Constructing a measure on $(S, \mathcal{S})$ with a initial distribution and a transitional probability

Question

Let $(S, \mathcal{S})$ be a measure space and $\mu$ is an initial distribution on $(S, \mathcal{S})$. Also, $p: S \times \mathcal{S} \to \mathbb{R}$ is a transitional probability satisfying

(i) For each $x \in S, A \to p(x, A)$ is a probability measure on $(S, \mathcal{S})$.

(ii) For each $A \in \mathcal{S}, x\to p(x,A)$ is a measurable function.

If we define $$\lambda(B_1 \times \cdots \times B_n) = \int_{B_1} \mu(dx_1)\int_{B_2} p(x_1, dx_2) \cdots \int_{B_n} p(x_{n-1}, dx_n)$$ where $B_i \in \mathcal{S}$, then is there a measure $\nu$ on $(S^n, \bigotimes_{i=1}^n\mathcal{S})$ such that $$\nu(B_1 \times \cdots \times B_n) = \lambda(B_1 \times \cdots \times B_n)$$? Can we show this by $\pi-\lambda$ system? Any help would be appreciated!

Answer 1

The set $\mathcal J_n = \{ B_1\times \ldots \times B_n: \forall i\ B_i \in \mathcal S\} $ forms a semi ring.

1. $\emptyset \in \mathcal J_n$
2. if $B, C\in \mathcal J_n$, then $B\cap C\in \mathcal J_n$, since $$ B\cap C = (B_1\times \ldots \times B_n)\cap (C_1\times \ldots \times C_n) = (B_1\cap C_1)\times \ldots \times (B_n\cap C_n) $$ and $B_i\cap C_i\in \mathcal S$ for all $i$.
3. For any $B,C \in \mathcal J$, $B\setminus C = \bigcup_{i=1}^{2^n-1} A_i$ with disjoint sets $A_i\in \mathcal S$, since $$ B\setminus C = \bigcup_{\mathbf a\in\{0,1\}^n\\ \sum_{i}a_i\geq 1} D_1^{a_1}\times\ldots \times D_n^{a_n} $$ where $$D_j^{a_j} = \begin{cases} B_j \cap C_j & \mbox{ if } a_j = 0 \\ B_j\setminus C_j & \mbox{ if } a_j =1. \end{cases} $$ Choose $A_{a} = D_1^{a_1} \times \ldots \times D_n^{a_n}.$

According to Carathéodory's extension theorem, $\lambda : \mathcal J_n\to [0,1]$, being a $\sigma$-additive premeasure, can be uniquely extended to the $\sigma$-algebra generated by $\mathcal J_n$. The generated $\sigma$-algebra coincides with $\mathcal S^{\otimes n}$ by definition.

