Cauchy principal value

Question

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.