I have a basic question about Gaussian measures on a Hilbert space: 


Let $\mu$ be a non-degenerate Gaussian measure on a Hilbert space $(H\_0,\left\langle \cdot,\cdot \right\rangle\_0)$. Then the covariance operator $S$ of $\mu$ is a bijective, non-negative, self-adjoint trace-class operator on $H\_0$.
Therefore,
$$
\left\langle x,y \right\rangle\_1 := \left\langle \sqrt{S} x, \sqrt{S} y \right\rangle\_0, \qquad x,y \in H\_0
$$
defines another inner product on $H\_0$ which is weaker than $\left\langle \cdot,\cdot \right\rangle\_0$, as for the induced norms we have that $\left\lVert \cdot \right\rVert\_1 \leq \left\lVert T \right\rVert\_{op} \left\lVert \cdot \right\rVert\_0$. Let $H\_1$ be the completion of $H\_0$ with respect to $\left\lVert \cdot \right\rVert\_1$. Then $H\_0 \subseteq H\_1$ densely and as a Borel set.


Let $\mathcal{B}(H\_0)$ and $\mathcal{B}(H\_1)$ denote the Borel sets of $H\_0$ and $H\_1$, with respect to the norms $\left\lVert \cdot \right\rVert\_0$ and $\left\lVert \cdot \right\rVert\_1$. Then $\mathcal{B}(H\_1) \cap H\_0 \subseteq \mathcal{B}(H\_0)$, where $\mathcal{B}(H\_1) \cap H\_0 = \{ A \cap H\_0 \colon A \in \mathcal{B}(H\_1) \}$


**My question is:**
If a probability measure $\nu$ defined on $(H\_0,\mathcal{B}(H\_0))$ agrees with $\mu$ on $\mathcal{B}(H\_1)\cap H\_0$, does it follow that $\nu=\mu$, i.e. that those measures also agree on $\mathcal{B}(H\_0)$? 



Is a Gaussian measure on a Hilbert space determined by the coarser topology induced by the covariance operator?
The answer is yes. Indeed, let $(e\_1,e\_2,\dots)$ be an orthonormal eigenbasis of $S$. For each natural $n$, let $V\_n$ be the linear span of $(e\_1,\dots,e\_n)$ and let $P\_n$ be the orthoprojector from $H\_1$ onto $V\_n$. Let $R\_n$ be the restriction of $P\_n$ to $H\_0$. Then the simple but crucial observation is that $R\_n$ is the orthoprojector from $H\_0$ onto $V\_n$. 


Take any $y\in H\_0$ and any real $r>0$, and let $B:=B\_r(y)$, the closed ball in $H\_0$ of radius $r$ (with respect to $\|\cdot\|\_0$) centered at $y$. Let then $B\_n:=R\_n B$, which is the closed ball in $V\_n$ of radius $r$ centered at $R\_n y$. Let $A\_n:=P\_n^{-1}(B\_n)$. Then $A\_n\in\mathcal B(H\_1)$, since $P\_n$ is bounded, and hence continuous and measurable. So, $A\_n\cap H\_0\in\mathcal B(H\_1)\cap H\_0$. 
Because $R\_n$ is the restriction of $P\_n$ to $H\_0$, we have $C\_n:=R\_n^{-1}(B\_n)=A\_n\cap H\_0\in\mathcal B(H\_1)\cap H\_0$, so that $\nu(C\_n)=\mu(C\_n)$, for all $n$. 


But the sequence $(C\_n)$ is decreasing and its intersection is $B$. So, $\nu(B)=\mu(B)$, for all closed balls in $H\_0$. So, $\nu=\mu$ on $\mathcal B(H\_0)$. 


