Let $(X,\Sigma)$ be a standard measurable space, let $\rho$ be a probability measure on $(X,\Sigma)$, and let $\mathcal{E}$ be a sub-$\sigma$-algebra of $\Sigma$. We will say that a stochastic kernel $(\rho_x^\mathcal{E})_{x \in X}$ on $X$ is a *regular conditional distribution of* $\rho$ *with respect to* $\mathcal{E}$ if

- the map $x \mapsto \rho_x^\mathcal{E}(A)$ is $\mathcal{E}$-measurable for all $A \in \Sigma$;
- for every $E \in \mathcal{E}$ and $A \in \Sigma$, $\rho(A \cap E) = \int_E \rho_x^\mathcal{E}(A) \, \rho(dx)$.

Is it necessarily the case that $\rho$-almost every $x \in X$ has the property that for all $E \in \mathcal{E}$, either $\rho_x^\mathcal{E}(E)=0$ or $\rho_x^\mathcal{E}(E)=1$?

Remark: In order for the above to be satisfied, I believe it is sufficient that there exists a family $(\rho_{x,y})_{x,y \in X}$ of probability measures on $X$ such that

- for $\rho$-almost every $x \in X$, $(\rho_{x,y})_{y \in X}$ is a rcd of $\rho_x^\mathcal{E}$ with respect to $\mathcal{E}$;
- the map $(x,y) \mapsto \rho_{x,y}(A)$ is $(\mathcal{E} \otimes \mathcal{E})$-measurable for all $A \in \Sigma$.

(Specifically, if we can find such a family, then I think we can show that for $\rho$-almost every $x$, for $\rho_x^\mathcal{E}$-almost every $y$, $\rho_{x,y}=\rho_x^\mathcal{E}$.)

*Some important remarks:*

The disintegration theorem guarantees that a rcd of $\rho$ with respect to $\mathcal{E}$ exists and is unique modulo $\rho$-null sets.

So of course (at least if we assume AC) there exists a family $(\rho_{x,y})_{x,y \in X}$ satisfying (1); but the question is whether there necessarily exists a family $(\rho_{x,y})_{x,y \in X}$ satisfying both (1) and (2).

It is worth emphasising: we do *not* necessarily have that for $\rho$-almost every $x \in X$, for all $E \in \mathcal{E}$, $\rho_x^\mathcal{E}(E)=\mathbf{1}_E(x)$. (For counterexamples, see the examples of "maximally improper rcd's" in the paper "Improper Regular Conditional Distributions" linked to by Jochen below.)

**Motivation - ergodic decomposition**: I'm keen to have a "nice" proof of the ergodic decomposition theorem for stationary probability measures of stochastic semigroups (jointly measurable in space and time) on standard measurable spaces. If I understand correctly, one can reduce the question of finding an ergodic decomposition of a stationary measure of a stochastic semigroup to the question of finding an ergodic decomposition of an invariant measure of a (deterministic) dynamical system, by considering the time-shift dynamical system on the space of $X$-valued functions of time. Already I'm not sure I'd deem this "nice", but even for dynamical systems I wonder whether there's a nicer proof of the ergodic decomposition theorem than the ones I've seen. For an invariant measure $\rho$ of a measurable dynamical system, the proofs that I've seen involve using Birkhoff's ergodic theorem to show that $\rho_x^\mathcal{I}$ is ergodic for $\rho$-almost all $x$, where $\mathcal{I}$ is the $\sigma$-algebra of invariant sets. But if the answer to my question is *yes*, then the ergodicity of $\rho_x^\mathcal{I}$ for $\rho$-almost all $x$ is immediate (once we have established the invariance of $\rho_x^\mathcal{I}$ for $\rho$-almost all $x$---but that is easy). I guess one could argue that Birkhoff's theorem is "nice enough" as it is, but if the answer to my question is yes, then the same proof will work directly for stochastic semigroups (so that we don't have to invoke the theorem of equivalence between ergodicity with respect to a stochastic semigroup and ergodic of the corresponding Markov measure under the time-shift dynamical system).

**A possible approach?** Perhaps I should mention a possible starting point that I've thought of, but have been unable to make into a full solution:

The difficulty behind the problem is that $\mathcal{E}$ might not be countably generated; however, as hinted at by Yuri below, perhaps it is possible to use the fact that $\mathcal{E}$ is countably generated $\bmod \rho$ to help. Of course, this fact cannot mean that all arguments for the countably generated case remain valid in the general case, since as we have said already, it is not necessarily the case that for $\rho$-almost every $x \in X$, for all $E \in \mathcal{E}$, $\rho_x^\mathcal{E}(E)=\mathbf{1}_E(x)$.

Nonetheless, perhaps we can proceed as follows: Let $\{E_n\}_{n \in \mathbb{N}} \subset \mathcal{E}$ be such that $\mathcal{E}$ is contained in the $\rho$-completion of $\sigma(E_n:n \in \mathbb{N})$. For each $n$, let $\mathcal{G}_n:=\sigma(E_i : 1 \leq i \leq n)$. Then I believe we have that

- for $\rho$-almost all $y \in X$, $(\rho_x^\mathcal{E})_{x \in X}$ is a rcd of $\rho_y^{\mathcal{G}_n}$ with respect to $\mathcal{E}$ for all $n$;
*[by the result mentioned in (2) of Conditional law as a random measure and convergence of random measures, combined with Levy's Upward Theorem]*for any fixed Polish topology on $X$, for $\rho$-almost all $y \in X$, $\rho_y^{\mathcal{G}_n} \to \rho_y^\mathcal{E}$ in the narrow topology as $n \to \infty$.

If I can somehow show that (1) and (2) together imply that

$\hspace{7mm}$ for $\rho$-almost all $y \in X$, $(\rho_x^\mathcal{E})_{x \in X}$ is a rcd of $\rho_y^\mathcal{E}$ with respect to $\mathcal{E}$

then I'm done! Any ideas??

yes, since clearly, for each $E \in \mathcal{E}$, $\rho_x^\mathcal{E}(E)=\mathbf{1}_E(x)$ a.e. (and then the $\pi$-$\lambda$ theorem gives the result, by considering a countable $\pi$-system generating $\mathcal{E}$). The difficulty in my questions isspecificallydue to the fact that I haven't assumed $\mathcal{E}$ to be countably generated. (I've only assumed $\Sigma$ to be countably generated, since $\Sigma$ is standard.) $\endgroup$ – Julian Newman Aug 11 '14 at 12:15