This is probably straight-forward by I'm lame with proofs and sets. How can i show that the intersection of any d-systems (Dynkin systems) is again a d-system?
Here is the definition of a Dynkin system.
A collection $\mathcal{D}$ of sub-sets of $S$ where:
(1) $S \in \mathcal{D}$
(2) If $E_1,E_2 \in \mathcal{D}$ where $E_1 \subset E_2$ then $E_2 \setminus E_1 \in \mathcal{D}$
(3) If $E_n \in \mathcal{D}$ and $E_n \subset E_{n+1}$ for any $n$, then $\bigcup_n E_n \in \mathcal{D}$
All such intersection arguments go the same way (i.e. the intersection of a set of $\sigma$-algebras is a $\sigma$-algebra, ditto for topologies etc.):
Let $\mathcal{D}_i$, $i \in I$ be any family of Dynkin-systems on a set $S$.
Then $\mathcal{D} = \bigcap_i \mathcal{D}_i$ is also a Dynkin-system:
We know that for all $i \in I$, $S \in \mathcal{D}_i$, as each $\mathcal{D}_i$ is a Dynkin system. So by the definition of intersection, $S \in \mathcal{D}$, as required.
Suppose $E_1 \subseteq E_2$ where $E_1, E_2 \in \mathcal{D}$. Let $i \in I$ be arbitary. We know that $E_1, E_2 \in \mathcal{D}_i$ (again definition of intersection), and as $\mathcal{D}_i$ is a Dynkin system, we know that $E_2\setminus E_1 \in \mathcal{D}_i$. As this works for all arbitary $i$, $E_2 \setminus E_1 \in \mathcal{D}$, as required.
Now suppose $E_n$ is an increasing sequence of subsets from $\mathcal{D}$. Then fix $i \in I$ again. We know that all $E_n$ are in particular in $\mathcal{D}_i$ and as this is given to be a Dynkin system, $\bigcup_n E_n \in \mathcal{D}_i$, and as $i$ was arbitary, $\bigcup_n E_n \in \mathcal{D}$, as required.
We checked all axioms so $\mathcal{D}$ is a Dynkin system.