B Radon-Nikodym derivatives and disintegration

Definition 49

✓

Let \(\kappa , \eta : \mathcal X \rightsquigarrow \mathcal Y\) be two finite kernels, with either \(\mathcal X\) countable or \(\mathcal{Y}\) countably generated. The Radon-Nikodym derivative of \(\kappa \) with respect to \(\eta \), denoted by \(\frac{d \kappa }{d \eta }\), is a measurable function \(\mathcal X \times \mathcal Y \to \mathbb {R}_{+, \infty }\) with \(\kappa = \frac{d \kappa }{d \eta } \cdot \eta + \kappa _{\perp \eta }\), where for all \(x\), \(\kappa _{\perp \eta }(x) \perp \eta (x)\).

Lemma B.0.1

✓

Let \(\kappa , \eta : \mathcal X \rightsquigarrow \mathcal Y\) be two finite kernels, with either \(\mathcal X\) countable or \(\mathcal{Y}\) countably generated. If for some \(f\) and \(\xi \), \(\kappa = f \cdot \eta + \xi \) with \(\xi (x) \perp \eta (x)\) for all \(x\), then for all \(x\), \(f(x, y) = \frac{d \kappa (x)}{d \eta (x)}(y)\) for \(\eta (x)\)-almost all \(y \in \mathcal Y\).

Proof ▶

Let \(f\) and \(\xi \) be such that \(\kappa = f \cdot \eta + \xi \) with \(\xi (x) \perp \eta (x)\) for all \(x\). Then for all \(x \in \mathcal X\), \(\kappa (x) = f(x, \cdot ) \cdot \eta (x) + \xi (x)\) with \(\xi (x) \perp \eta (x)\). By the uniqueness result for the Radon-Nikodym derivative of measures, \(f(x, \cdot ) = \frac{d \kappa (x)}{d \eta (x)}\) almost everywhere.

Corollary B.0.2

✓

For all \(x \in \mathcal X\), for \(\eta (x)\)-almost all \(y \in \mathcal Y\), \(\frac{d \kappa }{d \eta }(x, y) = \frac{d \kappa (x)}{d \eta (x)}(y)\).

Proof ▶

Apply Lemma B.0.1.

B.1 Composition-product

Lemma B.1.1

✓

Let \(\mu , \nu \) be two measures on \(\mathcal X\) with \(\mu \ll \nu \) and let \(\kappa , \eta : \mathcal X \rightsquigarrow \mathcal Y\) be two finite kernels with \(\kappa (x) \ll \eta (x)\) for \(\nu \)-almost all \(x\), with either \(\mathcal X\) countable or \(\mathcal{Y}\) countably generated. Then for \((\nu \otimes \eta )\)-almost all \((x, y)\), \(\frac{d (\mu \otimes \kappa )}{d (\nu \otimes \eta )}(x,y) = \frac{d\mu }{d\nu }(x)\frac{d \kappa }{d \eta }(x,y)\). This implies that the equality is true for \(\nu \)-almost all \(x\), for \(\eta (x)\)-almost all \(y\).

Proof ▶

It is sufficient to prove that the integrals of the two functions on all product sets \(s \times t\) for \(s\) and \(t\) measurable are equal (since these sets are a \(\pi \)-system). The computation of first integral is immediate :

\begin{align*} \int _{p \in s \times t}\frac{d (\mu \otimes \kappa )}{d (\nu \otimes \eta )}(p) \partial (\nu \otimes \eta ) & = (\mu \otimes \kappa )(s \times t) \: . \end{align*}

The second one uses Corollary B.0.2.

\begin{align*} \int _{p \in s \times t}\frac{d\mu }{d\nu }(p_X) \frac{d \kappa }{d \eta }(p) \partial (\nu \otimes \eta ) & = \int _{x \in s} \int _{y \in t} \frac{d\mu }{d\nu }(x) \frac{d \kappa }{d \eta }(x,y) \partial \eta (x) \partial \nu \\ & = \int _{x \in s} \frac{d\mu }{d\nu }(x) \int _{y \in t} \frac{d \kappa (x)}{d \eta (x)}(y) \partial \eta (x) \partial \nu \\ & = \int _{x \in s} \frac{d\mu }{d\nu }(x) \kappa (x)(t) \partial \nu \\ & = \int _{x \in s} \kappa (x)(t) \partial \mu \\ & = (\mu \otimes \kappa )(s \times t) \: . \end{align*}

Lemma B.1.2

✓

Let \(\mu , \nu \) be two measures on \(\mathcal X\) and let \(\kappa , \eta : \mathcal X \rightsquigarrow \mathcal Y\) be two finite kernels. Let \(\mu ' = \left(\frac{\partial \mu }{\partial \nu }\right) \cdot \nu \) and \(\kappa ' = \left(\frac{\partial \kappa }{\partial \eta }\right) \cdot \eta \). Then for \((\nu \otimes \eta )\)-almost all \(z\), \(\frac{\partial (\mu ' \otimes \kappa ')}{\partial (\nu \otimes \eta )}(z) = \frac{\partial (\mu \otimes \kappa )}{\partial (\nu \otimes \eta )}(z)\).

