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Let $(\Omega,\mathcal F,\mathbb P)$ is a standard Borel space (i.e. $\Omega$ is Polish and $\mathcal F = \mathcal B(\Omega)$).

Then $\mathcal F$ is separable and for every sub-sigma-algebra $\mathcal G \subset \mathcal F$, then there exists a regular conditional probability $\mathbb P^{\mathcal G} : \Omega \times \mathcal F \to [0,1]$ (i.e. satisfying a. for all $A \in \mathcal F$, $\mathbb P^{\mathcal G}(\cdot,A) : \Omega \to [0,1]$ is $\mathcal G$-measurable, b. for a.e. $\omega \in \Omega$, the map $\mathbb P^{\mathcal G}(\omega,\cdot) : \mathcal F \to [0,1]$ is a probability on $(\Omega,\mathcal F)$ and c. for all $A \in \mathcal F$ and $B \in \mathcal G$, we have $\mathbb P(A \cap B) = \int_\Omega 1_B(\omega) \mathbb P^{\mathcal G}(\omega, A) \mathbb P(d \omega)$).

I read here and there (Is the regular conditional probability of $\mathbb P$ dominated by $\mathbb P$ almost surely? ; https://mathoverflow.net/questions/354098/necessary-and-sufficient-conditions-for-almost-sure-absolute-continuity-of-regul ) that under this setting, for every $\mathcal G$, we have $\mathbb P^{\mathcal G}(\omega, \cdot) \ll \mathbb P$ almost surely (i.e. $\mathbb P({\omega : \mathbb P^{\mathcal G}(\omega, \cdot) \ll \mathbb P}) = 1$).

I must have a misunderstanding about r.c.p's because of the following:

take $(\Omega=[0,1],\mathcal F = \mathcal B([0,1]),\mathbb P = \mathrm{Leb})$ and $\mathcal G = \mathcal F = \mathcal B([0,1])$.

Then for $A \in \mathcal F$, we have $$\int_\Omega 1_A(\omega) P(d\omega) = \mathbb P(A) = \mathbb P(A \cap A) = \int_\Omega 1_A(\omega) \mathbb P^{\mathcal G}(\omega, A) \mathbb P(d\omega) \,,$$ which gives $1_A = 1_A \mathbb P^{\mathcal G}(\cdot, A)$, $\mathbb P$-almost surely. In addition, $$0 = \mathbb P(A \cap A^c) = \int_\Omega 1_{A^c}(\omega) \mathbb P^{\mathcal G}(\omega,A) \mathbb P(d\omega)\,,$$ thus $1_{A^c} \mathbb P^{\mathcal G}(\cdot,A) = 0$, $\mathbb P$-almost surely.

This implies that $\mathbb P^{\mathcal G}(\cdot,A) = 1_A$, $\mathbb P$-almost surely. (I can also directly check that $(\omega,A) \mapsto 1_A(\omega)$ satisfies a.,b.,c. from the definition of r.c.p above). Hence $\mathbb P^{\mathcal G}(\omega, \cdot) = \delta_\omega$.

Is there any reference for the fact that $\mathbb P(\omega : \mathbb P^{\mathcal G}(\omega,\cdot) \ll \mathbb P) = 1$ ? What is wrong with my above "counterexample" ?

Chill2Macht
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Cyril B.
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    I agree with you. Either the posts you link are mistaken, or there's something we're both missing. – Nate Eldredge May 23 '20 at 13:47
  • Under the first link, the only answer actually attempts to invalidate the claim, and the second link is broken. And indeed, the claim is false, and there is nothing wrong with your counterexample. Moreover, any $(\Omega,\mathcal F,\mathrm P)$, where $\mathrm P$ is not purely atomic, and $\mathcal G = \mathcal F$ is a counterexample. (In fact, it seems that the necessary and sufficient condition for the said fact to be true for any $\mathcal G$ is that $\mathrm P$ is purely atomic.) – zhoraster May 29 '20 at 08:40

1 Answers1

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Then for $A \in \mathcal F$, we have $$\int_\Omega 1_A(\omega) P(d\omega) = \mathbb P(A) = \mathbb P(A \cap A) = \int_\Omega 1_A(\omega) \mathbb P^{\mathcal G}(\omega, A) \mathbb P(d\omega) \,,$$ which gives $1_A = 1_A \mathbb P^{\mathcal G}(\cdot, A)$, $\mathbb P$-almost surely. In addition, $$0 = \mathbb P(A \cap A^c) = \int_\Omega 1_{A^c}(\omega) \mathbb P^{\mathcal G}(\omega,A) \mathbb P(d\omega)\,,$$ thus $1_{A^c} \mathbb P^{\mathcal G}(\cdot,A) = 0$, $\mathbb P$-almost surely.

You were neglecting, that $B=A$ in the formula

$$\mathbb P(A \cap B) = \int_\Omega 1_B(\omega) \mathbb P^{\mathcal G}(\omega, A) \mathbb P(d \omega)$$

needs to be a set in $\mathcal{G}$.

This implies that $\mathbb P^{\mathcal G}(\cdot,A) = 1_A$

This is correct, if $A\in\mathcal{G}$.

user408858
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