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.
 A: 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.

