# Cameron-Martin space has measure $1$ and also $0$?

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?

• 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? Jun 24, 2021 at 17:50
• 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 Jun 24, 2021 at 18:05
• Yeah, $i(H)$ is definitely not in the cylinder algebra of $X$. Jun 24, 2021 at 18:37

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.