# Uniqueness (a.s.) of regular conditional distributions

Let $$(\Omega, \mathcal{F}, \mathbf{P})$$ be a probability space, $$(\mathcal{X}, \mathcal{B})$$ a measurable space, and $$X : \Omega \to \mathcal{X}$$ a random element of $$\mathcal{X}$$. Also, let $$\mathcal{G}$$ be a sub-$$\sigma$$-algebra of $$\mathcal{F}$$.

Question. How unique are regular conditional distributions of $$X$$ given $$\mathcal{G}$$?

A regular conditional distribution of $$X$$ given $$\mathcal{G}$$ is a function $$P : \Omega \times \mathcal{B} \to [0, 1]$$ such that the following properties hold.

1. For all $$\omega \in \Omega$$, the map $$B \mapsto P(\omega, B)$$ from $$\mathcal{B}$$ into $$[0, 1]$$ is a probability measure on $$(\mathcal{X}, \mathcal{B})$$.
2. For all $$B \in \mathcal{B}$$, the map $$\omega \mapsto P(\omega, B)$$ from $$\Omega$$ into $$[0, 1]$$ is $$(\mathcal{G}, \mathcal{B}_{[0, 1]})$$-measurable (where $$\mathcal{B}_{[0, 1]}$$ denotes the Borel $$\sigma$$-algebra of $$[0, 1]$$).
3. For all $$B \in \mathcal{B}$$ and all $$G \in \mathcal{G}$$, we have $$\mathbf{P}(\{X \in B\} \cap G) = \int_G P(\cdot, B) \, d\mathbf{P}.$$

(Items 2. and 3. just say that, for each $$B \in \mathcal{B}$$, the random variable $$P(\cdot, B)$$ is a version of the conditional probability $$\mathbf{P}(X \in B\mid \mathcal{G})$$.)

Suppose $$P$$ and $$Q$$ are two regular conditional distributions of $$X$$ given $$\mathcal{G}$$.

On the one hand, it is not necessarily true that $$P(\omega, B) = Q(\omega, B)$$ for all $$\omega \in \Omega$$ and $$B \in \mathcal{B}$$. For example, for any $$\mathbf{P}$$-null set $$N \in \mathcal{F}$$ and any probability measure $$\mu$$ on $$(\mathcal{X}, \mathcal{B})$$, we can define $$P^\prime : \Omega \times \mathcal{B} \to [0, 1]$$ by $$P^\prime(\omega, B) = \begin{cases} P(\omega, B), & \text{if \omega \notin N,} \\ \mu(B), & \text{if \omega \in N.} \end{cases}$$ Then $$P^\prime$$ is another regular conditional distribution of $$X$$ given $$\mathcal{G}$$, but it might hold that $$P(\omega, B) \neq P^\prime(\omega, B)$$ for some $$\omega \in \Omega$$ and $$B \in \mathcal{B}$$.

On the other hand, suppose $$B \in \mathcal{B}$$ is fixed. Then we have $$\int_G P(\cdot, B) \, d\mathbf{P} = \mathbf{P}(\{X \in B\} \cap G) = \int_G Q(\cdot, B) \, d\mathbf{P}$$ for every $$G \in \mathcal{G}$$. Since $$P(\cdot, B)$$ and $$Q(\cdot, B)$$ are $$\mathcal{G}$$-measurable, this implies that there exists a $$\mathcal{P}$$-null set $$N \in \mathcal{F}$$ such that $$P(\omega, B) = Q(\omega, B)$$ for all $$\omega \in \Omega \setminus N$$. However, this null set depends on $$B$$, so we can't a priori conclude that there exists a $$\mathbf{P}$$-null set $$N^\prime \in \mathcal{F}$$ such that $$P(\omega, B) = Q(\omega, B)$$ for all $$\omega \in \Omega \setminus N^\prime$$ and all $$B \in \mathcal{B}$$.

More Precise Question. Suppose $$P$$ and $$Q$$ are two regular conditional distributions of $$X$$ given $$\mathcal{G}$$. Does there always exist a $$\mathbf{P}$$-null set $$N \in \mathcal{F}$$ such that $$P(\omega, B) = Q(\omega, B)$$ for all $$\omega \in \Omega \setminus N$$ and all $$B \in \mathcal{B}$$?

I think I remember reading that this is true somewhere, but I can't find a proof. I'm fine with assuming that any measurable spaces in question are standard Borel, if needed.

This is true if $$\mathcal{B}$$ is countably generated. Specifically, $$P(\omega,A)=Q(\omega,A) \quad\text{a.s.} \tag{1}\label{1}$$ for all $$A\in \mathcal{A}$$ (a countable algebra that generates $$\mathcal{B}$$). Therefore, there exists a $$\mathbf{P}$$-null set $$N$$ s.t. $$\eqref{1}$$ holds for all $$A\in\mathcal{A}$$ and all $$\omega\in \Omega\setminus N$$. Now extrapolate this result to $$\mathcal{B}$$.
• Perfect, thank you! The monotone class theorem does the trick to extend to $\mathcal{B}$ – Artem Mavrin Jan 29 at 16:52