Is Thomae's function Riemann integrable?

Solution 1:

Using that given $\epsilon$, $R(x) > \epsilon$ for only finitely many $x_i\in [0,1]$ is a step in the right direction. Think about how to make the intervals containing those $x_i$ arbitrarily small, and $\epsilon$ arbitrarily small at the same time (hint: the size of the intervals should depend on [be bounded by] $\epsilon$).

There is a complete version of this proof in Analysis by Steven Lay on page 278.

edit: I have left out many details and I can give more info about how to go about it if you get stuck.

Solution 2:

Since you asked for recommendations: It is Riemann integrable because it is continuous almost everywhere (at the irrationals) and bounded. Of course, that is not how that Riemann integrability of this function was originally proved (this is an important example/counter-example function), but the result about continuity almost everywhere is quite important.

Since this is homework: here is a hint using step functions. See Exercise 5.24.