TITLE: $L^\infty$ properties of an infinite-dimensional Gaussian semigroup
QUESTION [5 upvotes]: Let $W$ be a separable Banach space and $\mu$ a Gaussian Borel measure on $W$ which is centered and non-degenerate.  For $F : W \to \mathbb{R}$ bounded Borel and $t \ge 0$, let
$$P_t F(x) = \int_W F(x+\sqrt{t}y)\,\mu(dy)$$
be the transition semigroup of Brownian motion on $W$.

Is $P_t$ well defined as an operator from $L^\infty(W,\mu)$ to itself?  That is, if $F=0$ $\mu$-almost everywhere, must the same be true of $P_t F$?

Of course, if $W$ is finite dimensional the answer is yes, because the measures $\mu(x +  \cdot)$ are mutually absolutely continuous as $x$ ranges over $W$.  This suggests that the answer may be no in infinite dimensions, but I can't seem to find a counterexample.

REPLY [5 votes]: George Lowther's answer (thanks George!) set me on the right track.  Here's a note that the same argument can be used for arbitrary infinite-dimensional $W$.
Let $\mu_t(A) = \mu(\frac{1}{\sqrt{t}} A)$, so that $P_t F(x) = \int_W F(x+y)  \mu_t(dy)$.  Since $\mu_t$ is a convolution semigroup, we have $\int_W P_t F(x)\mu(dx) = \int_W F(x) \mu_{1+t}(dx)$.  (Or written another way, it's $P_{1+t}F(0)$.)
We can choose a sequence $f_1, f_2, \dots \in W^*$ which, viewed as random variables on the probability space $(W,\mu)$, are iid $N(0,1)$.  (Equip $W^*$ with the $L^2(\mu)$ inner product $\langle f ,g \rangle = \int_W f(x) g(x) \mu(dx)$ and use Gram-Schmidt.)  Then on $(W,\mu_t)$ they are iid $N(0,t)$.   Let
$$A_t = \left\{x \in W : \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n |f_k(x)|^2 = t\right\}.$$
Clearly the $A_t$ are Borel, pairwise disjoint, and by the strong law of large numbers $\mu_t(A_t)=1$.  If we take $F$ to be the indicator of $A_1^C$, then $F = 0$ $\mu$-a.e., but for every $t > 0$, $$\int_W P_t F(x) \mu(dx) = \mu_{t+1}(A^C) = 1.$$  Indeed, since $P_t F \le 1$, we have $P_t F = 1$ $\mu$-a.e.
This also works if $W$ is replaced by any reasonable infinite-dimensional TVS (say, Hausdorff and locally convex).