1
$\begingroup$

I have the following function (sorry for the vague title but I don't know if this kind of function has some special name) defined by:

$$f(x)= \begin{cases} 0 & \text{if } x\in\mathbb R \smallsetminus\mathbb Q, \\[6pt] \frac 1n & \text{if } x=\frac mn\in\mathbb{Q},\,\, \gcd(m,n)=1.\end{cases}$$

first I have to prove that this function is continuous only at the irrational points and I see why but I can't properly formalize it rigorously, then I have to prove that it is Riemann integrable on $[0,1)$ and that said integral is $0$ and I really don't know how to approach this.

$\endgroup$
  • $\begingroup$ This is variously known as Thomae's function or the popcorn function, or a number of other things. (No idea why Riemann's name would be attached to it.) $\endgroup$ – Chappers May 29 '17 at 16:47
  • $\begingroup$ @Chappers the question is to prove whether or not the function is Riemann integrable. The problem doesn't state that the function itself is associated with Riemann. $\endgroup$ – Tucker May 29 '17 at 16:51
  • $\begingroup$ @Tucker I'm commenting on one of the other names listed in the link. $\endgroup$ – Chappers May 29 '17 at 16:53
1
$\begingroup$

If $x\in\mathbb Q$, then take a sequence $(x_n)_{n\in\mathbb N}$ of irrational numbers such that $\lim_nx_n=x$. Then $\lim_nf(x_n)=0\neq f(x)$ and therefore $f$ is not continuous at $x$.

If $x\notin\mathbb Q$, then $f(x)=0$. Given $\varepsilon>0$, there are only finitely many points $x\in[0,1]$ such that $f(x)\geqslant\varepsilon$. So, take an interval $(x-\delta,x+\delta)$ so small that it contains no such number. This assures that $|y-x|<\delta\Longrightarrow|f(y)-f(x)|=|f(y)|<\varepsilon$.

Given $\varepsilon>0$, and given and partition $P$ of $[0,1]$, every lower sum $\underline\Sigma(f,P)$ is equal to $0$. If you prove that there is an upper sum $\overline\Sigma(f,P)$ smaller that $\varepsilon$, this will prove that the Riemann integral exits and that it is equal to $0$. Again, this follow from the fact that there are only finitely many points $x\in[0,1]$ such that $f(x)\geqslant\varepsilon$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I got the general idea but I can't fully understand the part when you say that "there are only finitely many points for which..." $\endgroup$ – Renato Faraone May 29 '17 at 16:57
  • $\begingroup$ I use that twice: in order to prove that $f$ is continuous in $\mathbb{R}\setminus\mathbb Q$ and in order to prove that $f$ is Riemann integrable. Which one do you have in mind? $\endgroup$ – José Carlos Santos May 29 '17 at 16:59

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.