# Showing $\frac{1}{x}-\left[\frac{1}{x}\right]$ is Riemann Integrable

Some days ago as i had asked as to how to test the Riemann Integrability of the function. Now i was recently given this question about proving that the given function is Riemann Integrable.

How can i show that this function is Riemann Integrable or not in the interval $[0,1]$. I tried using partitions, but it didn't work. I here don't want to use the Riemann Lebesgue lemma as i want to understand the methodology behind selecting the partitions.

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@Chandru1: Riemann-Lebesgue Lemma has nothing to do with this. – AD. Aug 23 '10 at 14:12

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– damiano Aug 23 '10 at 14:28

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– J. M. Aug 23 '10 at 14:29

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@Chandru1: Here's a hint. First, choose $\epsilon>0$ small. Then partition $[0,1]$ into $[0, \epsilon] \cup [\epsilon, 1]$. The difference of the lower and upper sums on the first interval is at most $\epsilon$ because this is a function taking values in the unit interval. On $[\epsilon, 1]$, you have finitely many jump discontinuities, so one may construct a fine partition of this where the upper and lower sums are close. – Akhil Mathew Aug 23 '10 at 14:30

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@damiano, usually Rieman integrals are defined w.r.t. finite partitions, and the function in the question has infinitely many discontinuities. – Mariano Suárez-Alvarez♦ Aug 24 '10 at 0:29
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## 3 Answers

The function is Riemann-integrable because it is bounded (it takes values in $[0,1]$) and has countably many discontinuities, namely, the points of the form $\frac{1}{n}$ and $0$.

This uses Lebesgue's criterion for Riemann integrability which you probably meant with Riemann-Lebesgue-Lemma and hence unfortunately didn't want to use.

As for doing it by hand with partitions, try Akhil Mathew's hint above.

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To show that the function in Riemann-integrable you have to show that it is bounded and has at most a countable set of discontinuities in the interval you are integrating. But this is the case of an improper integral, since $1/x$ is not defined in $x = 0$, you have to follow the criterions for improper integrals, which is the existence of the limit $\lim_{x\rightarrow 0} \int_x^1 \frac{1}{x} dx$ and its finiteness.

Assuming that by $[\frac{1}{x}]$ you mean the integer part of x (the closest integer less then or equal to x), we can always assume that $[\frac{1}{x}]$ is zero, because if x less than or greater than zero, you always have a number less than one, the integer part of which is always zero. So your integral is the same as $\int_0^1 \frac{1}{x} dx - \int_0^1 [\frac{1}{x}] dx = \int_0^1 \frac{1}{x} dx$, which does not converge on that interval.

EDIT: promt comments showed that I'm way over my head...Sorry for the wrong answer.

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If $x$ is in $(0,1]$ then $\frac{1}{x}$ is in $[1,\infty)$ and $[\frac{1}{x}]$ could be anything. – Rasmus Aug 23 '10 at 14:23
You're right, I'm an idiot...I'll correct – Andy Aug 23 '10 at 14:27

Let $\epsilon > 0$ be given. There is a positive integer $n_{0}$ such that $1/n < \epsilon/2$ for $n > n_{0}$. Choose a partition $P$ determined by $$0 = x_{0} < x_{1}= \frac{1}{n_{0}+1} < x_{2} < \cdots < x_{n'_{0}}= \frac{1}{n_0}$$

$$< x_{n'_{0}+1} < \cdots < x_{n_{1}}'= \frac{1}{n_{0}-1} < \cdots < x_{n_{n_{0}-1}'}=1$$ such that

$$\displaystyle x_{i} - x_{i-1} < \frac{\epsilon}{4n_{0}}$$ Then $$U(P,f) - L(P,f) = \frac{1}{n_{0}+1} + \sum\limits_{i=2}^{n_{0}'} (M_{i}-m_{i})(x_{i}=x_{i-1})$$

$$+ \sum\limits_{k=0}^{n_{0}-2} \sum\limits_{i=n_{k}'+1}^{n_{k+1}'}(M_{i}-m_{i})(x_{i}-x_{i-1})$$

$$< \frac{\epsilon}{2} + 2n_{0} \frac{\epsilon}{4n_{0}} = \epsilon$$

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