Is Thomae's function Riemann integrable?

Question

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?

Your "guess" sounds like a good approach to me. Hint: given $\epsilon$, choose an integer $n$ so large that $1/n < \epsilon$. For which $x$ is $R(x) \ge 1/n$? Try it for the first few values of $n$. Then, if there are only finitely many $x$ with $R(x) \ge \epsilon$, try constructing a step function $\psi$ such that $0 \le R \le \psi$ and $\int_0^1 \psi \le 2 \epsilon$. — Nate Eldredge, Jan 12 '11 at 20:16

AnonymousCoward · Accepted Answer · 2011-01-13T01:18:23.333

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.

score 2 · Answer 2 · edited Oct 16 '11 at 00:58

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.

edited Oct 16 '11 at 00:58

J. M. ain't a mathematician

76,540

answered Jan 12 '11 at 20:07

Aryabhata

83,877

Is there a name for this function? If so, I could try and find out more background information... – ghshtalt Jan 12 '11 at 20:11
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@user: Some people call it the Modified Dirichlet Function. – Aryabhata Jan 12 '11 at 20:15
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@user: Instead of trying to research this function, I suggest you think about it! It is worth thinking about and easy enough to be assigned as homework. – Aryabhata Jan 12 '11 at 20:42

score 1 · Answer 3 · answered Jan 12 '11 at 20:18

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.

Is Thomae's function Riemann integrable?

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