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

  1. for $\rho$-almost every $x \in X$, $(\rho_{x,y})_{y \in X}$ is a rcd of $\rho_x^\mathcal{E}$ with respect to $\mathcal{E}$;
  2. 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

  1. 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 $n$;
  2. [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??

  • 1
    $\begingroup$ I realise (thanks to Jochen Wengenroth) that I should perhaps emphasise: In the case that $\mathcal{E}$ is countably generated, the answer to both questions is 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 is specifically due 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.) $\endgroup$ Aug 11, 2014 at 12:15
  • $\begingroup$ I have deleted my answer (which in fact did not answer the question). Meanwhile, I think that the answer to you question is NO. It might be helpful to study (more carefully than I did) the following projecteuclid.org/… $\endgroup$ Aug 11, 2014 at 12:27
  • $\begingroup$ I think that this projecteuclid.org/euclid.aop/1015345764 answers your question. $\endgroup$ Aug 11, 2014 at 13:06
  • $\begingroup$ Thanks. I think this answers my second question, but not my first (unless I read the papers too quickly). $\endgroup$ Aug 11, 2014 at 14:56
  • $\begingroup$ I've "revamped" the question to take into account the papers you've informed me of. Thank you very much for your help. $\endgroup$ Aug 11, 2014 at 16:09

3 Answers 3


I've found the answer - it's NO!

The paper I found addressing the question is the following:

http://projecteuclid.org/euclid.aop/1175287757 ("0-1 Laws for Regular Conditional Probabilities")

A simple counterexample (which I've just slightly adapted from the counterexample in Example 2 of the above paper) is the following:

Let $X=[0,1] \times \{0,1\}$ (with $\Sigma=\mathcal{B}([0,1]) \otimes 2^{\{0,1\}}$), let $\rho$ be a probability measure on $X$ whose projection onto $[0,1]$ is atomless, and let $\mathcal{E} \subset \Sigma$ be the $\rho$-completion of $\mathcal{B}([0,1]) \otimes \{0,1\}$ relative to $\Sigma$. Then given any non-trivial probability measure $m$ on the binary set $\{0,1\}$, the stochastic kernel

$\hspace{5mm} \rho_{(x,i)}^\mathcal{E} \ := \ \delta_x \otimes m$

is a rcd of $\rho$ with respect to $\mathcal{E}$. Clearly, for any $x \in [0,1]$, $\{(x,0)\} \in \mathcal{E}$; and yet, for all $(x,i) \in X$,

$\hspace{5mm} \rho_{(x,i)}^\mathcal{E}(\{(x,0)\}) \ = \ m(0) \, \in \, (0,1).$

Regarding my motivation: Theorem 12 of the above paper claims to be a generalisation of the ergodic decomposition theorem (for measurable maps). However, I haven't yet managed to work out how to derive the ergodic decomposition theorem from Theorem 12 of the above paper.

  • $\begingroup$ My motivation was to have a "nice" proof of the ergodic decomposition theorem for measurable continuous-time Markov semigroups. However, I now realise I don't know a proof at all; I had had in mind to link Markov semigroups with deterministic dynamical systems via the Markov shift dynamical system---but I have read recently that this construction does not work in general for continuous time. So I've asked a separate question on ergodic decompositions in mathoverflow.net/questions/178700/…. $\endgroup$ Aug 16, 2014 at 23:09

Regarding your motivation, see Dynkin's paper Sufficient statistics and extreme points, Annals of Probability, 1978. He suggests a unified measure-theoretic approach to various extremal decomposition theorems, which I guess is not far from what you intend to do.

  • $\begingroup$ Thank you, I've had a quick browse. Theorem 9.1 (combined with Theorem 3.1) looks like it might be a general statement of the ergodic decomposition theorem for stationary measures of stochastic kernels, but I'd need to read the paper more carefully to be able to tell. $\endgroup$ Aug 12, 2014 at 1:29

The answer is "yes" if the sigma-algebra is countably generated, see Section 5.3 of "Ergodic Theory" by Einsiedler and Ward.

This seems to be a restriction but you still can prove the ergodic decomposition theorem in the general Borel space case, because for a Borel space $(X,\Sigma)$ equipped with a measure $P$ and a sigma-algebra $\mathcal{E}\subset \Sigma$, one can find a countably generated sigma-algebra $\mathcal{E}'$ such that $\mathcal{E}'=\mathcal{E}\pmod P$ (it's a lemma in the same Section)

  • $\begingroup$ Thank you for your reply. I am aware that the answer is "yes" when $\mathcal{E}$ is countably generated (see the first comment below the question). As for the ergodic decomposition: the reference you gave uses the fact that $\mathcal{E}$ is countably generated $\bmod P$ specifically to prove the "invariance" part [the easy part], not the "ergodicity" part [the part that I'm concerned about]. For the "ergodicity" part, the reference you gave still uses Birkhoff's ergodic theorem. $\endgroup$ Aug 12, 2014 at 0:33
  • $\begingroup$ Incidentally, the proof given for the "invariance" part seems a little overkill; see math.nus.edu.sg/~matsr/ProbII/Lec10.pdf for quite a direct proof. $\endgroup$ Aug 12, 2014 at 0:36

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