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Let $(x, A) \mapsto P(x, A)$ be a probability kernel whose "target" (wikipedia terminology) is a product space $Y \times Z$, and say both $Y$ and $Z$ are compact metric spaces. For every $x$ there is the disintegration $(\mu_{x,y})_{y \in Y}$ characterized by $$P(x,A_1 \times A_2)=\int_{A_1}\mu_{x,y}(A_2) P(x,\pi^{-1}(dy))$$ where $\pi \colon Y \times Z \to Y$ is the canonical map.

Well, but I need that $\mu_{x,y}$ measurably depends on $(x,y)$, or in other words I need a kernel $(x,y,A_2) \to \mu_{x,y}(A_2)$. How could I justify this measurability ?

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  • $\begingroup$ Not sure whether this answers your question, but Lemma 2 of "On the existence of good stationary strategies" by Sudderth may be of use. $\endgroup$ – Ilya Dec 5 '14 at 16:19
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In my own work I needed to disintegrate a probability kernel and I found a good reference. The following is a simplified version of Theorem 1.25 in Kallenberg's 2017 book, Random Measures, Theory and Applications.

Theorem: Consider a $\sigma$-finite kernel $\rho: S \to T \times U$, where $T$ and $U$ are Borel spaces. There exist $\sigma$-finite kernels $\nu: S \to T$ and $\mu: S \times T \to U$ such that $\rho = \nu \otimes \mu$. The assertion remains true for any fixed $\sigma$-finite kernel $\nu: S \to T$ such that $\nu_s \sim \rho_s(\cdot \times U)$ for all $s \in S$.

Here a Borel space is any measurable space isomorphic to a Borel subset of the real line, for example, a Polish space.

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