I posted this question on ME as "A Gaussian measure on $\mathcal{E}'(S^1)$ by Minlos Theorem and its value for $L^2(S^1)$", but it seems much more nontrivial than I expected... so, I post an extended version here.

Let $\mathcal{E}(S^1)$ be the space of smooth functions on the circle $S^1$ and denote its dual as $\mathcal{E}'(S^1)$.

Then, by the Minlos theorem, there exists a unique probability measure $\mu$ on $\mathcal{E}'(S^1)$ such that \begin{equation} e^{-\frac{1}{2}\lVert f \rVert^2_{L^2}}=\int_{\mathcal{E}'(S^1)} e^{iT(f)}d\mu(T) \end{equation} for all $f \in \mathcal{E}(S^1)$.

Also, we know that $\mathcal{E}(S^1) \subset H^{\alpha}(S^1) \subset \mathcal{E}'(S^1)$ for all $\alpha \in [0,\infty)$.

My question: is it true that $\mu\bigl(H^{\alpha}(S^1) \bigr) \neq 0$ for any $\alpha \in [0,\infty)$? Also, can we get an exact value of $\mu\bigl(\mathcal{E}(S^1) \bigr)$?

Could anyone please help me?