I've been stuck on the following questions for quite a while now, there doesn't seem to be much material to read on double integrals of piecewise functions. If someone could please help me out.
Let Q = [0, 1] × [0, 1] and define f : Q → R by letting f(x, y) = $\begin{cases} \frac{1}{q_x} &\text {if x,y $\in$ rationals}\\ 0, & \text {otherwise} \end{cases}$
We have for all y $\in$ [0,1] $\int_{0}^1 f(x,y)\,dx$ = 0 However, for x ∈ rationals ∩ [0, 1] the function f(x, y) is equal to $\frac{1}{q_x}$ for rational y and is 0 otherwise. Thus the integral $\int_{0}^{1} f(x,y) dy $ does not exist. My question is can't we just switch x and y in the above paragraph and the same argument can be applied to the dx integral to show that it too doesn't exist.
I don't have enough reputation so I will write my comment as an answer:
There is a mistake on your example. The integral $\int_0^1 f(x,y)dy$ exists because the set $\mathbb{Q}\cap [0,1]$ (where $f(x,y)\neq 0$) have measure zero in $[0,1]$. Futhermore, $\int_0^1 f(x,y)dy=0$.