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

Xavier Tolsa

Published 2014-08-29Version 1

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 $$\limsup_{r\to 0} \frac{\mu(B(x,r))}{r}>0\quad \text{ and }\quad \int_0^1\left|\frac{\mu(B(x,r))}{r} - \frac{\mu(B(x,2r))}{2r}\right|^2\,\frac{dr}r< \infty$$ 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 $C_\mu f(z) = \int \frac1{z-\xi}\,f(\xi)\,d\mu(\xi)$. Suppose that $\mu$ has linear growth, that is, $\mu(B(z,r))\leq c\,r$ 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 $$ \int_{z\in Q}\int_0^\infty\left|\frac{\mu(Q\cap B(z,r))}{r} - \frac{\mu(Q\cap B(z,2r))}{2r}\right|^2\,\frac{dr}r\,d\mu(z)\leq c\,\mu(Q) \quad\mbox{ for every square $Q\subset\mathbb C$.} $$