This is theorem 2.8.3 from Bogachev‘s book „Gaussian measures“:
Let $\gamma$ be a Gaussian measure on a locally convex space $X$ and let $q$ be a $\mathcal{E}(X)_{\gamma}$-measurable seminorm on $X$. Then the restriction of $q$ to the Cameron-Martín space $H(\gamma)$ is continuous.
I have read the proof and it looks perfectly plausible, but the Cameron-Martin space has measure zero. Can‘t I just redefine $q$ on $H(\gamma)$ to be the absolute value of a discontinuous linear functional and get away with it?
As was expected, the devil is in a rather subtle detail. A measurable seminorm is defined as a Lebesgue-completion-measurable function that is a seminorm on some linear subspace of full measure. But as it turns out, all linear subspaces of full measure contain the Cameron-Martin space (theorem 2.4.7 in the same book). Hence, while modifying a measurable seminorm on the Cameron-Martin space gives an object that is almost everywhere equivalent, the result will generally not be a measurable seminorm according to the above definition.