The $\sigma$-additivity of $\lambda$. Let $n=2$. Then $$ \lambda(B^1\times B^2) = \int_{B^1}\int_{B^2} p(x_1,dx_2) \mu(dx_1) $$ Let $ (C_n)_n = (B^1_n\times B_n^2)_n $ be a collection of pairwise disjoint sets, s.t. $\bigcup_n C_n \in \mathcal J_2$. We have to show that $$ \lambda(\bigcup_n C_n)= \sum_{n}\lambda(C_n). $$ The key here is that $\bigcup_n C_n \in\mathcal J_2 $ is a strong restriction on $(C_n)$, i.e. it has to be such that $\bigcup_n C_n = D^1 \times D^2$. Note that usually for general sets $(A^1 \times A^2) \cup (B^1\times B^2) \neq D^1 \times D^2$ for some $D^1,D^2$ does not hold, even if the two product sets are disjoint. If $C_n = A_n \times B_n$ is a family of disjoint sets and $\exists D^1, D^2 \in \mathcal S$ s.t. $$ \bigcup_n (B^1_n \times B^1_n) = D^1\times D^2 $$ then $$ (x,y)\in \bigcup_n C_n \Leftrightarrow \ \exists n\in \mathbb N: \ \ x\in B^1_n \wedge y\in B^2_n \Leftrightarrow x\in D^1 \wedge y \in D^2 $$ which means that $D^1 = \bigcup_n B^1_n $ and $D^2=\bigcup_n B_n $. Now assume that $C_n = B^1_n \times B_2$ or $C_n = B^1 \times B_n^2$. Let us consider the first case $$ \lambda(\bigcup_n C_n ) = \lambda(\bigcup (B^1_n \times B^2)) = \lambda((\bigcup_n B_n^1) \times B^2) = \int_{ \bigcup_n B_n^1} \int_{B_2} p(x_1,dx_2) \mu(dx_1) = \int_{\bigcup_n B_n^1} \bigg( \int_{B^2} p(x_1,dx_2) \bigg) \mu(dx_1) = \sum_n \int_{B_1^n} \int_{B^2} p(x_1,dx_2) \mu(dx_1) = \sum_n \lambda(B_n^1\times B^2) = \sum_n \lambda(C_n) $$ where the second last equality is due to the fact that $\mu$ is a measure. Now let consider the other case, i.e. $C_n = B_1\times B_2^n$. Then $$ \lambda(\bigcup_n C_n )= \lambda(\bigcup_n(B_1\times B_n^2)) = \lambda(B_1\times \bigcup_n B_n^2) = \int_{B_1} \int_{\bigcup_n B_n^2} p(x_1,dx_2) \mu(dx_1) = \quad \int_{B_1} \sum_{n} \int_{B_n^2} p(x_1,dx_2) \mu(dx_1) = \int_{B^1} \sum_{n} p(x_1,B_n^2) \mu(dx_1). $$ According to Beppo-Levi's theorem (monotone convergence) we have that $$ \int_{B^1} \sum_n p(x_1,B_n^2) \mu(dx_1) = \sum_n \int_{B^1} p(x_1,B_n^2) \mu(dx_1) = \sum_n \lambda(B^1\times B_n^2) = \sum_n \lambda(C_n) $$ Now consider the case $C_n = B^1_n \times B^2_n$. Then $\bigcup_n C_n \in \mathcal J_2$ implies that $\bigcup_n C_n = \bigcup_n B_n^1 \times \bigcup_n B_n^2$. The set $(B_n^j)_n$ ($j=1,2$) need not to be disjoint, but since the $C_n$'s are disjoint so for fixed $n\neq m$ $B_n^1\cap B_m^2= \emptyset $ or $B_n^2\times B_m^2 = \emptyset$. We also have $$ \forall m,n:\ m\neq n \quad B_n^j\cap B_m^j = \begin{cases} \emptyset \\ B_n^j. \end{cases} $$ The $\emptyset$ part is clear. Suppose $B_n^j \cap B_m^j \neq \emptyset$ for some $n\neq m$. Then $B_n^{3-j}\cap B_m^{3-j} = \emptyset$ has to hold, and $\exists (x,y)\in B_n^j\setminus B_m^j \times B_m^{3-j} \subset \bigcup_n B_n^1 \times \bigcup_n B_n^2 $, but $(x,y) \notin B_m^j \times B_m^{3-j}$, futhermore $(x,y)\notin B_k^1\times B_k^2 $ for any $k\neq m$. This would mean that $$ \bigcup_n B_n^1 \times B_n^2 \subsetneq \bigcup_n B_n^1 \times \bigcup_n B_n^2$$ which is a contradiction. This all means that $$ \bigcup_n B_n^1 \times \bigcup_n B_n^2 = \bigcup_n A_n^1 \times \bigcup_n A_n^2 = \bigcup_n \bigcup_m A_n^1\times A_m^2$$ where the sets $(A_n^j)_n \subset \mathcal S$ ($j=1,2$) are all disjoint. Now $$ \lambda(\bigcup_n C_n) = \lambda(\bigcup_n A_n^1 \times \bigcup_m A_m^2) = \lambda(\bigcup_n (A_n^1 \times \bigcup_m A_m^2)) = \sum_n \lambda(A_n^1 \times \bigcup_m A_m^2) = \sum_n \sum_m \lambda(A_n\times A_m) $$ due to what we have proven before.