Integration over a strange set w.r.t. Lebesgue measure of $\mathbb{R}^2$ Let $A=\{(x,y)\in[0,1]^2:x+y\not \in \mathbb Q \;\& \;xy \not \in \mathbb Q\}$ 
Find $\int_A y^{-1/2}\sin x \;dm^2$, where $m^2$ is the Lebesgue measure of $\mathbb{R}^2$. 
 A: Put $f(x, y) = x + y$ and $g(x, y) = xy$. These are continuous functions, and hence measurable. Thus $F := f^{-1}(\mathbb{Q})$ and $G := g^{-1}(\mathbb{Q})$ are measurable as well. Notice that $F = \cup_{q \in \mathbb{Q}} f^{-1}(\{q\})$, and $G = \cup_{q \in \mathbb{Q}} g^{-1}(\{q\})$. The sets 
$$
\{(x, y) \in [0, 1]^2 : x +  y = q\} \quad \text{and} \quad \{(x, y) \in [0, 1]^2 : xy = q\} 
$$
are each of measure zero (argue this by noting that you can reduce to the case $q = 0, 1$ respectively, and use covers with rectangles of $\epsilon/2^n$ measure, for arbitrary $\epsilon > 0$.), and thus the sets $F, G$ are of measure zero. 
Now put $h(x, y) = y^{-1/2} \sin x$, and notice that your set $A = [0, 1]^2 \setminus (F \cup G)$, hence 
$$
\int_{A} h(x,y)  = \int_{[0, 1]^2} h(x,y).
$$
Notice that $\int_{[0, 1]^2} |h(x, y)| \leq \int_{[0, 1]^2} 1/\sqrt{y} = \int_0^1 y^{-1/2} = 2 < +\infty$, so by Fubini's theorem, you may integrate in any order: 
$$\int_A h(x, y) = \int_0^1 \int_0^1 y^{-1/2} \sin x = \left(\int_0^1 y^{-1/2} \right)\left(\int_0^1 \sin x\right) = 2(1-\cos(1))$$
by Riemann integration, for example (these agree since $\sin x$ and $1/y^{1/2}$ are continuous). 
A: Show that $[0,1]^2 \setminus A$ is a nullset. In fact, we have $[0,1]^2 \setminus A = B_1 \cup B_2$ with $$B_1 = \{(x,y) \in [0,1]^2 \colon x+y \in \mathbb{Q} \}$$ and $$B_2 = \{ (x,y) \in [0,1]^2 \colon xy \in \mathbb{Q}\}.$$
First Question: Why $B_1$ and $B_2$, resp. $A$, are measurable sets?
Second Question: Using Fubini's theorem and change of variables, what measure has $B_1$, resp. $B_2$?
Third Question: How we can now determine the value of the integral? (Why the integrat is the same over $[0,1]^2$?)
