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.



Convex representation of a measure
The answer is no.


I understand the condition on $p\_f$ as belonging to the closure $\bar{P}$ for the weak-$\star$ topologie. And one can then ask whether $[\int p d\nu(p),\nu\in \mathcal{P}(\mathcal{P}([0,1]))]$ contain this closure. Consider the following counter example $$P=[\lambda \delta\_0+(1-\lambda)\delta\_1:0<\lambda<1]$$ Then $\delta\_0 \in \bar{P}$ but not to in the convex set.


Here we have $p\_ff=\hat{p}f$ which is a stronger condition than $\inf\_{p\in P}|pf-\hat{p}f|=0$. We therefore have to work a bit harder to construct the counter example.


Let $\hat{p}=\frac{1}{2}\delta\_0+\frac{1}{2}1\_{[0,1]}$ and choose $$P=\{q\_x :x\geq \frac{1}{2}\}\cup[p\in \mathcal{P}([0,1]):p1\_{\{0\}}<\frac{1}{2}]$$ with $q\_x=\frac{1}{2}\delta\_0+\frac{1}{2}\delta\_x$
We see that if $\hat{p}=\int\_P pd\nu(p)$ then $$\hat{p}1\_{\{0\}}=\frac{1}{2}=\frac{1}{2}\nu(\{q\_x : x\geq \frac{1}{2}\})+\int\_{P-\{q\}} p1\_{\{0\}}d\nu(p)$$ and therefore the support of $\nu$ is a subset of $\{q\_x :x\geq \frac{1}{2}\}$ which is impossible.


We now check that $P$ satisfies the condition. Let $f$ a bounded (measurable) function.


If $f(0)=a$ and $f(t)=b$ for all $t\in E\subset [0,1]$ a set of Lebesgue measure 1. Then there exist $x\geq \frac{1}{2}$ such that $f(x)=b$ and we have $\hat{p}f=\frac{1}{2}(a+b)=q\_x f$.


If $f$ is not constant on a set of Lebesgue measure 1. then there exists $\epsilon >0$ and $\tilde{p}$ with $\tilde{p}1\_{\{0\}}=0$ and $$\tilde{p}f>(1+\epsilon)\int f(t)dt $$
Suppose $f(0)\geq \int f(t)dt$. Then by continuity we can find $\lambda<\frac{1}{2}$ such that $\lambda f(0)+(1-\lambda)\tilde{p}f=\hat{p}f$.


Suppose $f(0)\leq \int f(t)dt$ then we use the same argument but with $$\tilde{p}f<(1-\epsilon)\int f(t)dt $$


