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).

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).

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 $\mu_t(A) = \mu(\frac{1}{t} A)$, 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).

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).