Radon-Nikodym for product measure

Question

Let $(E, \mathcal{E})$ be a measurable space, $\eta:\mathcal{E}\to [0,1]$ be a probability measure defined on it and $K:E\times\mathcal{E}\to [0, 1]$ a Markov Kernel from $(E, \mathcal{E})$ onto itself. Let $\mathcal{E}\otimes \mathcal{E}$ be the product sigma algebra of the cartesian product $E\times E$. Then define the product measure as $$ (\eta \times K)(A\times B) = \eta(A) K(x, B) \qquad \qquad \forall\, A, B\in \mathcal{E} \quad \text{and} \quad x\in E $$ This is a measure $\eta \times K: \mathcal{E}\otimes\mathcal{E} \to [0, 1]$ defined on $\mathcal{E}\otimes\mathcal{E}$ so now (assuming any regularity condition I need to assume) I would like to find the Radon-Nikodym derivative of $\eta \times K$. Suppose $\eta \ll \lambda$ and $K(x, \cdot) \ll \lambda$ have densities $$ \frac{d \eta}{d\lambda}(x) = p_\eta(x) \qquad \frac{d K(x, \cdot)}{d \lambda}(y) = k(y\mid x) $$ Can I say that the density of the product measure is now this? $$ \frac{d \eta \times K}{d \lambda \times \lambda}(x, y) = \frac{d \eta}{d \lambda}(x) \cdot \frac{d K(x, \cdot)}{d \lambda}(y) = p_\eta(x) k(y \mid x) $$

Answer 1

Here is the slightly version of the Theorem we discussed earlier. I worked on a proof to remember from some notes that I took in course in Random measures years ago, which I present here (there are some details that you may be able to fill). The ideas as taken from Doob's Stochastic processes book, but the main main ingredient is martingale convergence theorem. I still recommend to you to check the reference (Kallenberg) from my comment.

Theorem: (de Possel, Doob) Let $\mu$ and $\nu$ be $\sigma$--finite kernels from $(S,\mathscr{S})$ to $(T,\mathscr{T})$. If $\mathscr{T}$ is countably generated then, there is a measurable function $X:(S\times T,\mathscr{S}\otimes\mathscr{T})\rightarrow[0,\infty]$ such that for all $B\in\mathcal{T}$ \begin{align} \nu(s,B)=\int_B X(s,t)\mathbb{1}_{\{X<\infty\}}\mu(s,dt) + \nu\big(s,B\cap\{X=\infty\}\big)\tag{1}\label{dPD} \end{align} Moreover,

• For each $s\in S$, $X(s,\cdot)$ is unique $(\mu+\nu)_s$--a.s.
• The kernels $\nu^a:=(X\mathbb{1}_{\{X<\infty\}})\cdot\mu$ and >$\nu^s:=\mathbb{1}_{\{X=\infty\}}\cdot\nu$ are the absolutely continuous >and the singular part, respectively, of the Radon--Nikodym decomposition >of $\nu$ w.r.t. $\mu$
• The sets $\{r\in S: \nu(r,dt)\ll\mu(r,dt)\}$ and $\{r\in S: \nu(r,dt)\perp\nu(r,dt)\}$ belong to $\mathscr{S}$.

Sketch of of Proof:

Suppose $\mu$ and $\nu$ are $\sigma$--finite. Then $\rho:=\mu+\nu$ is also $\sigma$--finite; hence, there exists a function $f:(S\times T,\mathscr{S}\otimes\mathscr{T})\rightarrow(0,\infty)$ such that $\int f(s,t)\rho(s,dt)=\mathbb{1}_{\{\rho(s,T)\neq0\}}$. We may assume without loss of generality that $\rho(s,T)>0$ for all $s\in S$. It follows that $\rho'=f\cdot\rho$ is a stochastic kernel, and $\mu'=f\cdot\mu$, and $\nu'=f\cdot\nu$ are finite kernels.

