Let $(\Omega, \mathcal{A}, \mathbb P)$ be a probability space with $\mathcal{A}$ countably generated, and let $P: \mathcal{A} \times \Omega \to [0,1]$ be a random probability measure. By that I mean $P$ is a probability measure on $(\Omega, \mathcal{A})$ in its first argument and an a $\mathcal{A}$-measurable function in its second argument.


Assume that for all $A \in \mathcal{A}$
$$\mathbb P(A) = \int P(A,\omega)\mathbb P(d\omega), \tag{1}$$
and that for $\mathbb P$ almost every $\omega$ and all $A \in \mathcal{A}$
$$P(A, \omega) = \int P(A, \omega')P(d\omega', \omega).\tag{2}$$


For each $\omega \in \Omega$, let $C\_\omega = \{\omega' \in \Omega: P(\cdot, \omega') = P(\cdot, \omega)\}$.



> 
> Do (1) and (2) imply that for $\mathbb{P}$ almost every $\omega$
> $$P(C\_\omega, \omega)=1? \tag{3}$$
>  Does (2) imply (3) without (1)? Can the assumption that $\mathcal{A}$ is countably generated be dropped?
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> 
> 


It's easy to see that $(3)$ implies (2), and [this paper](https://www.researchgate.net/profile/Haim_Gaifman/publication/221551242_A_Theory_of_Higher_Order_Probabilities/links/55be62c208ae9289a099d712.pdf) (Theorem 2) asserts that the answer to my first question is affirmative, by "advanced ergodic theory." I have been unable to reconstruct the argument that (1) and (2) imply (3), with ergodic theory or otherwise.



Questions about some properties of random probabilities and random expectations
This is essentially Theorem 7 of 



> 
> Blackwell, David. "[Idempotent Markoff chains](https://www.jstor.org/stable/1968811)." Annals of Mathematics
>  (1942): 560-567.
> 
> 
> 


Blackwell shows there that if $(\Omega,\mathcal{A})$ is countably generated and (2) holds
for all $\omega$ and $A$, then (3) holds for all $\omega$ outside a measurable set $N$ such that $P(N,\omega)=0$ for all $\omega$. This clearly implies the desired result since (1) implies then that $\mathbb{P}(N)=0$ and we can always get rid of a null set on which (2) fails to apply Blackwell's result to the resulting smaller space.


