There is a paper "Conditioning as disintegration" by J. T. Chang and D. Pollard, which seems to construct the regular conditional probability from the disintegration. In particular, from Definition 1, Theorem 1 and Theorem 2.(iii) in that paper, we can summarize a theorem as follows:

Theorem. Let $\Omega$ be a Polish space, $\mathcal F = \mathcal B(\Omega)$ be the Borel $\sigma$-field for $\Omega$, and $\mathbf P$ be a probability measure on $(\Omega,\mathcal F)$. Let $(E,\mathcal E)$ be a measurable space, with $\mathcal E$ countably generated and containing all the singleton sets. Let $X:(\Omega,\mathcal F) \to (E,\mathcal E)$ be a random element. Denote by $P_X := X_*\mathbf P = \mathbf P\circ X^{-1}$ the pushforward measure of $X$ on $(E,\mathcal E)$. Then there is a family $\{\mathbf P^x\}_{x\in E}$ of probability measures on $(\Omega,\mathcal F)$, such that:

  • For every $x\in E$, the probability measure $\mathbf P^x$ concentrates on the event $\{X = x\}$.
  • For all $A\in\mathcal F$, the mapping $\mathbf P^\cdot(A): (E,\mathcal E)\to [0,1]$ is measurable.
  • For all $A\in\mathcal F$ and $B\in\mathcal E$, \begin{equation} \mathbf P\left(A\cap X^{-1}(B)\right) = \int_B \mathbf P^x(A) P_X (dx). \end{equation} Moreover, the family $\{\mathbf P^x\}_{x\in E}$ is uniquely determined up to an almost sure equivalence: if $\{\mathbf Q^x\}_{x\in E}$ is another family of probability measure on $(\Omega,\mathcal F)$ that satisfies above conditions, then \begin{equation*} P_X\{x\in E: \mathbf P^x \ne \mathbf Q^x\} = 0. \end{equation*}

Here is the problem.

Consider the special case that $E=\Omega$ and $\mathcal E$ is a sub-$\sigma$-field of $\mathcal F$ that contains all singletons. Since $\Omega$ is second countable, its Borel $\sigma$-field $\mathcal F$ must be countably generated and contain all singletons. As a sub-$\sigma$-field of $\mathcal F$, $\mathcal E$ is also countably generated. Let $X = \mathrm{Id}$. Then $P_\mathrm{Id} = \mathbf P$ and $\sigma(\mathrm{Id}) = \mathcal E$. Now all assumptions in the theorem are fulfilled. Hence, we get a $\mathbf P$-a.s. unique family of probability measures $\{\mathbf P^\omega\}_{\omega\in\Omega}$ on $(\Omega,\mathcal F)$ satisfying:

  1. For every $\omega\in\Omega$, the probability measure $\mathbf P^\omega$ concentrates on the singleton $\{\omega\}$.
  2. For all $A\in\mathcal F$, the mapping $\mathbf P^\cdot(A): (\Omega,\mathcal E)\to [0,1]$ is measurable.
  3. For all $A\in\mathcal F$ and $B\in\mathcal E$, \begin{equation} \mathbf P\left(A\cap B\right) = \int_B \mathbf P^\omega(A) \mathbf P (dx). \end{equation}

The statements 2 and 3 are completely the same as the formulation of conditional probability, that is, $\mathbf P^\omega(A) = \mathbf P(A\mid \mathcal E)(\omega)$. However, if we combine them with the statement 1, then there are something quite strange. Indeed, since $\mathbf P^\omega$ concentrates on $\{\omega\}$, we have $\mathbf P^\omega(A) = \mathrm{1}_A(\omega)$ for all $A\in\mathcal F$, while this should hold only for $A\in\mathcal E$ since $\mathbf P^\omega$ is the conditional probability by statement 3. Besides, the mapping $\mathbf P^\cdot(A) = \mathrm{1}_A: (\Omega,\mathcal E)\to [0,1]$ is measurable only for $A\in\mathcal E$, but not for all $A\in\mathcal F$ claimed in statement 2.

So where does it go wrong? Any comments or hints will be appreciated. TIA...

EDIT: Here are some further remarks:

  1. I just claimed that "as a sub-$\sigma$-field of $\mathcal F$, $\mathcal E$ is also countably generated". This is wrong. See e.g., here for a counterexample.
  2. Thanks to the comment by @aduh, the problem reduce to whether it must be $\mathcal E = \mathcal F$? Or does there exist a proper sub-$\sigma$-field of $\mathcal F$ that is countably generated and contains all singletons? I post this as another question in Math.SE.

Conclusion: Under my assumptions, $\mathcal E$ must coincide with $\mathcal F$. So the problem is trivial. See the accepted answer given by @GEdgar in the "another question" I mentioned for details.

  • $\begingroup$ It's not true that $\mathcal E$ is c.g. if $\mathcal F$ is. How do you know that your assumptions don't entail that $\mathcal E = \mathcal F$? $\endgroup$
    – aduh
    Sep 21 '20 at 3:35
  • $\begingroup$ @aduh Thank you for your comment. I am not sure if it must be $\mathcal E = \mathcal F$. I post another question for this in the site. Please see the EDIT part at the end. $\endgroup$
    – Dreamer
    Sep 21 '20 at 10:42
  • $\begingroup$ Right, good. And it looks like GEdgar's answer solves the problem. In your case, $\mathcal E = \mathcal F$. $\endgroup$
    – aduh
    Sep 21 '20 at 20:17
  • $\begingroup$ @aduh Yes, I see. Thank you and enjoy this discussion! $\endgroup$
    – Dreamer
    Sep 21 '20 at 20:43
  • $\begingroup$ You might consider writing an answer to this question so that it’s not left unanswered. $\endgroup$
    – aduh
    Sep 21 '20 at 22:43

I post here an answer for integrity.

As said in the Conclusion part at the end of the question, we can prove $\mathcal E = \mathcal F$ following the lines of @GEdgar. More precisely, we can prove the following theorem:

Theorem. Let $\Omega$ be a Polish space, $\mathcal F=\mathcal B(\Omega)$ be the Borel $\sigma$-field for $\Omega$. If $\mathcal E\subset \mathcal F$ is a countably generated sub-$\sigma$-field containing all the singleton sets, then $\mathcal E = \mathcal F$.

The theorem is trivial as long as we know the following lemma, which is adapted from Theorem 3 and Theorem 1 in the paper of D. Blackwell "On a Class of Probability spaces", as well as the two facts that a Polish space is analytic itself and that the atoms in a Polish space are nothing but singletons.

Lemma. Let $\Omega$ be a Polish space, $\mathcal F=\mathcal B(\Omega)$ be the Borel $\sigma$-field for $\Omega$. If $\mathcal E\subset \mathcal F$ is a countably generated sub-$\sigma$-field, then a set $A\in\mathcal F$ belongs to $\mathcal E$ if and only if $A$ is a union of singletons.


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