可求长测度、涉及密度的平方函数和柯西变换
Rectifiable measures, square functions involving densities, and the Cauchy transform
摘要：
本文给出了两个相关结果的证明。第一个定理断言，如果$μ$是$\mathbb R^d$中的Radon测度，满足$$\limsup_{r\to 0} \frac{μ(B(x,r))}{r}>0\quad \text{ and }\quad \int_0^1\left|\frac{μ(B(x,r))}{r} - \frac{μ(B(x,2r))}{2r}\right|^2\,\frac{dr}r< \infty$$，则$μ$ -a.e. $x\in\mathbb R^d$，则$μ$是可校正的。由于逆蕴涵是已知的，这产生了以下特征的可求积集：一个集$E\subset\mathbb R^d$有限$1$ -维Hausdorff测度$H^1$是可求积的当且仅当$$\int_0^1\left|\frac{H^1(E\cap B(x,r))}{r} - \frac{H^1(E\cap B(x,2r))}{2r}\right|^2\,\frac{dr}r< \infty \quad\mbox{ for $H^1$-a.e. $x\in E$.}$$第二个结果的文件涉及之间的关系，一个类似的平方函数在复平面和柯西变换$C_μf(z) = \int \frac1{z-ξ}\,f(ξ)\,dμ(ξ)$。假设$μ$具有线性增长，即，对于所有$z\in\mathbb C$和所有$r>0$，都是$μ(B(z,r))\leq c\,r$。证明了$C_μ$在$L^2(μ)$中有界当且仅当$$ \int_{z\in Q}\int_0^\infty\left|\frac{μ(Q\cap B(z,r))}{r} - \frac{μ(Q\cap B(z,2r))}{2r}\right|^2\,\frac{dr}r\,dμ(z)\leq c\,μ(Q) \quad\mbox{ for every square $Q\subset\mathbb C$.} $$
英文摘要：
This paper is devoted to the proof of two related results. The first one asserts that if $μ$ is a Radon measure in $\mathbb R^d$ satisfying $$\limsup_{r\to 0} \frac{μ(B(x,r))}{r}>0\quad \text{ and }\quad \int_0^1\left|\frac{μ(B(x,r))}{r} - \frac{μ(B(x,2r))}{2r}\right|^2\,\frac{dr}r< \infty$$ for $μ$-a.e. $x\in\mathbb R^d$, then $μ$ 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 $C_μf(z) = \int \frac1{z-ξ}\,f(ξ)\,dμ(ξ)$. Suppose that $μ$ has linear growth, that is, $μ(B(z,r))\leq c\,r$ for all $z\in\mathbb C$ and all $r>0$. It is proved that $C_μ$ is bounded in $L^2(μ)$ if and only if $$ \int_{z\in Q}\int_0^\infty\left|\frac{μ(Q\cap B(z,r))}{r} - \frac{μ(Q\cap B(z,2r))}{2r}\right|^2\,\frac{dr}r\,dμ(z)\leq c\,μ(Q) \quad\mbox{ for every square $Q\subset\mathbb C$.} $$
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