How to show that this modification of Thomae's function is Riemann integrable

I am dealing with the function $f(x)=\begin{cases} \frac{1}{n} & \text{if }\frac{1}{n+1}<x<\frac{1}{n},\:n\in\mathbb{N},\\ 0 & \text{ otherwise.} \end{cases}$ I want to show it is Riemann integrable without using the fact that every bounded function containing at most countably many discontinuities is Riemann integrable.

I have considered using the fact that the composition of a function $g$ that is Riemann integrable on $[a,b]$ with $g([a,b])\subset[c,d]$ and $h$ that is continuous on $[c,d]$ is Riemann integrable, ie. $h\circ g\in\mathcal{R}$. I think using Thomae's function as the $g$ is in the solution by I couldn't find a suitable $h$.

Other options include the squeeze theorem for integrals (although I am told the solution for this should be simple, and I find squeeze theorem is often lengthy), as well as linearity of the integral, the fact that the product of two Riemann integrable functions is Riemann integrable, and the additivity of the Riemann integrable.

• Note that the restriction of $f$ to $[\delta , 1]$ is Riemann integrable for all $\delta$. This fact together with the boundedness of $f$ implies that $f$ is Riemann integrable on $[0,1]$. – user99914 Feb 9 '16 at 3:52
• @JohnMa I understand that restricting the function to that domain makes it a step function on that interval and thus integrable, but how does the second part follow? – jofl Feb 9 '16 at 4:08

Theorem: Let $f : [a, b]\to \mathbb R$ be a bounded function. If for all $\delta >0$, $f$ is Riemann integrable on $[a+\delta , b]$, then $f$ is Riemann integrable on $[a,b]$.

To show that, let $\epsilon >0$. Then let $\delta < \epsilon/4C$, where $C$ is the bound of $f$. Since $f$ is integrable on $[a+\delta, b]$, there is a partition $P = \{\delta = x_1< x_2<\cdots< x_n = b\}$ of $[a+\delta, b]$ so that

$$U(P, f|_{[a+\delta, b]}) - L(P, f|_{[a+\delta, b]}) < \epsilon/2.$$

Let $\tilde P$ be the partition of $[a, b]$ given by

$$a = x_0< \delta = x_1 < x_2 <\cdots < x_n = b.$$

Then

$$\begin{split} U(P, f) - L(P, f) &= (M-m) \delta + U(P, f|_{[a+\delta, b]}) - L(P, f|_{[a+\delta, b]})\\ &< 2C\delta + \epsilon /2 \\ &< \epsilon/2 + \epsilon/2 = \epsilon. \end{split}$$

Since $\epsilon$ is arbitrary, $f$ is integrable.

Remark:

One can indeed show that

$$\int_a^b f (x) dx = \lim_{\delta \to 0} \int_{a+\delta} ^b f(x)dx.$$

It's kind of groady but you can proceed directly by constructing partitions.

Let $$P_n=\{0,{1\over n},{1\over n-1},\ldots,1/2,1\}$$ The upper and lower sums, if we ignore endpoints of intervals, are $$\left({1\over n}-0\right)\cdot{1\over n}+\left({1\over n-1}-{1\over n}\right)\cdot {1\over n-1} + \ldots + \left(1-{1\over 2}\right)\cdot 1$$ and $$\left({1\over n}-0\right)\cdot 0+\left({1\over n-1}-{1\over n}\right)\cdot {1\over n-1} + \ldots + \left(1-{1\over 2}\right)\cdot 1$$ respectively. They differ only in the first term, $1/n^2$, and so these "ignoring endpoints" upper and lower sums converge to each other as $n\to\infty$, and we have integrability under Baby Rudin 6.6.

Of course, if you religiously follow Baby Rudin, you can't ignore the endpoints of the intervals in partitions, and since the function as you have defined it is $0$ on all of them, the above proof doesn't work. You can fix it, however, by having a modified version of $P_n$ as follows: $$P'_n=\{0,{1\over n}-2^{-n},{1\over n}+2^{-n},{1\over n-1}-2^{-n},\ldots,1/2-2^{-n}, 1/2+2^{-n},1-2^{-n},1\}$$ That way the upper and lower sums differ from what is given above by at most $n2^{-n+1}$, the length of all the funny intervals $[1/i-2^{-n},1/i+2^{-n}]$, and so the upper and lower sums still converge to each other as $n\to\infty$.

Consider a partition $P$ with subintervals of the form

$$[1/(n+1), 1/(n+1) + \epsilon/2^{n}],\,\,[1/(n+1) + \epsilon/2^{n},1/n - \epsilon/2^{n}],\,\, [1/n - \epsilon/2^{n},1/n],$$

for $n = 1, 2, \ldots, m-1,$

along with $[0,\epsilon/2^{m}]$, $[\epsilon/2^{m}, 1/m - \epsilon/2^{m}]$, and $[1/m - \epsilon/2^{m},1/m]$

Then the upper sum is

$$U(P,f) = \sum_{n=1}^{m} \frac{1}{n}\left(\frac{1}{n}-\frac{1}{n+1}\right) + \frac{1}{m^2} \\ = \sum_{n=1}^{m} \frac{1}{n^2(n+1)} + \frac{1}{m^2},$$

and the lower sum is

$$L(P,f) = \sum_{n=1}^{m} \frac{1}{n}\left(\frac{1}{n}-\frac{1}{n+1} -\frac{2\epsilon}{2^n}\right) + \frac{1}{m}\left(\frac{1}{m} - \frac{2\epsilon}{2^m}\right) \\= \sum_{n=1}^{m} \frac{1}{n^2(n+1)} + \frac{1}{m^2} - \epsilon\sum_{n=1}^{m}\frac{1}{2^{n-1}}.$$

The difference satisfies

$$U(P,f) -L(P,f) < 2\epsilon.$$

Hence, $f$ is integrable by the Riemann criterion.