# Proving that Thomae's function is Riemann integrable

I am self-learning Real Analysis from the text Understanding Analysis by Stephen Abbott. Exercise 7.3.2 asks to show that Thomae's function is integrable. I would like a hint/clue for part (c) of this exercise problem, without giving away(revealing) the entire proof.

[Abbott 7.3.2] Recall that Thomae's function

$$\begin{equation*} t( x) =\begin{cases} 1 & \text{ if } x=0\\ 1/n & \ \text{if} \ x=m/n\in \mathbf{Q} \setminus \{0\} \ \text{is in its lowest terms with} \ n >0\\ 0 & \ \text{if } x\notin \mathbf{Q} \end{cases} \end{equation*}$$

has a countable set of discontinuities occurring at precisely every rational number. Follow these steps to prove that $$\displaystyle t( x)$$ is integrable on $$\displaystyle [ 0,1]$$ with $$\displaystyle \int _{0}^{1} t=0$$.

(a) First argue that $$\displaystyle L( t,P) =0$$ for any partition $$\displaystyle P$$ of $$\displaystyle [ 0,1]$$.

Proof.

Let $$\displaystyle P$$ be any arbitrary partition of $$\displaystyle [ 0,1]$$. Since the irrational numbers are dense in $$\displaystyle \mathbf{R}$$, every sub-interval of $$\displaystyle P$$ contains an irrational number. Thus:

$$\begin{equation*} L( t,P) =\sum _{k=1}^{n} m_{k} \Delta x_{k} =0\ \quad \{\because m_{k} =0,\forall k\} \end{equation*}$$

(b) Let $$\displaystyle \epsilon >0$$ and consider the set of points $$\displaystyle D_{\epsilon /2} =\{x\in [ 0,1] :t( x) \geq \epsilon /2\}$$. How big is $$\displaystyle D_{\epsilon /2}$$?

Proof.

Let $$\displaystyle N\in \mathbf{N}$$ be such that, for all $$\displaystyle n\geq N$$, we have:

$$\begin{equation*} \frac{1}{n} < \frac{\epsilon }{2} \end{equation*}$$ Thus, the set $$\displaystyle \{t( x) :t( x) \geq \epsilon /2\}$$ consists of :

$$\begin{equation*} \left\{\frac{1}{N-1} ,\frac{1}{N-2} ,\dotsc ,1\right\} \end{equation*}$$ Thus, the set $$\displaystyle D_{\epsilon /2}$$ consists of:

$$\begin{equation*} D_{\epsilon /2} =\left\{\frac{N-2}{N-1} ,\frac{N-3}{N-1} ,\dotsc ,\frac{1}{N-1} ,\frac{N-3}{N-2} ,\dotsc ,\frac{1}{N-2} ,\dotsc ,\frac{1}{2} ,1\right\} \end{equation*}$$

$$\displaystyle D_{\epsilon /2}$$ contains at most a finite number of points.

(c) Given $$\epsilon > 0$$, construct a partition $$P_\epsilon$$ for which $$U(f,P_\epsilon) < \epsilon$$.

• small quibble: you need to define $N$ to be the smallest natural number such that for $n \ge N, \frac 1n < \frac \epsilon 2$, not just any $N$ for which the statement is true. Only the smallest $N$ will make your next conclusion true. Mar 20 at 16:50
• To handle (c), surround each point of $D_{\epsilon/2}$ with intervals whose total length is $< \epsilon/2$. The regions between those intervals form the rest of the partition. Mar 20 at 16:53
• @PaulSinclair, I was able to use the clue you provided. Since $D_{\epsilon/2}$ is a finite non-empty set, we can denote it by an ordered set $\{q_1< \ldots <q_M\}$. Surround each $q_k$ by subintervals of length $\epsilon/2M$. Their total contribution to $U(f,P)$ is $\sum_{k=1}^{M}M_k \cdot \Delta x_k \leq \sum_{k=1}^{M}1\cdot \frac{\epsilon}{2M} = \epsilon/2$. The region between those intervals forms the rest of the partition. For this entire region, $t(x) < \epsilon/2$. And $\sum \Delta x_k < 1$. It's total contribution to $U(f,P)$ is smaller than $\epsilon/2$. Mar 22 at 21:36
• $0 \in D_{\epsilon/2}$ (when $\epsilon \le 2$), so $q_1 = 0$. Also, I suggest making your life a little easier by defining $\xi$ to be half the size. Then you can just add or subtract $\xi$, not $\xi/2$, But that is just a suggestion, not a problem with your current description. Mar 23 at 17:08
• Does this answer your question? Upper Riemann Sum of the Thomae function Apr 2 at 14:20

In total, you have $$N-2$$ points where $$t(x) = \frac{1}{N-1}$$. Suppose that we define a map $$t_{N-1}$$ always vanishing except on intervals centered on the points mentioned previously with length equal to $$l_{n-1}$$ where $$t_{N-1}$$ is equal to $$\frac{1}{N-1}$$.
Define in a similar way maps $$t_i$$ for $$i \in \{1, \dots, N-1\}$$. Then define $$\overline{t} = \sup (t_1, \dots, t_{N-1})$$. We will have
$$0 \le U(\overline{t}, P_{\overline{t}}) \le \sum_{i=1}^{N-1} l_i.$$ Now pick up the $$l_i$$ the right way!