Let $H$ be a finite measure on $(0,1)$. What conditions must $H$ fulfill, such that \begin{equation*} \frac{1}{x} \in L^1(H),\ \ \ \frac{1}{1 - x} \in L^1(H) \end{equation*} I'm trying to characterize those measures. Right now I have the space of all test functions that are zero in $(0,1)\(\epsilon, 1- \epsilon)$ for an arbitrary $\epsilon > 0$ as possible candidates, I yet refuse to believe that those are all choices..
Any idea?
I assume that $H$ is a positive measure. In this case, the function $x\mapsto 1/x$ will be integrable on $(0,1)$ for the measure $H$ as long as the series $\sum_{n=1}^{+\infty}2^n\cdot H\left(\left[2^{-(n+1)},2^{-n}\right)\right)$ converges. Indeed, we have the pointwise inequalities $$2^{n}\mathbf 1\left(\left[2^{-(n+1)},2^{-n}\right)\right)(x)\leqslant \mathbf 1\left(\left[2^{-(n+1)},2^{-n}\right)\right)(x)/x\leqslant 2^{n+1}\mathbf 1\left(\left[2^{-(n+1)},2^{-n}\right)\right)(x)$$ which can be integrated with respect to $H$. This gives that $$2^{n}H\left(\left[2^{-(n+1)},2^{-n}\right)\right)\leqslant \int_{\left[2^{-(n+1)},2^{-n}\right)}\frac 1x\mathrm dH(x)\leqslant 2^{n+1}H\left(\left[2^{-(n+1)},2^{-n}\right)\right),$$ and we conclude using the equivalences $$\sum_{n=1}^{+\infty}2^n\cdot H\left(\left[2^{-(n+1)},2^{-n}\right)\right)<+\infty \Leftrightarrow \sum_{n=0}^{+\infty}\int_{\left[2^{-(n+1)},2^{-n}\right)}\frac 1x\mathrm dH(x)<+\infty\Leftrightarrow \int_{(0,1))}\frac 1x\mathrm dH(x)<+\infty.$$
Of course, we can derive in a similar way a necessary and sufficient condition for the integrability of $x\mapsto 1/(1-x)$.