Gaussian rigidity and width of measures

Abstract

Given $0\le m<n$ and a positive Radon measure $\mu$ on $\mathbb R^n$ with at most polynomial growth, assume that it has finite $m$-dimensional Gaussian discrepancy, defined as $$\mathcal D(\mu):=\limsup_{R\to\infty}\Bigl[R^2\sup_{x\in\mathrm{spt}\mu}|G_R^\mu(x)-1|\Bigr],$$ where $G_R^\mu(x):=(4\pi R^2)^{-m/2}\int_{\mathbb R^n} e^{-|z-x|^2/(4R^2)}\,d\mu(z)$. Provided that $\mu$ admits a flat blow-down $\mathcal H^m\llcorner V$ (an assumption that cannot be dropped entirely), we prove that there exists a finite (smallest) $0\le\delta<\infty$, the so-called width of $\mu$, such that $$\mathrm{spt}\mu\subseteq \bar B_\delta(V+x_0),\quad \delta^2\le C(m,n)\mathcal{D}(\mu),$$ where $\bar B_\delta(V+x_0)$ denotes the closed $\delta$-neighborhood of a suitable translate of $V$. Note that the exponent is optimal because of scale invariance. In particular, when $\mathcal D(\mu)=0$, the support of $\mu$ is included in an $m$-dimensional affine plane. We use this fact to give a simpler proof of Preiss's rectifiability theorem: while still based on Gaussian averages of $\mu$, this new proof avoids showing uniqueness (and hence conicality) of blow-downs of $m$-uniform measures, as well as studying Gaussian-weighted moments beyond second order. The strategy of proof consists of an anisotropic blow-down of $\mu$, followed by an analysis of the resulting Young measure, using a rigidity lemma involving the heat semigroup.