Let $\mathcal P([0, 1])$ be the space of all probability measures on $[0, 1]$ endowed with the total variation metric. Let $P\subseteq \mathcal P([0, 1])$ be its closed convex subset, and $p'$ a measure that lies outside of it. Does there exist a bounded Borel-measurable function $f:[0,1]\to \Bbb R$ such that $$ p'f > \sup\limits_{p\in P} pf. $$ It seems that Hahn–Banach separation theorem can be applied here, but I am not sure in which way exactly.
Update: The answer to the original question is negative, unless we use a refined version of convexity that I had in mind initially, but did not put in the beginning, hoping that the usual definition of convexity would suffice. Unfortunately, it also means that Hahn-Banach separation theorem cannot be readily applied to my case.
First, why the original question has negative answer. Let $P$ be the closure (in total variation) of all discrete measures on $[0, 1]$ with finite support. No absolutely continuous measure belongs to $P$ since they are mutually singular with discrete measures. Yet, for any measure $p'$ and any given function $f$ we cna always construct a sequence of discrete measures $p_n$ such that $\sup_n p_nf \geq p'f$.
Second, instead of usual convexity I am working with barycentric convexity. A measurable set of measures $P\subseteq \mathcal P([0, 1])$ is said to be barycentrcially convex if for any measure $\nu \in \mathcal P(\mathcal P([0, 1]))$ such that $\nu(P) = 1$ it holds that $\int_P \mu \;\nu(\mathrm d\mu)\in P$. So, instead of only taking convex combinations with finite weights (like in usual convexity) we are allowed to take any possible linear combinations. So my question should read as: if $P$ is a barycentrically convex, closed (in TV) subset of $\mathcal P([0,1])$...
In this formulation the previous counterexample does not apply: the set of all discrete measures is not barycentrically closed, in fact its barycentric closure consists of all measures.