Rectifiable measures, square functions involving densities, and the Cauchy transform

This paper is devoted to the proof of two related results. The first one asserts that if $\mu$ is a Radon measure in ${\mathbb{R}}^d$ satisfying $$\underset{r\to 0}{\mathrm{lim\hspace{0.17em}sup}}\frac{\mu (B(x,r))}{r}>0\text{ and }{\int }_0^1{\mid \frac{\mu (B(x,r))}{r}-\frac{\mu (B(x,2r))}{2r}\mid }^2\frac{dr}{r}<\mathrm{\infty }$$ for $\mu$-a.e. $x\in {\mathbb{R}}^d$, then $\mu$ is rectifiable. Since the converse implication is already known to hold, this yields the following characterization of rectifiable sets: a set $E\subset {\mathbb{R}}^d$ with finite $1$-dimensional Hausdorff measure $H^1$ is rectifiable if and only \int_0^1\left|\frac{H^1(E\cap B(x,r))}{r} - \frac{H^1(E\cap B(x,2r))}{2r}\right|^2\,\frac{dr}r< \infty \quad for $H^1$-a.e. $x\in E$. The second result of the paper deals with the relationship between a similar square function in the complex plane and the Cauchy transform $C_\mu f(z) = \int \frac1{z-\xi},f(\xi),d\mu(\xi)$\mathrm{.}Supposethat$\mu$ has linear growth, that is, $\mu (B(z,r))\le cr$ for all $z\in \mathbb{C}$ and all $r>0$. It is proved that $C_{\mu }$ is bounded in $L^2(\mu )$ if and only if \int_{z\in Q}\int_0^\infty\left|\frac{\mu(Q\cap B(z,r))}{r} - \frac{\mu(Q\cap B(z,2r))}{2r}\right|^2\,\frac{dr}r\,d\mu(z)\leq c\,\mu(Q) \quad for every square $Q\subset\mathbb C$.