2
$\begingroup$

I need to prove or disprove that the next function is Riemann integrable on $[0,2]$:

$$ f(x) = \begin{cases} \dfrac{1}{x} &x > 0 \\ 0 &x = 0 \end{cases} $$

My intuition is that it's not, because $f$ is unbounded in that interval, so $U(f,Pn) = L(f,Pn) = \infty$

So I have two questions:

  1. Am I right?
  2. Can unbounded functions be Riemann integrable?
$\endgroup$
  • $\begingroup$ Well since a Riemann integrable function have to be bounded, then yes unbounded one cannot be Riemann integrable. $\endgroup$ – Azlif Jan 7 '18 at 1:05
  • $\begingroup$ math.stackexchange.com/questions/2246901/… $\endgroup$ – Azlif Jan 7 '18 at 1:11
  • $\begingroup$ in the title "If and only f" part is not true since bounded functton need not be Rieman Integrable $\endgroup$ – Azlif Jan 7 '18 at 1:15
2
$\begingroup$

It is not integrable, so your answer is right. As to whether an unbounded function can be Riemann integrable (and thus if the reasoning is as simple as that) it's important to distinguish between proper and improper Riemann integrability.

Regarding proper Riemann integrability, it is true that if a function is unbounded on $[a,b],$ it is not integrable. And we do have $U(f,P) = \infty$ for every partition of $[a,b].$ It is not the case that for $L(f,P)=\infty$ for all partitions, but it's enough that $U(f,P) = \infty$ to prove it the function is not integrable.

For improper Riemann integrability, the function need not be bounded. For instance $1/\sqrt{x}$ has an improper Riemann integral with value $2$ on $[0,1].$ However $1/x$ does not have an improper Riemann integral either on $[0,2]$ since $$ \int_a^2 \frac{1}{x}dx = \ln(2/a)\to_{a\to 0^+} \infty.$$

$\endgroup$
  • $\begingroup$ Why not for $L(f,P)$? $\endgroup$ – Pilpel Jan 7 '18 at 1:22
  • $\begingroup$ Why would $L(f,P) = \infty$? $\endgroup$ – spaceisdarkgreen Jan 7 '18 at 1:23
  • $\begingroup$ Same reason for $U$, no? $\endgroup$ – Pilpel Jan 7 '18 at 1:23
  • $\begingroup$ No.... $L(f,P)$ is the lower sum. In the case of $1/x$ on $[0,2],$ it will never be the case for any partition $P$ that $L(f,P) = \infty.$ $\endgroup$ – spaceisdarkgreen Jan 7 '18 at 1:25
  • $\begingroup$ $L(f) = \sup_{P}L(f,P)$ may well be infinity (and is in this case). $\endgroup$ – spaceisdarkgreen Jan 7 '18 at 1:26

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.