Is this property of a sequence of random probabilities well-known? Let $(\Omega, \mathcal{F}, (\mathcal{F}_n)_n, P)$ be a probability space equipped with a filtration, and let $(P_n)_n$ be a sequence of random probabilities adapted to $(\mathcal{F}_n)_n$. By that I mean that $P_n: \mathcal{F} \times \Omega \to [0,1]$ is a probability measure on $(\Omega, \mathcal{F})$ in its first argument and a $\mathcal{F}_n$-measurable random variable in its second argument. Assume that $P_1(A, \omega) = P(A)$.
I'm wondering if the following property is well-known and if it has any interesting consequences:

For all $n$, all $m > n$, ($P$-almost) all $\omega \in \Omega$, and all $A \in \mathcal{F}$, 
  $$P_n(A, \omega) = \int P_m(A, \omega')P_n(d\omega', \omega).$$

It seems plausible to me, for instance, that this property could guarantee that $(P_n(A))_n$ has an almost sure limit for every $A$, but I haven't been able to prove it yet. 
Mainly, I'm just wondering if this property has been studied, and if so, what a good reference is.
 A: I'll address a slightly different question, namely, about a similar family satisfying 
$$
P_n(A, \omega) = \int_\Omega P_m(A, \omega')P_n(d\omega', \omega) \tag{1}
$$
for all $A\in\mathcal F$, almost all $\omega\in \Omega$, and all $m\ge n$ (so I assume that the equality also holds for $m=n$).
I claim that in this case $P_n(A,\omega) = P(A\mid \mathcal G_n)$ for some $\sigma$-algebra $\mathcal G_n$ such that $\mathcal G_n \subset \mathcal F_n\cap \mathcal G_{n+1}$. 
For $X\in L^1(\Omega)$ define
$$
E_n X = \int_{\Omega} X(\omega')P_n(d\omega',\omega).
$$
Then $(1)$ is equivalent to:
$$
E_n E_m  = E_{n}, m\ge n. \tag{2}
$$
Also note that $E_1 X = E[X]$.
Setting $n=m$ in $(2)$, we get $E_n^2 = E_n$, so $E_n$ is a projection. Note also that $E_n$ is positive and
$$
E[|E_n X|]\le E[E_n |X|] = E_1 E_n |X| = E_1|X| = E[|X|],
$$
so $E_n$ is contractive in $L^1(\Omega,\mathcal F,P)$. Finally, since $P_n$ is a probability measure, $E_n \mathbf{1} = \mathbf{1}$. Therefore, by Douglas' Theorem (Corollary 4 here), there exists some $\sigma$-algebra $\mathcal G_n$ such that $E_n X = E[X\mid \mathcal G_n]$. Owing to the assumption that $P_n$ is $\mathcal F_n$-measurable in the second argument, we have $\mathcal G_n\subset \mathcal F_n$. The fact that $\mathcal G_{n}\subset \mathcal G_{n+1}$ follows from $(2)$ applied to $m=n+1$.
