Questions about some properties of random probabilities and random expectations 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?

It's easy to see that $(3)$ implies (2), and this paper (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.
 A: This is essentially Theorem 7 of 

Blackwell, David. "Idempotent Markoff chains." 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.
A: First of all, it seems easier to me to reformulate the question in Markov terms. Then your "random probability measures" become just the transition probabilities of a Markov chain on $\Omega$, condition (1) means that the measure $\mathbb P$ is stationary, and condition (2) means that almost surely one step transition probabilities coincide with the two step transition probabilities. Assuming the measure $\mathbb P$ is ergodic, this implies (by the ergodic theorem), that almost all transition probabilities coincide with $\mathbb P$. In the general case one decomposes $\mathbb P$ into the ergodic components with respect to the chain (this is where one needs the separability condition) and applies the above argument component-wise.
