Show that $\mathcal{D}$ is a Dynkin-System (concerning transition measure) In our lecture we had the following definition for "transition measure".
Let $(\Omega_1,\mathcal{A}_1), (\Omega_2,\mathcal{A}_1)$ be two measurable spaces. A function $K\colon \Omega_1\times\mathcal{A}_2\to [0,\infty)$ is called transition measure from $(\Omega_1,\mathcal{A}_1)$ to $(\Omega_2,\mathcal{A}_2)$ if
(1) $A\mapsto K(\omega_1,A)$ is a measure on $\mathcal{A}_2$ for all $\omega_1\in\Omega_1$
(2) $\omega_1\mapsto K(\omega_1,A)$ is a $\mathcal{A}_1-\mathcal{B}$-measurable function for all $A\in\mathcal{A}_2$.
Now the following exercise is given:


Show that it is enough to claim the condition (2) in the definition of transition mesaure only for sets $A$ out of a generator $\mathcal{C}$ of $\mathcal{A}_2$ which is closed under intersections, because then (2) is achieved for all $A\in\mathcal{A}_2$.


I already had some thoughts about that and going through the script, I think the strategy is to show that 
$$
\mathcal{D}:=\left\{A\in\mathcal{A}_2: \omega_1\mapsto K(\omega_1,A)\text{ is }\mathcal{A}_1-\mathcal{B}-\text{measurable}\right\}
$$
is a Dynkin-System with $\mathcal{C}\subset\mathcal{D}$. Because then it follows
$$
\mathcal{A}_2=\sigma(\mathcal{C})=\delta(\mathcal{C})\subset\mathcal{D},
$$
where $\delta(\mathcal{C})$ is the Dynkin-system generated by $\mathcal{C}$.
So far so good. But now I have problems to show that $\mathcal{D}$ is indeed a Dynkin-system.
There are three things to show in order to prove that $\mathcal{D}$ is a Dynkin-system:
(a) $\Omega_2\in\mathcal{D}$
(b) $A,B\in\mathcal{D}, A\subset B\implies B\setminus A\in\mathcal{D}$
(c) $A_1,A_2,\ldots\in\mathcal{D}$, pairwise disjoint $\implies \bigcup_{n=1}^{\infty}A_n\in\mathcal{D}$
Furthermore it is to show that
(d) $\mathcal{C}\subset\mathcal{D}$.

As I said I have unexpected problems to show (a), (b), (c) and (d).
I start with (a) to show you what f.e. is my problem.
(a)
It is to show that $\omega_1\mapsto K(\omega_1,\Omega_2)$ is $\mathcal{A}_1-\mathcal{B}$-measurable. To do so i take f.e. $(-\infty, c]\in\mathcal{B}$ and I have to show that the preimage is in $\mathcal{A}_1$. But how can I do so?
With greetings
math12
 A: a) You have to assume something in addition (as I will show below), such as $K(\cdot,\Omega)$ being constant, say when $K(\omega,\cdot)$ is always a probability measure.  
b) You can write $K(\cdot,B\backslash A)$ as the difference of two measurable functions.
c) Pick some real number $r$. By assumption, $K(\omega,\cdot)$ is a measure for all $\Omega$. So the set of $\omega$ such that $K\Big(\omega,\bigcup_n A\Big)>r$ is the set of all $\omega$ such that $\sum_{n=1}^\infty K(\omega,A_n)>r$, which equals the set of all $\omega$ such that $\sum_{n=1}^N K(\cdot,A_n)>r$ for some $N$. Let $Q^n$ be the countable set of $n$-tuples of nonnegative rationals with sum larger than $r$. Let $E_N$ be the event that $\sum_{n=1}^N K(\cdot,A_n)>r$. We can write it as $$E_N=\bigcup_{(q_1,\ldots,q_N)\in Q^N}\bigg(K(\cdot,A_1)^{-1}\Big([q_1,\infty)\Big)\cap\ldots\cap K(\cdot,A_N)^{-1}\Big([q_N,\infty)\Big)\bigg),$$
which is clearly measurable. So the set of $\omega$ such that $K\Big(\omega,\bigcup_n A_n\Big)>r$ is $\bigcup_N E_N$, which is measurable. Since $r$ was arbitrary, we have proven this part.
d) Holds by assumption.

Here is an example where the result fails. Let $(\Omega_1,\mathcal{A}_1)$ be a measurable space that contains a nonmeasurable subset $N$. Let $\Omega_2$ be nonempty and let $\mathcal{A_2}=\{\Omega_2,\emptyset\}$. We can take $\mathcal{C}=\emptyset$ here, so that (2) holds for all sets in $\mathcal{C}$ vacuously. Now a measure on $(\Omega_2,\mathcal{A}_2$ is determined by the measure assigned to $\Omega_2$. So let $K(\omega,\Omega_2)=1$ for $\omega\in N$ and $K(\omega,\Omega_2)=2$ for $\omega\notin N$. Then $\omega\to K(\omega,\Omega_2)$ is not measurable.
