Let $(\Omega, \Sigma, \mu)$ be a probability space that has whatever properties necessary so that disintegrations exist. For a measurable partition $\mathcal{P}$ of $\Omega$, call $\\{ \mu_P \\}_{P \in \mathcal{P}}$ a disintegration of $\mu$ if
- $\mu_P(P) = 1$ for $\mu$-a.e. $P$
- The map $\phi: A \mapsto \mu_P(E)$ is measurable for every $E \in \Sigma$
- $\mu(E) = \int_{\mathcal{P}} \mu_P(E) d\mu $
(abusing notation, let $\mu$ denote its own pushforward on the measure space defined by the partition).
Question: Let $\mathcal{P}, \mathcal{Q}$ be two measurable partitions and let $\mathcal{R}$ denote their common refinement. Is it possible to find a family of disintegrations $\\{ \mu_P \\}_{P \in \mathcal{P}}$, $\\{ \mu_Q \\}_{Q \in \mathcal{Q}}$, $\\{ \mu_R \\}_{R \in \mathcal{R}}$ such that
- $\mu_P(E) = \int_{\mathcal{R} \cap P} \mu_R(E) d\mu_P $
- $\mu_Q(E) = \int_{\mathcal{R} \cap Q} \mu_R(E) d\mu_Q $
for all $\mu$-a.e. $P \in \mathcal{P}$ and $Q \in \mathcal{Q}$ .
In words, so that $\\{ \mu_R \\}_{R \in \mathcal{R}, R \subseteq P}$ is a disintegration of $\mu_P$, and the analogous thing holds for $Q \in \mathcal{Q}$.
Clearly, if you just have a $\mathcal P$ and a refinement $\mathcal R$, this is straightforward, as you can just disintegrate each $\mu_P$ individually. My concern comes from requiring that the "two different directions" of integration interchange: for a given family of disintegrations for each $\mu_P$ there is no reason that integrating over $\mathcal Q$ would end up with $\mu$.
A follow up: what if we do not have three partitions but a general family of them. Is it still possible to find a corresponding family of disintegrations that nest in the same way?