# Measurability for disintegration of a kernel

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 ?

• Not sure whether this answers your question, but Lemma 2 of "On the existence of good stationary strategies" by Sudderth may be of use. – Ilya Dec 5 '14 at 16:19

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 $$s \in S$$.