Application of Dynkin's Lemma: Why is this function measurable? Let $(\Omega_1, \mathcal{A}_1)$, $(\Omega_2, \mathcal{A}_2)$ be two measurable spaces. 
In the lecture note it is written that
for all $A \in \mathcal{A}_1 \times \mathcal{A}_2$ we have
$$\omega_1 \mapsto \int_{\Omega_2} 1_A(\omega_1,\omega_2) K(\omega_1, d \omega_2)$$
is $\mathcal{A}_1$-measurable,
where $K$ is a stochastic kernel from $(\Omega_1, \mathcal{A}_1)$ to $(\Omega_2, \mathcal{A}_2)$, i.e.


*

*for all $A \in \mathcal{A}_2$ the map $\omega_1 \mapsto K(\omega_1, A)$ is $\mathcal{A}_1$-measurable and

*for all $\omega_1 \in \Omega_1$ the map $A \mapsto K(\omega_1, A) \in [0,1]$ is a  probability measure on $(\Omega_2, \mathcal{A}_2)$. 

 A: Hint: Show that
$$\mathcal{D} := \{A \in \mathcal{A}_1 \times \mathcal{A}_2; \omega_1 \mapsto \int_{\Omega_2} 1_{A}(\omega_1,\omega_2) \, K(\omega_1,d\omega_2) \, \text{is measurable}\}$$
is a $\sigma$-algebra and that $A_1 \times A_2 \in \mathcal{D}$ for any $A_1 \in \mathcal{A}_1$, $A_2 \in \mathcal{A}_2$.
A: Let $\mathcal{D}$ denote the collection of sets $A\in\mathcal{A}_{1}\times\mathcal{A}_{2}$
for wich the function is measurable.
For $A=A_{1}\times A_{2}$ with $A_{i}\in\mathcal{A}_{i}$ the function
is $\omega_{1}\mapsto\int_{\Omega_{2}}1_{A}\left(\omega_{1},\omega_{2}\right)K\left(\omega_{1},d\omega_{2}\right)=1_{A_{1}}\left(\omega_{1}\right)K\left(\omega_{1},A_{2}\right)$
hence is measurable.
Observe that:


*

*$\Omega=\Omega_{1}\times\Omega_{2}\in\mathcal{D}$.

*$\int_{\Omega_{2}}1_{A^{c}}\left(\omega_{1},\omega_{2}\right)K\left(\omega_{1},d\omega_{2}\right)=K\left(\omega_{1},\Omega_{2}\right)-\int_{\Omega_{2}}1_{A}\left(\omega_{1},\omega_{2}\right)K\left(\omega_{1},d\omega_{2}\right)$
so $\mathcal{D}$ is closed under complementation.

*If sets $A^{\left(i\right)}\in\mathcal{D}$ are disjoint and $A=\bigcup_{i=1}^{n}A^{\left(i\right)}$
then $A\in\mathcal{A}$ and $\int_{\Omega_{2}}1_{A}\left(\omega_{1},\omega_{2}\right)K\left(\omega_{1},d\omega_{2}\right)=\sum_{i=1}^{\infty}\int_{\Omega_{2}}1_{A^{\left(i\right)}}\left(\omega_{1},\omega_{2}\right)K\left(\omega_{1},d\omega_{2}\right)$
is measurable.
Proved is now that $\mathcal{D}$ is a Dynkinsystem and this with
$\left\{ A_{i}\times A_{2}\mid A_{1}\in\mathcal{A}_{1},A_{2}\in\mathcal{A}_{2}\right\} \subset\mathcal{D}$.
Collection $\left\{ A_{1}\times A_{2}\mid A_{1}\in\mathcal{A}_{1},A_{2}\in\mathcal{A}_{2}\right\} $
is closed under finite intersection.
Now Dynkins lemma (the Dynkin system generated by π-system is a
σ-algebra.) can be applied and we conclude that 
$\mathcal{A}_{1}\times\mathcal{A}_{2}=\sigma\left(\left\{ A_{1}\times A_{2}\mid A_{1}\in\mathcal{A}_{1},A_{2}\in\mathcal{A}_{2}\right\} \right)\subseteq\mathcal{D}$.
