[0,1]$ and has measure zero too?
Besides, what about its complement? Is it dense in $[0,1]\times[0,1]$ and has measure zero?

Any countable set is of zero Lebesgue measure, and the product of (finitely many) countable sets is countable. Also, a product of finitely many dense sets is dense. Thus, the answer is yes.
