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

Abstract

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 $$\underset{r\to 0}{\mathrm{limsup}}\frac{\mu (B(x,r))}{r}>0\text{ and }{\int }_0^1{\left|\frac{\mu (B(x,r))}{r}-\frac{\mu (B(x,2r))}{2r}\right|}^2\frac{dr}{r}<\infty$$ 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 $$\int_0^1\left|\frac{H^1(E\cap B(x,r))}{r} - \frac{H^1(E\cap B(x,2r))}{2r}\right|^2\,\frac{dr}r< \infty \quad\mbox{ for $H^1$-a.e. $x\in E$.}$$ The second result of the paper deals with the relationship between a similar square function in the complex plane and the Cauchy transform $C_{\mu }f(z)=\int \frac{1}{z-\xi }f(\xi )d\mu (\xi )$. Suppose that $\mu$ 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 $$ \int_{z\in Q}\int_0^\infty\left|\frac{μ(Q\cap B(z,r))}{r} - \frac{μ(Q\cap B(z,2r))}{2r}\right|^2\,\frac{dr}r\,dμ(z)\leq c\,μ(Q) \quad\mbox{ for every square $Q\subset\mathbb C$.} $$

Minor corrections and adjustments