# From disintegration to conditioning

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 $$B\in\mathcal E$$, $$$$\mathbf P\left(A\cap X^{-1}(B)\right) = \int_B \mathbf P^x(A) P_X (dx).$$$$ Moreover, the family $$\{\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 $$B\in\mathcal E$$, $$$$\mathbf P\left(A\cap B\right) = \int_B \mathbf P^\omega(A) \mathbf P (dx).$$$$

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.

• 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 $\mathcal E = \mathcal F$?
Sep 21 '20 at 3:35
• @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. Sep 21 '20 at 10:42
• Right, good. And it looks like GEdgar's answer solves the problem. In your case, $\mathcal E = \mathcal F$.
Sep 21 '20 at 20:17
• @aduh Yes, I see. Thank you and enjoy this discussion! Sep 21 '20 at 20:43
• You might consider writing an answer to this question so that it’s not left unanswered.
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 $$\mathcal E = \mathcal F$$.
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.