# Countable Dense Set of Discontinuities and Riemann-Integrability (Rudin Question) [duplicate]

Possible Duplicate:
Series with fractional part of $nx$

I am working through Question 10 from Chapter 7 in Baby Rudin and I am very confused.

Letting (x) denote the fractional part of the real number $x$, consider the function $$f(x)=\sum_{n=1}^{\infty} \frac{(nx)}{n^2}, \; x \in \mathbb{R}$$ Find all discontinuities of $f$, and show that they form a countable dense set. Show that $f$ is nevertheless Riemann-integrable on every bounded interval.

I have looked at the solution (here) and my question is the following:

• Why is $f(x)$ discontinuous on all rationals and continuous on all irrationals?

If I can get a cogent explanation for this fact then I understand the question and the proof. Any help is greatly appreciated!

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## marked as duplicate by Norbert, Brian M. Scott, tomasz, Nate Eldredge, BenjaLim Oct 6 '12 at 17:54

There you can find the answer to your question – Norbert Oct 2 '12 at 7:16
@Norbert The answer in that question uses Lebesgue's criterion which is usually unfamiliar to someone who's just studied Riemann integrability. In Baby Rudin, the criterion is introduced in a later chapter. – Ayman Hourieh Oct 2 '12 at 9:26

Here is a plot of the fractional part of $x$:
Notice how it's discontinuous when $x$ is an integer, and continuous everywhere else. To prove this formally, notice that on $[n, n + 1) : n \in \mathbb{N}$, this function is equal to $x - n$. Only at integer values does this function make a jump.
$(nx)$ is only slightly different from $(x)$. Instead of being discontinuous at integers $\mathbb{Z}$. It's discontinuous at $\frac{1}{n}\mathbb{Z}$.
For a fixed $n$ and bounded interval, $f_n = \dfrac{(nx)}{n^2}$ has a finite number of such discontinuities. Thus, it's Riemann integrable. It follows that the uniform limit of $\displaystyle \sum_{n=1}^\infty f_n$ is also Riemann integrable on a bounded interval.