Let $(\mathcal X,\Sigma_X,p_X)$ be a probability space. Define the following set $S=\{ p : (\mathcal X,\Sigma_X,p)\text{ is a probability space, } p\ll p_X \}$ where $\ll$ denote absolute continuity. I equip this space with the total variation metric : $\| p-q \|=\sup_{A\in\Sigma_X} |p(A)-q(A)|$ (I am not too attached to that metric if something else would work better). I want to make statements on that space of the kind of the Krein-Milman theorem on that set, I know it is not compact but following another question of mine I think it could be enough for me to have that $S$ to be a topological subset of some reflexive TVS $E$, hence my question
Is there a reflexive topological vector space $E$ such that $S$ is a topological subset of $E$.
I wonder if we have to add a condition that $p_X$ is atomic, if so I think the extreme points of $S$ may be something like $\{ 1_A : A\in\mathcal A \}$ where $\mathcal A$ are the atoms of $p_X$.
I think there is a problem with that, there is a chance that if $\mathcal X=[0,1]$, $\Sigma_X$ is the set of Borel subsets of $\mathcal X$ and $p_X$ is the Lebesgue measure on $[0,1]$ then $S$ doesn't have any extreme points right ? But having $S$ in a reflexive space would mean that $S$ is the closed convex hull of it's extreme points which it could not be. Am I wrong there ?
To long for a comment.
Yes, if $S$ does not possess extremal points, it cannot be compact in any locally convex vector space topology (due to Krein-Milman).
It is also possible to apply Krein-Milman to the weak-* topology in duals of Banach spaces. I think in most cases, the set of probability measures is weak-* compact in the dual of $C_0(\mathcal X)$ (continuous functions on $\mathcal X$ vanishing at infinity). However, with your additional absolute continuity constraint you loose the weak-* closedness of the set.
Finally, if $p_X$ is atomic, your set $S$ should be isomorphic to a subset of (a weighted version of?) $\ell^1$, which is the dual of $c_0$ (convergent sequences with limit $0$).