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?

I'll post a slick proof here.
Let $d(r)$ denote the denominator of the rational number $r$. In other words, if $r=p/q$ holds for positive integer $q$ and integer $p$, where $p,q$ are co-prime, we have $d(r)=q$. Then $d(r)\ge1$ for all $r\in\Bbb Q$.
Suppose that $D(m)=\left\{\,x\,\big|\,0\le x\le1\land d(x)=m\,\right\}$, we have $D(1)=\{0,1\}$, and $0<|D(m)|<m$ for $m>1$, therefore
$$\sum_{x\in D(m)}\frac1{d(x)^4}=\sum_{x\in D(m)}m^{-4}<m^{-3}$$
Thus
$$\sum_{x\in [0\mathinner{..}1]}\frac1{d(x)^4}=2+\sum_{m>1}\sum_{x\in D(m)}\frac1{d(x)^4}<2+\sum_{m>1}m^{-3}=C$$
For all $0\le x,y\le1$ and $x,y\in\Bbb Q$, we draw a circle whose center is $(x,y)$ and radius is $(d(x))^{-2}(d(y))^{-2}\epsilon$. The total area is
\begin{align}
\sum_{\substack{0\le x,y\le1\\x,y\in\Bbb Q}}\frac{\pi\epsilon^2}{d(x)^4d(y)^4}
&=\pi\epsilon^2\sum_{\substack{0\le x,y\le1\\x,y\in\Bbb Q}}\frac1{d(x)^4d(y)^4}\\
&=\pi\epsilon^2\sum_{\substack{0\le x\le1\\x\in\Bbb Q}}\frac1{d(x)^2}\sum_{\substack{0\le y\le1\\y\in\Bbb Q}}\frac1{d(y)^2}\\
&\le\pi\epsilon^2C^2
\end{align}
Therefore the measure is not greater than $\pi\epsilon^2C^2$. Let $\epsilon\to0$, we get the answer.
Note The summation-interchanging works well because the terms are all nonnegative.
