Local integrability of a Cauchy transform in the plane Let $\mu$ be a Radon measure in the plane, typically with support included in a small neighborhood of the origin. Let $h(z)=\int \frac{d\mu(y)}{z-y}$.
I am wondering when it can be said that $\frac{h(z)}{z}$ is locally integrable near the origin. Obviously when $\mu$ is a dirac mass at the origin $h(z)/z=\frac{1}{z^2}$ is not locally integrable. On the other hand, if $\mu$ is the Lebesgues measure on $[0,1]$ (or even absolutely continuous with respect to it) then $\frac{h(z)}{z}$ is integrable near 0.
Question : Does there exist non-atomic Radon measures such that $\frac{h(z)}{z}$ is not integrable near 0 ? 
If so, can we find a nice sufficient condition to rule this out ? Intuition suggests that if such measures exist they must concentrate a lot of mass near 0, so the "worst possible case" is the dirac. The other extreme is the 2-dimensional Lebesgues measure : then $h$ is even bounded near 0. 
I guess the question amounts to this : is it possible to find a function $h$ such that :


*

*$h(z)/z$ is not integrable near 0

*$h(z)=o(1/z)$

*$\overline{\partial} h$ is a Radon measure ?

 A: If $\int \log|y|\,d\mu(y)<\infty$, then $h(z)/z$ is integrable near the origin. Indeed, writing $dm$ for the Lebesgue measure and $D$ for the unit disk, we have 
$$
\begin{split}
\int_{D} \frac{1}{|z|}|h(z)|\,dm(z) &\le 
\int_{D} \int_D \frac{1}{|z||z-y|}\,d|\mu|(y)\,dm(z) \\ &= 
\int_D \left(\int_{D}\frac{1}{|z||z-y|}\,dm(z)\right)\,d|\mu|(y)<\infty
\end{split} \tag1$$
because the integral in parentheses is comparable to $\log(1/|y|)$. 
On the other hand, you can find measures such that $\int \log|y|\,d\mu(y)=\infty$ and for which $h(z)/z$ is not locally integrable. One possibility is 
$$d\mu(x) = \frac{dx}{|x|\log^{3/2}(1/|x|)}\tag2 $$ supported on the segment $[-1,0]$. When $|\arg z|\le \pi/4$, we have 
$$
\operatorname{Re}\frac{1}{z-x} \ge \frac{1}{\sqrt{2}}\frac{1}{|z-x|}
\tag3$$
which implies that no substantial cancellation occurs in the integral defining $h$:
$$
|h(z)|\ge \operatorname{Re} h(z) \ge \frac{1}{\sqrt{2}}\int_{-1}^1 \frac{1}{|z-x|}\,d\mu(x)
\tag4 $$
This brings us back to (1) but from the other side:
$$
\begin{split}
\int_{D} \frac{1}{|z|}|h(z)|\,dm(z) &\ge 
\frac{1}{\sqrt{2}} \int_{-1}^0 \left(\int_{|z|<1, |\arg z|<\pi/4}\frac{1}{|z||z-x|}\,dm(z)\right)\,d|\mu|(x)\\ &\ge c \int_{-1}^0 \log\frac{1}{|x|}\,d\mu(x) =\infty
\end{split} \tag5$$
