Let $\mathcal P(X)$ denote the space of all probability measure defined on a measurable space $X$. We canonically endow the former with its own measurability structure, generated by evaluation maps. Let $P \subseteq \mathcal P([0,1])$ be a measurable subset of probability measures, and let $\hat p\notin P$ be such that for every bounded measurable $f:[0,1] \to \Bbb R$ there exists $p_f\in P$ satisfying $$ \int_{[0,1]} f(x) \hat p(\mathrm dx) = \int_{[0,1]} f(x) p_f(\mathrm dx), $$ or $\hat p f = p_f f$ in a short form. Does it necessarily means that there exists a probability measure $\nu \in \mathcal P(\mathcal P([0,1]))$ such that $\nu(P) = 1$ and $\hat p = \int_P p\,\nu(\mathrm dp)$?

I think this result is quite easy to show for finite $X$ where $\mathcal P(X)$ is just a subset of $\Bbb R^n$, however I am not sure whether it still holds true in my more general case. Obviously, from $[0,1]$ it would generalize to any Borel space.

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