# Is a functional bounded by a measurable seminorm also measurable?

Let $$\mu$$ be a centred Radon Gaussian measure on a locally convex space $$X$$ and $$q : X \to \mathbb{R}$$ a seminorm that is $$\mathcal{B}(X)_\mu$$-measurable, where $$\mathcal{B}(X)_\mu$$ is the Lebesgue completion of the Borel $$\sigma$$-algebra on $$X$$. Consider a linear functional $$\phi : X \to \mathbb{R}$$ with $$\vert \phi \vert \le q$$. Is it also necessarily $$\mathcal{B}(X)_\mu$$-measurable?

It follows from theorem 2.10.11 in Bogachev's monograph "Gaussian measures" that there is a unique (up to equivalence $$\mu$$-almost everywhere) linear functional $$\psi : X \to \mathbb{R}$$ with $$\phi = \psi$$ on the Cameron-Martin space of $$\mu$$. But I do not see how to prove whether the equality would hold $$\mu$$-almost everywhere.

I have found an answer exploiting the equivalence of Lusin and Borel measurability as stated in Radon Measures on Arbitrary Topological Spaces and Cylindrical Measures on page 6, theorem 5:

Let $$H : X \to Y$$. If $$H$$ is Lusin $$\mu$$-measurable, then $$H$$ is Borel $$\mu$$-measurable, and conversely, if $$Y$$ is metrizable and separable, then every Borel $$\mu$$-measurable function is also Lusin $$\mu$$-measurable.

By assumption $$q : X \to \mathbb{R}$$ is Borel $$\mu$$-measurable such that $$q$$ is also Lusin $$\mu$$-measurable. Consequently, there is a compact set $$K$$ of positive measure on which $$q$$ is continuous and in particular, bounded. Hence, $$\phi$$ is also bounded on $$K$$ and the proof is now actually given in theorem 3.11.3 of Bogachev's book.

The idea is to prove that $$\phi$$ has measurable minorants and majorants that are linear.

For completeness, the definitions of measurability from Schwartz's book on page 25:

Definition 9. The mapping $$H$$ is said to be Lusin $$\mu$$-measurable if for every compact set $$K \subseteq X$$ and every $$\delta > 0$$, there exists a compact set $$K_\delta \subseteq K$$ with $$p(K \setminus K_\delta) < \delta$$ and such that $$H$$ restricted to $$K_\delta$$ is continuous.

We recall here the weaker notion of Borel $$\mu$$-measurable functions: $$H$$ is Borel $$\mu$$-measurable if, the inverse image under $$H$$ of every Borel set in $$Y$$, is a $$\mu$$-measurable set.

• Just wanted to add that Schwartz's book can be found online; the Tata Institute for Fundamental Research has a scanned version on its publications page. (The quoted theorem is on pages 26 to 27 of the text, or pp. 42–43 of the PDF.) Commented Jan 25 at 2:44