Let $f= \sin(\frac 1x)$. I would now like to know whether $f$ is Lebesgue-integrable on $(0,1)$.

I know that $f$ must be measurable, since it it continuous. I am not quite sure how to check integrability...

I know $f$ is Lebesgue integrable if for $$\int_{(0,1)}f\ d\lambda = \int_{(0,1)}f^+\ d\lambda - \int_{(0,1)}f^-\ d\lambda$$ the two (or atleast one for semi integrability) summands are smaller than infinity,i.e: $$\int_{(0,1)}f^-\ d\lambda < \infty \ \ \ \hbox{and}\ \ \ \int_{(0,1)}f^+\ d\lambda < \infty$$

I figured that I can bound $|\sin(\frac 1x)|\le 1$, so I thought maybe, for a sequence $f_n \nearrow f$, I could just chose a sequence $g_n \nearrow g$, where $g$ denotes the constant function $x\mapsto 1$, with $g_n\ge f_n$. Then it would follow that $f$ is Lebesgue integrable. Is this argument valid, or do I need to show this in another way?

Any help would be greatly appreciated!

  • 1
    $\begingroup$ A bounded function on a bounded interval is integrable if and only if it is measurable. $\endgroup$ – Jonas Meyer Jun 7 '17 at 18:57

The function $f$ is continuous, except at $0$, and bounded. It is Riemann integrable on $[0,1]$ so certainly Lebesgue integrable on $(0,1)$.

(A function $[a,b]\to\Bbb R$ is Riemann integrable iff it is bounded and its set of discontinuities has Lebesgue measure zero.)

| cite | improve this answer | |
  • $\begingroup$ This does make sense in a way, but can I show it without using Riemann integrability? Say, I have never heard of the Riemann integral before, what would I do then? $\endgroup$ – Jack4t3 Jun 7 '17 at 18:56
  • 1
    $\begingroup$ @Jack4t3 It is bounded and Lebesgue measurable. $\endgroup$ – Angina Seng Jun 7 '17 at 18:57

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.