Question on Borel Set $X = \Bbb{Q} \times (\Bbb{R} \setminus \Bbb{Q})$

Question

I came across this exercise, but can't find a way to do it :

Argue that the set $X = \Bbb{Q} \times (\Bbb{R} \setminus \Bbb{Q})$, which consists of points in $\Bbb{R}^2$ with the first coordinate rational and the second coordinate irrational, is Borel and satisfies λ2(X) = 0.

You are not allowed to use Tonelli's or Fubini's theorems

Thanks for the help !

Answer 1

Hint:

We can write $X$ as a countable union of sets:$$X=\bigcup_{q\in\mathbb Q}\left(\{q\}\times(\mathbb R\setminus\mathbb Q)\right)$$

So it is enough to prove that $\{q\}\times(\mathbb R\setminus\mathbb Q)\subseteq\mathbb R^2$ is a Borel set having Lebesgue measure $0$ for every $q\in\mathbb Q$.

Answer 2

$X=(\mathbb Q \times \mathbb R) \setminus (\mathbb Q \times \mathbb Q)$. $\mathbb Q \times \mathbb R$ is a Borel set because it is the union of the closed sets $\{q\}\times \mathbb R $ and $ \mathbb Q \times \mathbb Q$ because it is the union of the closed sets $\{q\}\times \{q\} $( $q$ varying over $\mathbb Q$).

Now $\lambda_2 (\{q\}\times \mathbb R) =(0)(\infty) =0$ for all $q$. Taking union over $q$ we see $\lambda_2 (\mathbb Q\times \mathbb R) =0$. that Since the given set is as subset of $\mathbb Q\times \mathbb R%$ it is also of measure $0$.