Let $(X,Y,b)$ be a pairing, i.e. $X$ and $Y$ are real vector spaces and $b:X\times Y\to\mathbb R$ is bilinear. $\sigma(X,Y,b)$ is the weakest topology on $X$ such that $b(\cdot, y)$ is continuous for all $y\in Y$. Let $\Sigma$ be the Borel sigma algebra generated by $\sigma(X,Y,b)$.
Let $Z=\{ b(\cdot, y) :y\in Y \}$ and let $\Sigma'$ be the smalest $\sigma$-algebra such that all $z\in Z$ is measurable.
Under what conditions is it true that $\Sigma=\Sigma'$ ?
It is quite clear that $\Sigma'\subseteq \Sigma$, indeed all element of $z$ are $\sigma(X,Y,b)$-continuous and therefore $\Sigma$-measurable.
I am a little worried about the other direction, indeed suppose we have an open set $A=\bigcup_{i\in\mathcal I} z_i^{-1}(B_i)$ where $\mathcal I$ is uncountable, $z_i\in Z$ and $B_i$ some non empty open intervals. Then it feels like because the union is uncountable, this may not be in $\Sigma'$.
My goal is to try to define $\Sigma$ without appealing to topology in a document I am writing, I would like to avoid imposing a topological background on the people that might read it, hopefully it is doable in some way.
Any input on one of those two things would be most welcome.
An alternative definition without having to dwell too much into topology would be the following.
For any $A\subseteq X\times Y$, let $U_A=\{ z\in X : \forall (x,y)\in A, b(x,y)\leq b(z,y) \}$. Then let $\mathcal F=\{ U_A : A\subseteq X\times Y \}$, it is not too hard to show that $\sigma(\mathcal F)=\Sigma$.
I think this could do for my purpose but anything simpler would be most welcome.