Does this Riemann integral over $[0,1]^2$ exist? While waiting for responses to this question, I did some research and came across this function on $[0,1]^2$: $f(x,y) = 0$ if $x$ or $y$ is irrational and $f(x,y) = 1/q$ if $x$ and $y$ are rational and $x = p/q$ in lowest terms.
Its claimed that the double Riemann integral $\int_{[0,1]^2}f $ exists since $f$ is continuous almost everywhere, but if $x$ is rational then $\int_0^1 f(x,y)dy $ does not exist as a Riemann integral.
I understand the second part since $f(x,y)$ looks like the Dirichlet function (when $x =p/q$ fixed) alternating between $1/q$ and $0$ for rational and irrational $y$.  Just because $f$ alternates between $0$ and a variable non-zero value off and on a rational grid does not make it completely obvious  about the continuity. 
So I would like to see how to prove the first part directly using Darboux sums.  
 A: Clearly, for any partition $P$ of $[0,1]^2$, the lower sum satisfies $L(P,f) = 0$ since irrationals are dense.
Given $\epsilon >0$, choose a positive integer $N > 1/\epsilon$. 
The set  $A_N = \{x \in \mathbb{Q} \cap [0,1]: x = p/q, (p,q) = 1, q \leqslant N \}$ is finite.  Here we have the rational numbers in $[0,1]$ where in lowest terms the denominator in $x = p/q$ is no bigger than $N$.  Let $m = \#(A_N)$.
An upper sum can be split into a sum over subrectangles (1) including and (2) excluding points $(x,y)$ with $x \in A_N$:
$$U(P,f) = \sum_{(1)}M_j \, vol(R_j) + \sum_{(1)}M_j \, vol(R_j) $$
where $M_j$ is the supremum of $f$ over the subrectangle $R_j$.
For the first sum, there are at most $4m$ subrectangles $R_j$ including points where $f(x,y) = 1/q \geqslant 1/N$. Thus $1/N \leqslant M_j \leqslant 1$. Choosing a partition where the largest subrectangle has content less than $\epsilon/m$  we have 
$$\sum_{(1)} \leqslant 4m \cdot 1 \cdot \sup_{R_j} vol(R_j) < 4\epsilon$$
For the second sum, the subrectangles contain no points with $x \in A_N$. Thus $M_j \leqslant 1/N < \epsilon$ and 
$$\sum_{(2)} \leqslant \epsilon \sum_{(2)} vol(R_j) < \epsilon.$$
Therefore, there is a partition $P$ such that $U(P,f) < 5\epsilon$ proving that $f$ is Riemann integrable and
$$\int_{[0,1]^2} f = 0$$ 
