
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 $t$.
Analysis on real and complex Lie groups, Gaussian semigroups of measures, Trotter approximation theorem, Probability measures on groups or semigroups, Fourier transforms, factorization, Zero-one laws, Gaussian semigroups, 60B15, convolution semigroup of probability measures, 22E30
Analysis on real and complex Lie groups, Gaussian semigroups of measures, Trotter approximation theorem, Probability measures on groups or semigroups, Fourier transforms, factorization, Zero-one laws, Gaussian semigroups, 60B15, convolution semigroup of probability measures, 22E30
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