Let $A=\{(x,y) : x \in \mathbb{Q}, y \in \mathbb{R} \}$. Show that $m(A)=0$.

Solution 1:

Your set is a countable union of copies of the real line (indexed by $\mathbb{Q}$). The two-dimensional measure of a line in $\mathbb{R}^2$ is zero so your total measure is zero by countable additivity

Solution 2:

You have to conisder $(x_n-\frac {\epsilon} {2^{k+n}},x_n+\frac {\epsilon} {2^{k+n}})\times (k-1,k+1)$. Both $n$ and $k$ are varying.

If $q \in \mathbb Q$ and $ y \in \mathbb R$ then there exist $k$ such that $y \in (k-1,k+1)$ and $q=x_n$ for some $n$. Hence, $(q,y) \in (x_n-\frac {\epsilon} {2^{k+n}},x_n+\frac {\epsilon} {2^{k+n}})\times (k-1,k+1)$. Now check that the total measure of all these intervals is less than $4\epsilon$.