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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!

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    $\begingroup$ 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
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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.)

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  • $\begingroup$ 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
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    $\begingroup$ @Jack4t3 It is bounded and Lebesgue measurable. $\endgroup$ – Lord Shark the Unknown Jun 7 '17 at 18:57

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