Let $(\Omega,\mathcal A,P)$ be a probability space and $\mathcal F$ a sub-$\sigma$-algebra of $\mathcal A$.
Then the map $A\mapsto P[A|\mathcal F]$ is not a probability measure on $(\Omega,\mathcal A)$, even almost surely. For example, although
$$P[\cup_{i=1}^\infty A_i|\mathcal F]=\sum_{i=1}^\infty P[ A_i|\mathcal F] \quad \quad P\text{-almost surely}$$
for any sequence $(A_i)$ of disjoints sets in $\mathcal A$ (by the conditional DCT), the null set depends on the chosen sequence $(A_i)$, and there are possibly uncountably many such sequences.
On the other hand, a markov kernel $\kappa:\mathcal A \times \Omega \to [0,1]$ with source $(\Omega,\mathcal F)$ and target $(\Omega,\mathcal A)$ will satisfy the properties
- For every $A\in\mathcal A$, the map $\omega\mapsto \kappa(A,\omega)$ is $\mathcal F$-measurable.
- For every $\omega\in\Omega $, the map $A \mapsto \kappa(A,\omega)$ is a probability measure on $(\Omega,\mathcal A)$.
Question: Can I find a markov kernel as above such that $\kappa(A,\cdot)$ is a version of $P[A|\mathcal F]$ for all $A\in \mathcal A$?
If I understand correctly Wikipedia, we need the target space $(\Omega,\mathcal A)$ to be Polish with its Borel $\sigma$-algebra. Can we dispense with this assumption?
An argument for the case of a countable generator partitioning $\Omega$:
Suppose there exists a countable partition $\mathcal C= \{ C_n: n \in \mathbb{N}\}$ of $\Omega$ such that $\mathcal A=\sigma(\mathcal C)$. WLOG we may assume that each $C_n$ is nonempty. Then $ \sigma(\mathcal C) = \{\cup_{i \in I} C_i: I \subseteq \mathbb{N}\}$.
Consider the real random variable $X=\sum_{n=1}^{\infty} n1_{C_n}$ on $(\Omega,\mathcal A,P)$. Then $X$ has a regular conditional distribution given $\mathcal F$, i.e. there exists a markov kernel $\kappa_{X,\mathcal F}:\mathcal B(\mathbb R) \times \Omega \to [0,1]$ with source $(\Omega,\mathcal F)$ and target $(\mathbb R,\mathcal B(\mathbb R))$ such that $\kappa_{X,\mathcal F}(B,\cdot)$ is a version of $P[X\in B|\mathcal F]$ for all $B\in\mathcal B(\mathbb R)$.
Now, given $\cup_{i \in I} C_i\in\mathcal \sigma(\mathcal C)$ and $\omega\in\Omega$ put
$$\kappa(\cup_{i \in I} C_i,\omega):=\kappa_{X,\mathcal F}(\cup_{i \in I}\{i\},\omega)$$
Since $\mathcal C$ is a partition of $\Omega$ into nonempty sets the representation $\cup_{i \in I} C_i$ is unique and $\kappa$ is well-defined. We also check that $\kappa$ is a markov kernel with source $(\Omega,\mathcal F)$ and target $(\Omega,\mathcal A)$. Moreover,
$$\kappa(\cup_{i \in I} C_i,\cdot)=\kappa_{X,\mathcal F}(\cup_{i \in I}\{i\},\cdot)=P[X\in \cup_{i \in I}\{i\}|\mathcal F]=P[\cup_{i \in I} C_i|\mathcal F]$$
$P$-almost surely for every $\cup_{i \in I} C_i\in\mathcal \sigma(\mathcal C)$. Therefore $\kappa$ is a markov kernel with the desired property.
Is this correct? Thanks a lot for your help.
