Prove $f:[0,1]\times[0,1]\to \mathbb R$ is integrable. Let $f:[0,1]\times[0,1]\to \mathbb R$,
$$f(x,y)=
\begin{cases}
\frac1q+\frac1n,  & \text{if $(x,y)=(\frac mn,\frac pq) \in \Bbb Q\times\Bbb Q,$   $  (m,n)=1=(p,q)$ } \\
0, & \text{if $x$ or $y$ irrational$ $ or $0,1$}
\end{cases}
$$
Prove that f is integrable over $R=[0,1]\times[0,1]$ and find the value of the integral (I know its value is zero, because every lower sum is zero).
I'm trying to find the set of discontinuities of $f$ over $R$ and prove that it has measure zero, so that $f$ is integrable.
I remember doing this for the one dimensional case (Thomae´s function), proving that $f$ was continuous over the irrationals and discontinuous over the rationals, but I can't prove it this time, so I need some help, it will be really appreciated.
 A: Let $u\mapsto T(u)$ $\>(0\leq u\leq1)$ be Thomae's function. Then
$$0\leq f(x,y)\leq T(x)+T(y)\qquad\bigl((x,y)\in Q:=[0,1]^2\bigr)\ .$$
By "Fubini's theorem" for Riemann integrals one obtains
$$\int_Q T(y)\>{\rm d}(x,y)=\int_0^1 \int_0^1 T(y) dy\ dx=0\ ,$$
and similarly for $(x,y)\mapsto T(x)$. It follows that $\int_Q f(x,y)\>{\rm d}(x,y)=0$.
A: If $a,b\in[0,1]\setminus\mathbb{Q}$, then $f$ is continuous in $(a,b)$. Given $\epsilon>0$, the number of rationals in $[0,1]$ with denominator (when written as $p/q$ an irreducible fraction) larger than than $1/\epsilon$ is finite. Then there exists $\delta>0$ such that if $|x-a|,|y-b|<\delta$, then $f(x,y)=0$ (if one of $x$ or $y$ is irrational) or less than $2\,\epsilon$ (if both $x$ and $y$ are rational.) In any case, $|f(x,y)-f(x,a)|\le2\,\epsilon$. The set of discontinuities of $f$ is contained in
$$
D=\bigcup_{r\in\mathbb Q}\Bigl(\{r\}\times[0,1]\cup[0,1]\times\{r\}\Bigr).
$$
$D$ is the countable union of sets of measure $0$, so it has also measure $0$.
