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.
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
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.