Abstract
Let $(\mu_t)_{t>0}$ be a Gaussian semigroup on a metric, separable, complete group $G$. If $H$ is a Borel measurable normal subgroup of $G$ such that $\mu_t(H) > 0$ for all $t$, then $\mu_t(H) = 1$ for every $t$. If, in addition, $\mu_t$ are symmetric, then $\mu_t(H) > 0$ for a single $t$ implies $\mu_t(H) = 1$ for all $t$.
Citation
T. Byczkowski. A. Hulanicki. "Gaussian Measure of Normal Subgroups." Ann. Probab. 11 (3) 685 - 691, August, 1983. https://doi.org/10.1214/aop/1176993513
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