 Solution

Because $V$ is measurable for the cylinder sigma-algebra, membership in $V$ depends on only countably many coordinates $T_0=\{t_1,t_2,\dots \}$. Its image in ${\mathbb{R}}^{T_0}$ is a measurable linear subspace $V_0$, and

By successively applying the finite-dimensional Gaussian regression formula, realize this Gaussian sequence as a lower-triangular linear transform of independent standard normal variables:where every coordinate of the sum contains only finitely many terms. If some deterministic column $a_k$ does not belong to $V_0$, then, after conditioning on every $g_j$ except $g_k$, at most one value of $g_k$ can put the sum in $V_0$. The continuous normal distribution gives probability zero. If every $a_k$ belongs to $V_0$, changing finitely many $g_k$ does not change the membership event. It is then a tail event, and the Kolmogorov zero-one law gives probability zero or one. This proves the Gaussian zero-one law for measurable linear subspaces.

Now let $X(t)=\sqrt{t},g_t$ for independent standard normal variables $g_t$. DefineBoth are cylinder-measurable infinite-dimensional linear subspaces. For every $\epsilon >0$,so the Borel-Cantelli lemmas imply $g_t/\sqrt{t}\to 0$ almost surely and hence $\mathbb{P}(X\in V)=1$. On the other hand, $g_t^2\ge 1$ infinitely often almost surely, again by Borel-Cantelli, soalmost surely. Therefore $\mathbb{P}(X\in W)=0$.

Solved by gpt-5.6-sol high.