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I strongly believe the following statement : "If $g\varphi$ is integrable for any smooth $\varphi$ with compact support then $g$ is locally integrable." But then I also know that $\text{vp}(\frac{1}{x}) : \varphi \mapsto \lim_{\varepsilon \rightarrow 0} \int_{|x|>\varepsilon} \frac{\varphi(x)}{x}\,\text{d}x$ is a (tempered) distribution. And even though this definition bears all the signs of precaution, it's merely the Riemann integral on $\mathbb{R}^*$ of $\varphi g$ with $g(x)=\frac{1}{x}$ isn't it ? Which then is the Lebesgue integral : $\int_{\mathbb{R}} \varphi \widetilde{g}\,\text{d}m$ with $\widetilde{ g}(0) = \text{anything}$. So basically, I have a well defined function that is not locally integrable but which is integrable against any smooth function with compact support. So where am I wrong ? I would like to think that my first statement is correct. I think I am wrong about it being the Lebesgue integral. I actually think that even though this limit exists, the function is not Lebesgue integrable because not absolutely Riemann-convergent.

James Well
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  • I don't think the problem is in what type of integral you are using because we are always using Lebesgue integration. The problem lies in how you take the limit as $\epsilon\rightarrow 0$. – Jacky Chong Oct 15 '16 at 21:52
  • Well that's where I think we're both wrong, I mean the integral let's say from 0 to $t$ of $\frac{sin(x)}{x}$ is convergent as $t \rightarrow \infty$ which does not make the function Lebesgue integrable. – James Well Oct 15 '16 at 21:54
  • I think being Lebesgue integrable is a stronger statement than being Riemann integrable when dealing with piecewise continuous funcions and stuff like that. – James Well Oct 15 '16 at 21:56
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    Lets get this straight. p.v. $\frac{1}{x}$ is a linear functional, which I will denote by $L$, that doesn't come from any function, i.e. there doesn't exist a function $f \in L^1$ such that $L(\phi) = \langle f, \phi\rangle$. – Jacky Chong Oct 15 '16 at 22:13
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    The reason I said we are always using Lebesgue integration is to avoid ill-defined linear functionals of the form $L(\phi) = \langle f, \phi\rangle$. – Jacky Chong Oct 15 '16 at 22:14

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