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Show that

$ f(x) = \begin{cases} x^{\frac{-1}{2}}\sin(\frac{1}{x}) , & 0 <x\leq 1 \\ 0 , & x=0 \end{cases}$

is integrable on $[0,1]$. Intregable means that $\displaystyle\int_{[0,1]} |f(x)|\,\mathrm dx < \infty$.

My attempt:

First, I checked to see if $f(x)$ had a bounded derivative on $[0,1]$. If it did, then it would be a Lipschitz function, hence absolutely continuous and hence Integrable. Unfortunately, the derivative was not bounded.

Okay, so then I noticed that if $0 < a \leq 1$ then $f(x)$ is a well defined continuous function and so Riemann integrable on $[0,1]$, and hence Lebesgue integrable on $[0,1]$.

Next, I need to figure out how to prove that $f$ is continuous. I wouldn't mind some help with this part. Then I have that $\lim_{a \rightarrow 0^+} f(a)=f(0)$.

So, $$f(1)-f(0) = \lim_{a \rightarrow 0^+} f(1)-f(a) = \lim_{a \rightarrow 0^+} \int_{[a,1]} f(x)\,\mathrm dx = \lim_{a \rightarrow 0^+} \int_a^1 f(x)\,\mathrm dx.$$

Ugh idk honestly I'm lost.. Would appreciate help. I'm pretty slow so lots of details appreciated! Thanks!

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1 Answer 1

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You can just do it like $\displaystyle\int_{0}^{1}|f(x)|\,\mathrm dx\leq\int_{0}^{1}x^{-1/2}\,\mathrm dx=2<\infty$.

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    $\begingroup$ $|\sin u|\leq 1$ no matter what. $\endgroup$
    – user284331
    Dec 3, 2019 at 2:33
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    $\begingroup$ I missed the absolute sign to the $f$. It suffices to check for the absolute integrand, and bound it by less than $\infty$, that's it. $\endgroup$
    – user284331
    Dec 3, 2019 at 2:35
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    $\begingroup$ I don't understand your question? Can you paraphrase? $\endgroup$
    – user284331
    Dec 3, 2019 at 4:08
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    $\begingroup$ Note that Riemann integrable function is itself Lebesgue integrable and the values are equal. Also that nonnegative improper Riemann interable function is also Lebesgue integrable with the same value again. $\endgroup$
    – user284331
    Dec 3, 2019 at 16:11
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    $\begingroup$ For nonnegative functions, Lebesgue and Improper Riemann are the same, so feel free to do anything to the integrals. $\endgroup$
    – user284331
    Dec 3, 2019 at 18:02

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