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Let $H$ be a separable Hilbert space and $\nu$ its canonical cylinder measure. By the construction of Gross there exists a separable Banach space $X$ s.t. $i: H \hookrightarrow X$ and a measure on $X$ defined by

$$\mu(C) = \nu(i^{-1}(C)) = \nu(C \cap H), \forall C \in \mathcal{C} $$

where $\mathcal{C}$ denotes the cylinder algebra on $X$, and the $i(H)$ coincides with the Cameron-Martin subspace of $(X, \mu)$.

Now, on the one hand, the definition of the measure $\mu$ implies that

$$\mu(H) = \nu(H \cap H) = \nu(H) = 1$$

be the definition of being a cylinder measure. But on the other hand, one can show that the Cameron-Martin subspace has measure $0$ unless $X$ is finite-dimensional.

Where is my mistake here?

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  • $\begingroup$ Your "definition" of $\mu$ can't be right because $\nu(i^{-1}(C))$ wouldn't be defined for $C \in \sigma(\mathcal{C})$, only for $C \in \mathcal{C}$ itself. Where did you read this definition, and is it possible you misread it? $\endgroup$ Jun 24, 2021 at 17:50
  • $\begingroup$ Correct. It should only be the cylinder algebra. Though I am still confused about $\mu(H)$ supposedly being $1$. Maybe $H \not\in \mathcal{C}$ after all $\endgroup$ Jun 24, 2021 at 18:05
  • $\begingroup$ Yeah, $i(H)$ is definitely not in the cylinder algebra of $X$. $\endgroup$ Jun 24, 2021 at 18:37

1 Answer 1

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When $X$ is infinite-dimensional, the Cameron-Martin space $H$ is not a cylinder set of $X$ (exercise), and so the first equation does not imply $\mu(H)=1$, and there is no contradiction.

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