$f(x) = {\rm sgn}({\rm sin}(\frac{\pi}{x}))$ if $x \neq 0$ and $f(0) = 0$

where ${\rm sgn}(x) = 1$ if $x > 0$, ${\rm sgn}(x) = −1$ if $x < 0$ and ${\rm sgn}(0) = 0$.

Show $f$ is Riemann integrable on $[0,1]$

Is there a way to do this without actually solving the upper and lower sums? If not, then how do you evaluate the upper and lower sums?

  • $\begingroup$ Can you prove the function is continuous on [0,1] $\endgroup$ – user113353 Dec 4 '13 at 8:40
  • $\begingroup$ Certainly not. That would imply that $\sin(\frac\pi x) \neq 0 \qquad\forall\ x\in[0,1]$ $\endgroup$ – AlexR Dec 4 '13 at 8:41
  • $\begingroup$ A function does not have to be continuous to be Riemann integrable? $\endgroup$ – user113353 Dec 4 '13 at 8:46
  • $\begingroup$ Yes, that is correct. Take a characteristic function of an interval, for example. Then with $$\chi_{[a,b]}(x) := \begin{cases}1 & x\in[a,b]\\0& \text{else}\end{cases}$$ You have $$\int_{-\infty}^{\infty} \chi_{[a,b]}(x) dx = b-a$$ And as soon as your partition has a node at $a$ and one at $b$, upper and lower sums will be equal (to $b-a$). $\endgroup$ – AlexR Dec 4 '13 at 8:50
  • $\begingroup$ of course! Do you know how to calculate the riemann sums? $\endgroup$ – user113353 Dec 4 '13 at 9:04

You can prove in general that if you have a bounded function on $[a,b]$ which is integrable on $[a+\varepsilon,b]$ for all $\varepsilon >0$ then it is integrable on $[a,b]$.

To prove this claim just consider a partition of $[a,b]$ where the first interval is $[a,a+\varepsilon]$. The area on that interval is bounded by $\varepsilon L$ where $L$ is the bound of the function. On the rest you know that the upper and lower sums are adjacent.


$ f(x)=-1$ on $[\frac{1}{2n},\frac{1}{2n-1}]$

and $ f(x)=1$ on $[\frac{1}{2n-1},\frac{1}{2n-2}]$

So $$\int_0^1 f(x)\ dx = \sum_{n=1}^\infty \frac{1}{2n}-\frac{1}{2n-1} +\sum_{n=2}^\infty \frac{1}{2n-2} - \frac{1}{2n-1}$$ $$=\sum_{n=1}^\infty \frac{-1}{(2n-1)2n} +\sum_{n=2}^\infty \frac{1}{(2n-1)(2n-2)}=\sum_{n=1}^\infty \frac{(-1)^n}{n(n+1)} $$

$ |\sum_{n=1}^\infty \frac{(-1)^n}{n(n+1)} |<\infty$. So integrable

Detail : Given $\epsilon >0$ there exists $N$ s.t. $$ \sum_{n \geq N}^\infty \frac{(-1)^n}{n(n+1)} <\epsilon $$

Let $\delta = \frac{\epsilon}{N}$ and consider partition $P,\ P_N$ s.t. $$ \| P\| < \delta,\ P_N \ :\ 0 < \frac{1}{N} < \frac{1}{N-1}< \cdots < 1 $$

Then let $P_0=P\cup P_N$ In this partition, riemann sum $I$ satisfies $$ |I- \sum_{n=1}^\infty \frac{(-1)^n}{n(n+1)} | < N\delta + \epsilon $$

Surely if $Q$ is finer than $P_0$,

$$ |I(Q)- \sum_{n=1}^\infty \frac{(-1)^n}{n(n+1)} | < N\delta + \epsilon $$

  • $\begingroup$ Could you explain to me what you did? $\endgroup$ – user113353 Dec 4 '13 at 9:04
  • $\begingroup$ I add some explanation. $\endgroup$ – HK Lee Dec 4 '13 at 9:46
  • $\begingroup$ I suspect that you are using the $\sigma$-additivity of the integral... which is not granted for Riemann-integral. $\endgroup$ – Emanuele Paolini Dec 4 '13 at 15:39

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