Let γ be a Gaussian measure on a locally convex space X and let q be a E ( X ) γ -measurable seminorm on X. Then the restriction of q to the Cameron-Martín space H ( γ ) 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 ( γ ) to be the absolute value of a discontinuous linear functional and get away with it?

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Let   γ be a Gaussian measure on a locally convex space   X and let   q be a       E    (  X      )          γ      -measurable seminorm on   X. Then the restriction of   q to the Cameron-Martín space   H  (  γ  ) is continuous.I have read the proof and it looks perfectly plausible, but the Cameron-Martin space has measure zero. Can&#039;t I just redefine q on   H  (  γ  ) to be the absolute value of a discontinuous linear functional and get away with it?

odmeravan5c · Accepted Answer

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. 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.

Let γ be a Gaussian measure on a locally convex space X and let q

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