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

  • $\begingroup$ 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
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    $\begingroup$ 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

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

  • $\begingroup$ 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

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