Notation question about probability kernels

Question

I'm looking at a theorem in Kallenberg's Foundations of Modern Probability (1st ed). It's talking about probability kernels from $(T, \mathcal{T})$ to $(S, \mathcal{S})$, being functions $\nu : T \times \mathcal{S} \to [0,1]$ so that $\nu(t, \bullet)$ is a measure for all $t$ and $\nu(\bullet, B)$ is $\mathcal{T}$-measurable for all $B \in \mathcal{S}$.

He then states a theorem on p. 85 (first edition), which I am paraphrasing:

Fix two measurable spaces $(S, \mathcal{S})$ and $(T, \mathcal{T})$, a $\sigma$-field $\mathcal{F} \subset \mathcal{A}$, and a random element $\xi$ in $S$ such that $P(\xi \in \bullet \mid \mathcal{F}): \Omega \times \mathcal{S} \to \mathbb{R}$ is a kernel $\nu$. Further consider an $\mathcal{F}$-measurable random element $\eta$ in $T$ and a measurable function $f$ on $S \times T$ with $E|f(\xi, \eta)| < \infty$. Then

$$E(f(\xi, \eta) \mid \mathcal{F}) = \int \nu(ds)f(s,\eta) \quad \text{a.s.}$$

What is he talking about with $\nu(ds)$? I thought $\nu$ was a function $\Omega \times \mathcal{S} \to [0,1]$...

Answer 1

In the theorem, $\nu$ is defined to be the probability density kernel $P(\xi|\mathcal{F})$. This means that $\nu(ds)$ is the probability measure of $ds\in \Omega$. So, if you integrate over the entire domain space, $\int_{\Omega}{\nu(ds)} = 1$. So in the Theorem, it's the measure theoretic approach towards defining an expected value.

Answer 2

The dependence on $\omega \in \Omega$ is hidden not just for $\nu$ but throughout that equation (and frequently beyond). A more explicit version would be

$$ E[f(\xi,\eta) \mid \mathcal{F}](\omega) = \int_S \nu(\omega, ds) f(s, \eta(\omega)) \text{ for almost every $\omega \in \Omega$.}$$

Essentially in each “location” / “case” / “atom” (represented by $\omega$) that can be discerned by $\mathcal{F}$ in the probability space, we know that $\xi$ is distributed according to a restricted law $\nu(\omega, \mathord{\cdot})$, and we further know the value of $\eta$ is fixed because it is $\mathcal F$-measurable. So we can then just integrate $f(\cdot, \eta(\omega))$ against $\nu(\omega, \mathord{\cdot})$ over $S$ to find the conditional expectation.

(For search purposes, this is the "disintegration" theorem in Kallenberg.)