Convex representation of a measure

Question

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.

Answer 1

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 $$\tilde{p}f<(1-\epsilon)\int f(t)dt $$