# Prove a function with infinite discontinuous points is Riemann integrable

$$$$f(x)=\begin{cases} x& x=\frac{1}{n}&n=\mathbb Z/\{0\}\\ \\ 1 & \text{others} \end{cases}$$$$

The Riemann integration is $$\displaystyle \int^1_{-1} f$$

I actually don't know whether this funtion is integrable or not, but I suspect that it is integrable, so I try to prove it is integrable. Below is my attempt

The domain of the integration is from -1 to 1, and this function is bounded on $$[-1,1]$$. $$\sup_{[-1,1]} f-\inf_{[-1,1]} f=1-(-1)=2$$

Then for partition $$P,\{t_0=1<..., let $$|P|<\dfrac{\epsilon}{2n},$$for all $$\epsilon>0$$

For each interval $$[t_{i-1},t_i]$$ let $$\omega_i=\sup_{[t_{i-1},t_i]} f-\inf_{[t_{i-1},t_i]} f$$

Then by the Riemann criterion:

$$\displaystyle \sum^n_{i=1} \omega_i\Delta t_i\leq\displaystyle \sum^n_{i=1}2\Delta t_i\leq\displaystyle \sum^n_{i=1} 2\cdot \dfrac{\epsilon}{2n}=2\cdot\dfrac{\epsilon}{2n}\cdot n=\epsilon$$

Thus the function is intergrable on $$[-1,1]$$ by Riemann criterion.

Is this proof right?

Edit:

Divide the integration into two parts $$\displaystyle \int_{-1}^0f+\displaystyle \int_0^1 f$$

Then consider the $$[0,1]$$ interval will be sufficient, if the function is integrable on $$[0,1]$$, then it will be integrable on $$[-1,1]$$

the discontinuous points are $$\frac{1}{n}$$ and for any interval $$[\frac{1}{n},\frac{1}{n-1}]$$ let $$|P|<\frac{1}{n-1}-\frac{1}{n}=\dfrac{1}{n(n-1)}$$ for $$n>\epsilon>0$$

and on $$[0,1]$$ interval $$\omega=\sup_{[0,1]} f-\inf_{[0,1]} f = 1- \frac{1}{n}=\frac{n-1}{n}$$ and $$n>\epsilon>0$$

Thus $$\displaystyle \sum^n_{i=1} \omega_i\Delta t_i\leq\displaystyle \sum^n_{i=1}\frac{n-1}{n}\cdot \frac{1}{n(n-1)}=\dfrac{1}{n}<\epsilon$$ based on Archemidean property

Edition 2

Regard the function on $$[0,1]$$, if it is integrable on $$[0,1]$$ then it should also be integrable on $$[-1,1]$$

The discontinuous points on $$[0,1]$$ are $$\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots, \frac{1}{n}$$ for $$n=1,2,3,...$$

Then regard interval $$[\frac{1}{2},1]$$, since only $$\frac{1}{2}$$ is the point that is discontinuous on this interval, the function is integrable on $$[\frac{1}{2},1]$$

Thus there exists a partition $$P_1$$ such that $$\displaystyle \sum^{n_1}_{i=1} \omega_i \Delta t_i <\epsilon$$

Also for interval $$[\frac{1}{3},\frac{1}{2}]$$, since the function is only discontinuous at two end points, then it is integrable on it.

Thus there also exists a partition $$P_2$$ such that $$\displaystyle \sum^{n_2}_{i=1} \omega_i \Delta t_i <\epsilon$$

Similarly, for all intervals $$[\frac{1}{n+1},\frac{1}{n}]$$

There always exists a partition $$P_n$$ such that $$\displaystyle \sum^{n_n}_{i=1} \omega_i \Delta t_i <\epsilon$$

Then let $$P=\min \{P_1,P_2,...,P_n\}$$ for $$n=1,2,3,...$$

Thus on each interval it has

$$\displaystyle \sum^{m_1}_{i=1} \omega_i \Delta t_i <\epsilon/n$$ (on $$[\frac{1}{2},1]$$)

$$\displaystyle \sum^{m_2}_{i=1} \omega_i \Delta t_i <\epsilon/n$$ (on $$[\frac{1}{3},\frac{1}{2}]$$)

...

