0
$\begingroup$

Suppose $f:[0,1]\rightarrow\mathbb{R}$. defined by $f(x)= n $ when $x\in(1/(n+1),1/n]$ and $f(0)=0$. Show that $f$ is not Riemann integrable.

My answer is, NO, it is Riemann integrable because the function defined in closed bounded interval, So, if the set of all points when $f$ is discontinuous has measure zero. Then, $f$ is Riemann integrable in my example $f$ has countable many points which make $f$ is discontinuous. Am I wrong? Any help will appreciated

$\endgroup$
  • $\begingroup$ Did you mean $f(x) = 0$ for all the other points? $\endgroup$ – Luca Oct 30 '16 at 18:43
  • $\begingroup$ No, just on zero will be zero $\endgroup$ – Gob Oct 30 '16 at 18:52
  • $\begingroup$ It is defined because take $n=3$ as example so $f(x) =1$ when $x\in(1/2,1]$ and $f(x)=2$ when $x\in(1/3,1/2]. $\endgroup$ – Gob Oct 30 '16 at 19:04
  • $\begingroup$ Yes, I answered below: that was posted by mistake $\endgroup$ – Luca Oct 30 '16 at 19:05
  • 2
    $\begingroup$ it has to be a bounded function on a compact interval s.t the sets of discontinuities have measure zero in order to conclude that it is Riemann integrable $\endgroup$ – Andres Mejia Oct 30 '16 at 19:07
0
$\begingroup$

No, this is not integrable: $$ \int_0^1 f(x) = \sum_{n=1}^{\infty}n \cdot (\frac 1{n} - \frac 1{n+1}) = \sum_{n=1}^{\infty}n \cdot (\frac 1{n(n+1)}) = \sum_{n=1}^{\infty}\frac 1{n+1} = \infty $$

So, the integral diverges

$\endgroup$
  • $\begingroup$ Thank you, Could give me name for the book that compute the integral by series my text book does not have. $\endgroup$ – Gob Oct 30 '16 at 19:18
  • $\begingroup$ Gob, I just used the definition. $\endgroup$ – Luca Oct 30 '16 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.