Is Thomae's function Riemann integrable? Let $\displaystyle f: [0,1] \rightarrow \mathbb{R}$ given by 
$$f(x) = \begin{cases}
0 & x \notin \mathbb{Q} \\
\\
0 & x = 0 \\
\\
\frac{1}{q_x} & x = \frac{p_x}{q_x} \in \mathbb{Q} \backslash \{0\}, \ p_x \in \mathbb{Z}, \  q_x \in \mathbb{N}, \ \text{gcd}(|p_x|, q_x) = 1
\end{cases}$$
Is $\displaystyle f$ Riemann integrable?
I am trying to use the equivalent statements $\displaystyle g:[a,b] \rightarrow \mathbb{R}$ is Riemann integrable and $\displaystyle \forall \epsilon >0 \ \exists$ step functions $\displaystyle \rho, \psi$ with $\displaystyle \rho \leq g \leq \psi$ such that $\displaystyle \int_a^b (\psi - \rho) \leq \epsilon$.
I guess that means I would have to somehow show that given $\displaystyle \epsilon$, there exists only a finite amount of $\displaystyle x \in [0,1]$ with $\displaystyle f(x) \geq \epsilon$? 
Is it recommended that I consider something else instead? If not, how should I do this?
 A: 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.
A: 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.
A: For any $\epsilon > 0$, there are finitely many points in $[0,1]$ such that $R(x) \ge \epsilon / 2$ (specifically, the rational numbers $x$ with denominators $q_x \le 2 / \epsilon$).  You can construct a step function $\psi$ that is equal to $1$ when sufficiently near one of these points, equal to $\epsilon / 2$ otherwise, and whose total area is $\le \epsilon$.  That should let you conclude that $R$ is Riemann-integrable.