Let $\mathscr{B}_n\subset\mathscr{B}$ be a sequence of finite partitions of $T$ such that $\mathscr{B}_{n+1}$ is a refinement of $\mathscr{B}_n$. For each $n$, let \begin{align} X_n(s,t) = \sum_{B\in\mathscr{B}_n} \frac{\nu'(s,B)}{\rho'(s,B)}\mathbb{1}_B(t)\mathbb{1}_{(0,\infty)}(\rho'(s,B))\tag{2}\label{prod-mart} \end{align} Notice that $0\leq X_n\leq 1$. For each $s\in S$, the sequence of functions $t\mapsto X_n(s,t)$ is a bounded martingale with respect to $\{\sigma(\mathscr{B}_n):n\in\mathbb{N}\}$ and the measure $\rho'_s(\cdot):=\rho'(s,\cdot)$. Hence, $X_n(s,\cdot)$ converges $\rho'_s$-a.s. (and in $\mathcal{L}_1(\rho'_s)$ by dominated convergence) to some $\mathscr{T}$--measurable function $0\leq X'(s,\cdot)\leq 1$. $X'$ admits a product--measurable version given by $$ X'(s,t)=\limsup_nX_n(s,t),\qquad (s,t)\in S\times T. $$ From~\eqref{prod-mart} we have that $$ \nu'(s,B)=\int_B X_n(s,t)\rho'(s,dt),\qquad B\in\mathscr{B}_n $$ By $\mathcal{L}_1(\rho'_s)$ convergence, $\nu'(s,dt)= X'(s,t)\cdot\rho'(s,dt)$ in $\sigma(\mathscr{B}_n)$ for each $n\in\mathbb{N}$, which also extends to $\mathscr{B}=\sigma\Big(\bigcup_n\sigma(\mathscr{B}_n)\Big)$ by the monotone class theorem. Any other function $Y$ such that $Y\cdot\rho=X'\cdot\rho$ satisfies $Y(s,\cdot)=X'(s,\cdot)$ $\rho_s$-a.s. for each $s\in S$. From $\rho'=\mu'+\nu'$, we obtain $\big((1-X')\cdot f\big)\cdot \nu=(X'\cdot f)\cdot \mu$. Hence \begin{align} f\cdot\nu&= (\mathbb{1}_{\{X'<1\}}\cdot f\big)\cdot \nu + \big(\mathbb{1}_{\{X'=1\}}\cdot f)\cdot \nu\\ &= \Big(\mathbb{1}_{\{X'<1\}}\frac{X'}{1-X'}\cdot f\Big)\cdot\mu + \big(\mathbb{1}_{\{X'=1\}}\cdot f\big)\cdot \nu\\ &=\big(\mathbb{1}_{\{X<\infty\}}X\cdot f\big)\cdot\mu + (\mathbb{1}_{\{X=\infty\}}f\big)\cdot\nu \end{align} where $X=\frac{X'}{1-X'}$ with $1/0:=\infty$. Consequently $\nu=\big(\mathbb{1}_{\{X<\infty\}}X\big)\cdot\mu + (\mathbb{1}_{\{X=\infty\}}\big)\cdot\nu$.

Clearly $\nu^a:=\big(\mathbb{1}_{\{X<\infty\}}X\big)\cdot\mu\ll\mu$, and for any $B\in\mathscr{T}$ the map $$s\mapsto \nu^a(s,B)=\int_BX(s,t)\mathbb{1}_{\{X(s,\cdot)<\infty\}}(t)\mu(s,dt)$$ is $\mathscr{S}$--measurable. Since $\{X=\infty\}=\{X'=1\}$ and $$ \mu'(s,X'(s,\cdot)=1)=\int_{\{X'(s,\cdot)=1\}}X'(s,t)\mu'(s,dt)=\int_{\{X'(s,\cdot)=1\}}(1-X'(s,t))\nu'(s,dt)=0, $$ we conclude that $\mu\perp \nu^s:=(\mathbb{1}_{\{X=\infty\}}\big)\cdot\nu$. The measurability of $r\mapsto \nu^s(r,B)$ follows from $\nu^s=\nu-\nu^s$. The uniqueness of $X$ follows from the uniqueness of $X'=1-\frac{1}{1+X}$.