# Is a Gaussian measure on a Hilbert space determined by the coarser topology induced by the covariance operator?

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 $$\mathcal{B}(H_0)$$?

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 $$\mathcal B(H_0)$$.
• Thank you very much, a very clear and elegant answer as always! So if I understand correctly, your argument is completely independent of the Gaussian structure, so it actually proves my question when starting from any non-degenerate measure $\mu$. Nov 26, 2018 at 10:39