Product Spaces, Transition Kernel and Rubini's Theorem

Lemma 1. Sectioning sets Sections of measurable sets are measurable.  If $A\in \mathcal{B}_1 \times \mathcal{B}_2$, then for all $\omega_1 \in \Omega_1$,
$$A_{w_1}\in \mathcal{B_2}$$
Corollary 1. Sections of measurable functions are measurable.  That is if
$$X: (\Omega_1\times \Omega_2, \mathcal{B}_1 \times \mathcal{B}_2) \mapsto (S,\mathcal{S})$$ then $$X_{\omega_1} \in \mathcal{B}_2.$$  We say $X_{\omega_1}$ is $\mathcal{B}/\mathcal{S}$ measurable.

Define the transition (probability) kernel

$$K(\omega_1,A_2):\Omega_1 \times \mathcal{B}_2 \mapsto [0,1]$$

if it satisfies the following

1. for each $\omega_1, K(\omega_1,\cdot)$ is a probability measure on $\mathcal{B}_2$, and
2. for each $A_2\in \mathcal{B}_2, K(\cdot, A_2)$ is $\mathcal{B}_1/\mathcal{B}([0,1])$ measurable.

Transition kernels are used to define discrete time Markov processes where $K(\omega_1,A_2)$ represents the conditional probability that, starting from $\omega_1$, the next movement of the system results in a state in $A_2$.

Theorem 1.  Let $P_1$ be a probability measure on $\mathcal{B}_1$, and suppose

$$K:\Omega_1 \times \mathcal{B}_2 \mapsto [0,1]$$ is a transition kernel.  Then $K$ and $P_1$ uniquely determine a probability on $\mathcal{B}_1 \times \mathcal{B}_2$ via the formula

$$P(A_1\times A_2)= \int_{A_1}K(\omega_1,A_2)P_1(dw_1),$$ for all $A_1\times A_2$ in the class of measurable rectangles.

Theorem 2. Marginalization  Let $P_1$ be a probability measure on $(\Omega_1,\mathcal{B}_1)$ and suppose $K: \Omega_1\times \mathcal{B_2} \mapsto [0,1]$ is a transition kernel.  Define $P$ on $(\Omega_1 \times \Omega_2, \mathcal{B_1}\times \mathcal{B_2})$ by
$$P(A_1\times A_2)= \int_{A_1} K(\omega_1,A_2)P_1(d\omega_1).$$  Assume
$$X:(\Omega_1\times \Omega_2, \mathcal{B}_1\times \mathcal{B}_2) \mapsto (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ and furthermore suppose $X$ is integrable.  Then
$$Y(\omega_1)=\int_{\Omega_2} K(\omega_1,d\omega_2)X_{\omega_2}(\omega_2)$$ has the properties

1. $Y$ is well defined.
2. $Y \in B_1$
3. $Y \in L_1(P_1)$ and furthermore

$$\int_{\Omega_1\times \Omega_2}XdP = \int_{\Omega_1} Y(\omega_1)P_1(d\omega_1) = \int_{\Omega_1} [ \int_{\Omega_2} K(\omega_1,d\omega_2) X_{\omega_1}(d\omega_2)]P_1(\omega_1).$$

Theorem 3.  Fubini Theorem  Let $P=P_1\times P_2$ be a product measure.  If $X$ is $\mathcal{B}_1 \times \mathcal{B}_2$ measurable and is either non-negative or integrable with respect to $P$ then

$$\begin{aligned} \int_{\Omega_1\times \Omega_2}XdP &= \int_{\Omega_1}[ \int_{\Omega_2}X_{\omega_1}(\omega_2) P_2(d\omega_2) ] P_1(d\omega_1) \\                                                                 &= \int_{\Omega_2}[ \int_{\Omega_1}X_{\omega_2}(\omega_1) P_1(d\omega_1) ] P_2(d\omega_2) \end{aligned}$$