Proof ▶

Lemma B.1.3

✓

Let \(\mu , \nu \) be two measures on \(\mathcal X\) and let \(\kappa , \eta : \mathcal X \rightsquigarrow \mathcal Y\) be two finite kernels, with either \(\mathcal X\) countable or \(\mathcal{Y}\) countably generated. Let \(\mu ' = \left(\frac{d \mu }{d \nu }\right) \cdot \nu \) and \(\kappa ' = \left(\frac{d \kappa }{d \eta }\right) \cdot \eta \). Then for \((\nu \otimes \eta )\)-almost all \((x, y)\), \(\frac{d\mu '}{d\nu }(x)\frac{d \kappa '}{d \eta }(x,y) = \frac{d\mu }{d\nu }(x)\frac{d \kappa }{d \eta }(x,y)\).

Proof ▶

Lemma B.1.4

✓

Let \(\mu , \nu \in \mathcal M(\mathcal X)\) and let \(\kappa : \mathcal X \rightsquigarrow \mathcal Y\) be a finite kernel. Then for \((\nu \otimes \kappa )\)-almost all \((x, y)\), \(\frac{d (\mu \otimes \kappa )}{d (\nu \otimes \kappa )}(x,y) = \frac{d\mu }{d\nu }(x)\).

Proof ▶

TODO: wlog \(\mu \ll \nu \), and we can consider products of sets.

\begin{align*} \int _{p \in s \times t} \frac{d (\mu \otimes \kappa )}{d (\nu \otimes \kappa )}(p) \partial (\nu \otimes \kappa ) & = (\mu \otimes \kappa ) (s \times t) \: . \end{align*}

\begin{align*} \int _{p \in s \times t} \frac{d \mu }{d \nu }(p_1) \partial (\nu \otimes \kappa ) & = \int _{x \in s} \int _{y \in t} \frac{d \mu }{d \nu }(x) \partial \kappa (x) \partial \nu \\ & = \int _{x \in s} \frac{d \mu }{d \nu }(x) \kappa (x)(t) \partial \nu \\ & = \int _{x \in s} \kappa (x)(t) \partial \mu \\ & = (\mu \otimes \kappa ) (s \times t) \: . \end{align*}

Lemma B.1.5

✓

Let \(\mu , \nu \) be two measures on \(\mathcal X\) and let \(\kappa , \eta : \mathcal X \rightsquigarrow \mathcal Y\) be two finite kernels, with either \(\mathcal X\) countable or \(\mathcal{Y}\) countably generated. Then for \((\nu \otimes \eta )\)-almost all \((x, y)\), \(\frac{d (\mu \otimes \kappa )}{d (\nu \otimes \eta )}(x,y) = \frac{d\mu }{d\nu }(x)\frac{d \kappa }{d \eta }(x,y)\). This implies that the equality is true for \(\nu \)-almost all \(x\), for \(\eta (x)\)-almost all \(y\).

Proof ▶

Use Lemma B.1.2 for \(\mu ', \kappa '\) (defined in previous lemmas) then B.1.1 and B.1.3.

Corollary B.1.6

✓

Let \(\mu \) be a measures on \(\mathcal X\) and let \(\kappa , \eta : \mathcal X \rightsquigarrow \mathcal Y\) be two finite kernels, with either \(\mathcal X\) countable or \(\mathcal{Y}\) countably generated. Then for \((\mu \otimes \eta )\)-almost all \((x, y)\), \(\frac{d (\mu \otimes \kappa )}{d (\mu \otimes \eta )}(x,y) = \frac{d \kappa }{d \eta }(x,y)\).

Proof ▶

B.2 Composition

Lemma B.2.1

✓

Let \(\mu , \nu \in \mathcal M(\mathcal X)\) with \(\mu \ll \nu \), \(g : \mathcal X \to \mathcal Y\) a measurable function and denote by \(g^* \mathcal Y\) the comap of the \(\sigma \)-algebra on \(\mathcal Y\) by \(g\). Then \(\nu \)-almost everywhere,

\begin{align*} \frac{d g_*\mu }{d g_*\nu }(g(x)) = \nu \left[ \frac{d \mu }{d \nu } \mid g^* \mathcal Y\right](x) \: . \end{align*}

Proof ▶

We show that the integrals of the two functions agree on all \(g^* \mathcal Y\)-measurable sets. It suffices to show equality on all sets oc \(\mathcal X\) of the form \(g^{-1}(t)\) for \(t\) a measurable set of \(\mathcal Y\). TODO: also show measurability.

\begin{align*} \int _{x \in g^{-1}(t)}\frac{d g_*\mu }{d g_*\nu }(g(x)) \partial \nu & = \int _{y \in t}\frac{d g_*\mu }{d g_*\nu }(y) \partial (g_*\nu ) \\ & = g_*\mu (t) \: . \end{align*}

\begin{align*} \int _{x \in g^{-1}(t)}\nu \left[ \frac{d \mu }{d \nu } \mid g^* \mathcal Y\right](x) \partial \nu & = \int _{x \in g^{-1}(t)}\frac{d \mu }{d \nu }(x) \partial \nu \\ & = \mu (g^{-1}(t)) \\ & = g_*\mu (t) \: . \end{align*}

