Let we have the function $f:[0,1]\times[0,1]\rightarrow\mathbb{R}$ where $f(x,y)=1$ when $x\in \mathbb{Q}\cap[0,1], y\in [0,1]$ and $f(x,y)=0$ in the other cases. First of all, I have to prove that it is Lebesgue measurable, and then, I have to calculate $\iint_{[0,1]\times[0,1]}f$.
This is what I've done:
$\{(x,y): x\in \mathbb{Q}\cap[0,1], y\in [0,1] \}=\{(x,y): x\in \mathbb{Q}\cap[0,1], y\in \mathbb{Q}\cap[0,1] \} \cup\{(x,y):x\in \mathbb{Q}\cap[0,1], y\in \mathbb{I}\cap[0,1]\}$
If the measure of those two sets is $0$, I could say that $f\equiv 0$ almost everywhere. So $\iint_{[0,1]\times[0,1]}f=0$.
The measure of $\{(x,y): x\in \mathbb{Q}\cap[0,1], y\in \mathbb{Q}\cap[0,1] \}$ is $0$ owing to the fact that the measure of $\mathbb{Q}$ is $0$ $(m(\mathbb{Q})=0)$.
But how can I demostrate that the measure of the second set is $0$??? I mean, how I can demostrate that $m(\{(x,y):x\in \mathbb{Q}\cap[0,1], y\in \mathbb{I}\cap[0,1]\})=0$???
We have $f(x) = I_A(x)$ where $I_A(x)$ is an indicator function and $A = \{ (x,y) \in [0,1]^2 | x \in \mathbb{Q} \}$. Hence $\int_{[0,1]^2}f(x) dx = \mu(A)$. Consider $[0,1] \cap \mathbb{Q} = \{x_1, x_2, \ldots \}$. Then $\mu(A) = \mu(\bigsqcup_{i}^{\infty} x_i \times [0,1] ) = \sum_{i}^{\infty} \mu(x_i \times [0,1]) = 0$ because $\mu(y \times [0,1])= \mu(\{ y\})\cdot \mu([0,1]) = 0 \cdot 1 = 0$ as a measure of rectangle for all $y \in \mathbb{R}$.