$$\displaystyle \sum^{m_n}_{i=1} \omega_i \Delta t_i <\epsilon/n$$ (on $$[\frac{1}{n+1},\frac{1}{n}]$$)

As a result, on the entire $$[0,1]$$

$$\displaystyle \displaystyle \sum^{M}_{i=1} \sup f_i \Delta_i-\displaystyle \sum^{M}_{i=1}\inf f_i \Delta t_i =\sum^{M}_{i=1} \omega_i \Delta t_i=\sum^{m_1}_{i=1} \omega_i \Delta t_i +\sum^{m_2}_{i=1} \omega_i \Delta t_i+...+\sum^{m_n}_{i=1} \omega_i \Delta t_i < n \cdot \epsilon/n=\epsilon$$

Thus, the function is integrable on $$[0,1]$$

No, it is not correct.

First of all, you seem to think that $$\inf_{[-1,1]}f=-1$$. Actually, $$\inf_{[-1,1]}f=0$$.

After writing that$$\sup_{[-1,1]}f-\inf_{[-1,1]}f=2,\tag1$$you use no other property of the function in your proof. Therefore, if your proof was correct, it would prove that any function for which $$(1)$$ holds is Riemann-integrable. That's not the case. If, say,$$f(x)=\begin{cases}1&\text{ if }x\in\Bbb Q\\-1&\text{ otherwise,}\end{cases}$$then $$f$$ is not Riemann-integrable, although $$(1)$$ holds.

Your error lies in assuming that, for every $$\varepsilon>0$$ and every $$n\in\Bbb N$$, there is a partition $$P$$ of $$[-1,1]$$ into $$n$$ intervals with $$|P|<\frac\varepsilon{2n}$$.

Your function is actually Riemann-integrable. To see why, take $$\varepsilon>0$$. Since $$\frac\varepsilon2>0$$ and since $$f(x)=1$$ only for finitely many points of $$\left[\frac\varepsilon2,1\right]$$, it's not hard to see that there is a partition $$P=\left\{a_0\left(=\frac\varepsilon2\right),a_1,\ldots,a_n(=1)\right\}$$ of $$\left[\frac\varepsilon2,1\right]$$ such that$$\overline\sum\left(f_{\left[\frac\varepsilon2,1\right]},P\right)-\underline\sum\left(f_{\left[\frac\varepsilon2,1\right]},P\right)<\frac\varepsilon2.$$But then, if $$P^\star=\left\{-1,0,a_0,a_1,\ldots,a_n\right\}$$, then$$\overline\sum\left(f_{\left[-1,1\right]},P^\star\right)-\underline\sum\left(f_{\left[-1,1\right]},P^\star\right)<\varepsilon.$$

• Thank you for pointing that. I edit my proof a little bit. Does it now look better?
– M_k
Feb 2, 2022 at 4:00
• I don't see a reason for thinking that there is, for any $n\in\Bbb N\setminus\{1\}$, a partition $P$ of $[0,1]$ into $n$ intervals such that $|P|<\frac1{n(n-1)}$. Feb 2, 2022 at 7:47
• Basically, I want to make sure the partition is as small as possible. Since the discontinuous points at $[0,1]$ are $\frac{1}{2},\frac{1}{3},\frac{1}{4},...$, I let the distance of any partitioned interval $[t_{i-1},t_i]$ be smaller than the distance of any $[\frac{1}{n},\frac{1}{n-1}]$. As a result, I give |P| < $\frac{1}{n-1}-\frac{1}{n}=\frac{1}{n(n-1)}$. Then I also think that the largest difference between sup and inf on $[0,1]$ will be $1-1/n$ since function value can't be zero on that interval.
– M_k
Feb 2, 2022 at 19:12
• How do you know that, for each $n\in\Bbb N\setminus\{1\}$, there is a partition $P$ of $[0,1]$ into $n$ intervals such that $|P|<\frac1{n(n-1)}$? Feb 2, 2022 at 19:17
• Can I say that based on Archimedean property, there always exists a sufficientlt large integer q such that q > $n(n-1)>0$. Thus $0<\frac{1}{q}<\frac{1}{n(n-1)}$. Then I let the interval $[0,1]$ to be evenly partitioned into q intervals, each of them have length $\frac{1}{q}$?
– M_k
Feb 2, 2022 at 19:47