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 $\mathcal{B}(H_0)$?

up vote 3 down vote accepted

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 $\mu$. – r_faszanatas Nov 26 at 10:39
  • That is right. I wanted to mention it that the argument works for any measure with a bijective trace-class covariance operator, but then forgot about that. Thank you for your comment. – Iosif Pinelis Nov 26 at 13:55

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