Lemma B.2.2

✓

Let \(\mu , \nu \in \mathcal M(\mathcal X)\) with \(\mu \ll \nu \) and let \(\kappa , \eta : \mathcal X \rightsquigarrow \mathcal Y\) be finite kernels with \(\kappa (x) \ll \eta (x)\) \(\nu \)-a.e.. Let \(\mathcal B\) be the sigma-algebra on \(\mathcal X \times \mathcal Y\) obtained by taking the comap of the sigma-algebra of \(\mathcal Y\) by the projection. Then for \((\nu \otimes \eta )\)-almost every \((x,y)\),

\begin{align*} \frac{d(\kappa \circ \mu )}{d(\eta \circ \nu )}(y) & = (\nu \otimes \eta )\left[ \frac{d (\mu \otimes \kappa )}{d (\nu \otimes \eta )} \mid \mathcal B \right](x,y) \: . \end{align*}

Proof ▶

Let \(\pi _Y : \mathcal X \times \mathcal Y \to \mathcal Y\) be the projection \(\pi _Y(x,y) = y\). Remark that \(\kappa \circ \mu = \pi _{Y*}(\mu \otimes \kappa )\) and similarly for \(\nu , \eta \). By Lemma B.2.1, \((\nu \otimes \eta )\)-almost everywhere,

\begin{align*} \frac{d(\kappa \circ \mu )}{d(\eta \circ \nu )}(y) & = \frac{d \pi _{Y*}(\mu \otimes \kappa )}{d \pi _{Y*}(\nu \otimes \eta )}(\pi _Y((x,y))) \\ & = (\nu \otimes \eta )\left[ \frac{d (\mu \otimes \kappa )}{d (\nu \otimes \eta )} \mid \mathcal B\right](x,y) \: . \end{align*}

Lemma B.2.3 [ Csi63 ]

✓

Let \(\mu , \nu \in \mathcal M(\mathcal X)\) with \(\mu \ll \nu \) and let \(\kappa : \mathcal X \rightsquigarrow \mathcal Y\) be a finite kernel. Let \(\mathcal B\) be the sigma-algebra on \(\mathcal X \times \mathcal Y\) obtained by taking the comap of the sigma-algebra of \(\mathcal Y\) by the projection. Then for \((\nu \otimes \kappa )\)-almost every \((x,y)\),

\begin{align*} \frac{d(\kappa \circ \mu )}{d(\kappa \circ \nu )}(y) & = (\nu \otimes \kappa )\left[ (x, y) \mapsto \frac{d \mu }{d \nu }(x) \mid \mathcal B \right](x,y) \: . \end{align*}

Proof ▶

Apply Lemma B.2.2 to get, \((\nu \otimes \kappa )\)-almost everywhere,

\begin{align*} \frac{d(\kappa \circ \mu )}{d(\kappa \circ \nu )}(y) & = (\nu \otimes \kappa )\left[ \frac{d (\mu \otimes \kappa )}{d (\nu \otimes \kappa )} \mid \mathcal B\right](x,y) \: . \end{align*}

Finally, by Corollary B.1.4, \(\frac{d (\mu \otimes \kappa )}{d (\nu \otimes \kappa )} = \frac{d \mu }{d \nu }\) a.e.

B.3 Sigma-algebras

For \(\mathcal A\) a sub-\(\sigma \)-algebra and \(\mu \) a measure, we write \(\mathcal\mu _{| \mathcal A}\) for the measure restricted to the \(\sigma \)-algebra.

Lemma B.3.1

✓

Let \(\mu , \nu \) be two finite measures on \(\mathcal X\) with \(\mu \ll \nu \) and let \(\mathcal A\) be a sub-\(\sigma \)-algebra of \(\mathcal X\). Then \(\frac{d \mu _{| \mathcal A}}{d \nu _{| \mathcal A}}\) is \(\nu _{| \mathcal A}\)-almost everywhere (hence also \(\nu \)-a.e.) equal to \(\nu \left[ \frac{d \mu }{d \nu } \mid \mathcal A\right]\).

Proof ▶

TODO: the current Lean proof is not the proof presented here (but a direct computation).

The restriction \(\mu _{| \mathcal A}\) is the map of \(\mu \) by the identity, seen as a function from \(\mathcal X\) with its \(\sigma \)-algebra to \(\mathcal X\) with \(\mathcal{A}\). We can thus apply Lemma B.2.1.

Lemma B.3.2

\begin{align*} \frac{d g_*\mu }{d g_*\nu }(g(x)) = \frac{d \mu _{| g^* \mathcal Y}}{d \nu _{| g^* \mathcal Y}}(x) \: . \end{align*}

Proof ▶

Combine Lemma B.2.1 and Lemma B.3.1.