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Hi I have been attempting given in the link below. I am confused about the argument used to show the function is not Lebesgue integrable.

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What each person has used to answer is the fact that $ \displaystyle\int_0^{\infty} \big|\frac{sin(x)}{x}\big|\; dx$ (Riemann integral!!) does not converge implies $\displaystyle\frac{sin(x)}{x}$ is not Lebesgue integrable. I have not seen any theorems that say you can do this. I have a theorem in my notes which says;

If $f$ and $|f|$ are integrable on an interval I (bounded or unbounded)in the improper Riemann sense, then it is Lebesgue integrable on I and the integrals coincide.

This theorem is not iff and so this argument makes no sense to me, its like saying if we have A implies B then not A implies not B. I have looked around for other theorems but they are all in this direction. Please can someone explain rigorously why the above argument works and in general when you can switch $d\lambda(x)$ for $dx$ because people just seem to do it whenever they want and its confusing. Thanks

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th0masb
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  • $L_1$ is the space of all Lebesgue integrable functions. $sinc$ does not belong in the $L_1$ space, therefore it is not integrable. For example here: http://press.princeton.edu/chapters/s9627.pdf (pdf) it is defined as an iif. Am I missing something? – user3371583 Mar 24 '15 at 17:13
  • do you mean sin(x) does not belong to $L_1$ and so sin(x)/x does not belong to it? If so why is that true? – th0masb Mar 24 '15 at 17:18
  • On what page is it defined as an iff? – th0masb Mar 24 '15 at 17:20
  • On page 2: When $p=1$ .... The Lebesgue integral of a function is defined as the difference of the Lebesgue integral of its positive part minus $\int$ of its negative part (both must be $<\infty$). Since $\int (sinc(x))^+ ,dx = +\infty$ then $sinc$ is not in $L_1$ therefore not L-integrable. $sinc(x) = sin(x)/x$. – user3371583 Mar 24 '15 at 17:27
  • Yes but you have just evaluated $sinc(x)$ as a Riemann integral, why can you just do that, the condition is that $\int(sinc(x))^+ d\lambda(x)<\infty$ notice the $d\lambda(x)$ not $dx$. do you take it as an integral on a bounded interval then let one end tend to infinity? – th0masb Mar 24 '15 at 17:36
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    You take then on compacts of the shape $[0, a]$ and you take $a$ to infinity. The $R-\int$ and the $L-\int$ on such compacts of $(sinc(x))^+$ coincide. So in the limit, they also coincide. – user3371583 Mar 24 '15 at 17:55
  • Ok thank you, that makes sense now – th0masb Mar 24 '15 at 17:58

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If the Lebesgue integral exists, it must be finite. That is, we need $$\int_0^{\infty}\frac{|\sin x|}{x}<\infty$$ By Lebesgue's Monotone Convergence Theorem, we have that $$\int_0^{\infty}\frac{|\sin x|}{x}=\sum_{n=1}^{\infty}\int_{(n-1)\pi}^{n \pi}\frac{|\sin x|}{x}\ dx\geq \sum_{n=1}^{\infty}\frac{1}{n\pi}\int_0^{\pi}|\sin x|\ dx=\infty$$

Hence the Lebesgue integral is not finite and so it does not exist.

Remark: We have equality on the first step because if $f:[a,b]\to\mathbb{R}$ is a bounded Riemann Integrable function, then $f$ is also Lebesgue measurable on $[a,b]$ and the two integrals agree.

Sujaan Kunalan
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  • I see now, thank you. The way he says not Lebesgue integrable because $\displaystyle\int_0^{\infty} \frac{|sin(x)|}{x} ; dx$ does not converge is incorrect and misleading. It should be $d\lambda(x)$. – th0masb Mar 24 '15 at 17:50