Do isonormal Gaussian processes have measurable sample paths?

Let $H$ be a real separable Hilbert space. Let $W=\{W(h):h\in H\}$ be a real-valued stochastic process defined on a complete probability space $(\Omega,\mathcal{F},P)$. Assume that $W$ is a centered Gaussian family of random variables such that $E[W(h)W(g)]=\langle h,g\rangle_H$ for all $h,g\in H$. In other words, $W$ is an isonormal Gaussian process on $H$.

For $\omega\in\Omega$, the sample path of $W$ is the function $h\mapsto W(h,\omega)$ from $H$ to $\mathbb{R}$. Is it true that for $P$-a.e. $\omega\in\Omega$, the sample path is $(\mathcal{B}(H),\mathcal{B}(\mathbb{R}))$-measurable, where $\mathcal{B}(H)$ and $\mathcal{B}(\mathbb{R})$ are the Borel $\sigma$-algebras on $H$ and $\mathbb{R}$, respectively? Or, perhaps more precisely, is it true that $W$ has a modification that has this property?

A related question, I suppose, is whether or not $W$ is a measurable process, i.e., is $(h,\omega)\mapsto W(h,\omega)$ a $(\mathcal{B}(H)\otimes\mathcal{F},\mathcal{B}(\mathbb{R}))$-measurable function. I assume this is well-known, but I could not find the answer in any of my usual sources. Hopefully someone out there can help me with a reference. Thanks.

EDIT: Here is something I tried. Fix $\omega\in\Omega$. We want to show that $W(\cdot,\omega)$ is a measurable function on $H$. Let $\{e_i\}$ be an orthonormal basis of $H$ and define $$W_n(h,\omega) = \sum_{i=1}^n W(e_i,\omega)\langle h,e_i\rangle_H.$$ Clearly, $W_n(\cdot,\omega)$ is a measurable function on $H$. Since the pointwise limit of measurable functions is measurable, it will suffice to show that $W_n(h,\omega)\to W(h,\omega)$ for all $h\in H$. Since I want this to be true for $P$-a.e. $\omega\in\Omega$, I want to show that $$P(W_n(h) \to W(h) \text{ for all } h\in H) = 1.$$ Unfortunately, all I know how to prove is $$P(W_n(h) \to W(h)) = 1 \text{ for all } h\in H.$$ In fact, I believe the probability with the quantifier on the inside is 0, since $W_n(h) \to W(h)$ for all $h\in H$ seems to imply $W(e_i)\to0$, whereas $\{W(e_i)\}$ are i.i.d. standard normals.

• Hi Jason, to let your question open, I removed my answer since it was not complete. – Fabrice Baudoin Jun 5 '14 at 22:41
• Hi Jason, can't you just set $\tilde W(h,\omega) = \lim_n W_n(h,\omega)$ for those $h,\omega$ for which the limit exists and $\tilde W(h,\omega) = 0$ otherwise? This should give you the measurable version you're after. – Martin Hairer Jun 16 '14 at 19:54
• Yes, as Martin points out, this gave at least a measurable version. Related to the question, the paper projecteuclid.org/euclid.aop/1176997026 studies continuity sets of W, that is the subsets of H on which W is continuous for some natural topology (such sets may for instance be obtained by looking at the uniformity in h of the convergence $W_n \to W$) – Fabrice Baudoin Jun 16 '14 at 23:55
• Thanks, Martin. I can see it now. I figured I was overlooking something elementary. If you want to write this as an answer, then I can accept it. – Jason Swanson Jun 19 '14 at 20:16