1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ $f$ is nonnegative and is zero a.e. so its integral is zero a.e. $\endgroup$ – Math1000 Feb 29 '16 at 1:46
3
$\begingroup$

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 $\int_Q f(x,y)\>{\rm d}(x,y)=0$.

$\endgroup$
1
$\begingroup$

If $a,b\in[0,1]$ are irrational, then $f$ is continuous in $(a,b)$. Given $\epsilon>0$, the number of rationals in $[0,1]$ with denominator (when written as $p/p$ an irreducible fraction) larger than than $1/\epsilon$ is finite. Then the exusts $\delta>0$ 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 conutable union of sets of measure $0$, so it has also measure $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.