# How to prove that this function is integrable on $[0,1]$ Here I tried to find two step functions, one of them is less than $f$ on $[0,1]$ whereas one of them is greater than $f$ on the same closed interval, to prove this function is Riemann-integrable on this interval, however, I could not find such two step functions to help me show this function is integrable. May I get some hint about the solution to this problem?

• What does the symbol $[1/x]$ mean? – Antoine Aug 29 '15 at 8:31
• @Antoine It has been used to denote greatest integer function in my book and lecture notes so far. – frosh Aug 29 '15 at 8:33
• Fix $\varepsilon > 0$. Choose $n$ such that $\frac{1}{n} < \frac{\varepsilon}{2}$. Consider the partition $0, \frac{1}{n}, \frac{1}{n-1}, \frac{1}{n-2},\dotsc, \frac{1}{3}, \frac{1}{2}, \frac{1}{1}$. – Daniel Fischer Aug 29 '15 at 8:49
• @DanielFischer You should write this hint as an answer. – frosh Aug 29 '15 at 9:06
• Please do not use pictures for critical portions of your post. Pictures cannot be searched and are inaccessible to those using screen readers. – Lord_Farin Aug 29 '15 at 9:41

## 3 Answers

Hint:

$$\int_0^1 (-1)^{[1/x]} \, dx = \lim_{n \to \infty}\sum_{k=1}^n\int_{1/(k+1)}^{1/k} (-1)^{[1/x]} dx = \lim_{n \to \infty}\sum_{k=1}^n\int_{1/(k+1)}^{1/k} (-1)^k dx \\ = \sum_{k=1}^{\infty} (-1)^k \left[\frac1{k}-\frac1{k+1} \right]=\sum_{k=1}^{\infty} \frac{(-1)^k}{k(k+1)}$$

Alternatively consider a partition $P = (0, 1/n, \ldots, 1/2, 1)$.

The difference between upper and lower Riemann sums is $1/n$ since

$$\sup_{(1/(k+1),1/k]} f(x) - \inf_{(1/(k+1),1/k]} f(x) = (-1)^k - (-1)^k = 0,$$

and

$$\sup_{[0,1/n]} f(x) - \inf_{[0,1/n]} f(x) = 1.$$

By choosing $n$ sufficiently large, this difference can be made smaller than any $\epsilon > 0$ and $f$ is Riemann integrable.

• I am sorry but in the book writer did not mention relation between lim and integral operator. – frosh Aug 29 '15 at 8:51
• Yes I know, in calculus course it is treated so but in analysis course we just learned epsilon and step functions definition and I have to proceed in that way on this example. – frosh Aug 29 '15 at 8:59
• Ok then follow @Daniel Fischer's suggestion. I'll finish this to show how you evaluate the integral, in case you are interested. – RRL Aug 29 '15 at 9:02
• Please, continue to answer :) – frosh Aug 29 '15 at 9:03

I think if you use $‎\dfrac{1}{k+1}\leq{x}\leq{\dfrac{1}{k}}$ and cover interval $[0,1]$. Then it shows that discountinuty of $f$ is in countable point.

First try to sketch the graph.

Denote the upper and the lower Riemann sums for a partition $P$ of a function $f$ by $U(P,f)$ and $L(P,f)$ respectively.

We say that $f$ is Riemann integrable if for any $\varepsilon >0$ there is a partition $P$ such that $$\big| U(P,f)-L(P,f)\big|< \varepsilon$$

Observe that the upper and the lower sums are always equal for a constant function.

Pick $\varepsilon >0$. There is a natural number $N$ such that $\frac 2 {N} < \varepsilon$

Consider the partition $0 < \frac 1 N < \frac 1 {N-1} < \frac 1 {N -2} < \cdots < 1$

Now since $f$ is constant on $( \frac 1 2 , 1]$ and $( \frac 1 3, \frac 1 2 ]$ and $\dots$ and $( \frac 1 N,\frac 1 {N-1}]$, upper and lower sums will be equal in these intervals.

Also, $f$ is bounded above by $1$ and below by $-1$ gives that

$$\big| U(P,f) - L(P,f) \big| \leq \bigg| \big(1 \cdot \frac 1 N + \text{ some other terms} \big) - \big( -1 \cdot \frac 1 N + \text{ some other terms} \big) \bigg| = \frac 2 N < \varepsilon$$

Many steps are missing in the proof above.

• Why is there plus sign after $-1 \cdot \frac 1 N$? – frosh Aug 29 '15 at 10:47
• Would you mind answering @ThePortakal ? – frosh Aug 29 '15 at 12:28
• Ooops, sorry. I fell asleep. What I meant there was $U(P,f) = 1 \cdot \frac 1 N + something$ and $L(P,f) = -1 \cdot \frac 1 N + something$ and so that when you subtract $L(P,f)$ from $U(P,f)$ the "something" terms will cancel each other. – ThePortakal Aug 30 '15 at 0:44
• The thing that I did not understand is if there should be minus sign before second $something$ or not? – frosh Aug 30 '15 at 5:43
• No. It is fine the way it is above. – ThePortakal Aug 30 '15 at 5:54