Assume we have two arbitrary probability measures $\mathbb P_1, \mathbb P_2$ on the same arbitrary measurable space $(\Omega, \mathcal B)$. (Or more generally, $n$ such probability measures).
I want to generate a new probability space that captures the same information, but that captures the information in $\mathbb {P_1,P_2}$ as conditional distributions generated by information $I_1,I_2$ on the same prior distribution $\mathbb P$.
My intuition is that we can generate a new probability space $(\Omega',\mathcal B',\mathbb P)$, and for any variable $V:\Omega\to X$, we can define $V':\Omega'\to X$, and then choose $I_1, I_2$ to partition $\Omega'$ such that for any event in the original space $E\in \mathcal B$, there is a natural corresponding $E'\in\mathcal B'$, where we have $$\mathbb P_i(E)=\mathbb P(E'|I_i)$$
Edit: I want the information in $\mathbb P_1,\mathbb P_2$ to be represented in $I_1,I_2$, not in $\mathbb P$. That is $\mathbb P$ should induce a "uniform" distribution on $\Omega$: e.g. if $\Omega=[0,1]^{10}$, then $\mathbb P$ should induce a uniform distribution on $[0,1]^{10}$, and the $I_1,I_2$ should capture the "non-uniformity" in $P_1,P_2$.
This would mean that we can fully represent the two probability distributions as conditional distributions with the same prior.
Is this indeed possible? I am not sure how to actually do this. It seems to me that $\Omega'$ needs to be much bigger than $\Omega$ in order to capture the two distributions.
What is the cleanest way to do this?
If I've understood your question, yes, this is possible.
Let $\Omega' = \Omega \times \{i_1,i_2\}$. Let $\mathcal{B}' = \mathcal{B} \otimes 2^{\{i_1,i_2\}}$.
Let $A \in \mathcal{B}$, and define $Q(A \times \{i_1\})=1/2 P_1(A)$ and $Q(A \times \{i_2\}) = 1/2 P_2(A)$. You can verify that $Q$ so defined extends to all of $\mathcal{B}'$.
Associate with each event $A \in \mathcal{F}$, the event $A'= A \times \{i_1, i_2\} \in \mathcal{B}'$. Let $I_j = \Omega \times \{i_j\}$, $j=1,2$.
Finally, verify that $Q(A' \mid I_j) = P_j(A)$, $j=1,2$.
Does this answer your question?