Dynkin System Generated by a Set Given a set $X$ and a collection of subsets of $X$, $\mathcal{E}$, which is closed under finite intersections, we use $\mathcal{D}(\mathcal{E})$ to denote the Dynkin system generated by $\mathcal{E}$ (that is, the intersection of all Dynkin systems on $X$ containing $\mathcal{E}$).
My question is that, for any $B\in \mathcal{D}(\mathcal{E})$, if we define the set $\mathcal{L}_B$ as the following:
$$
\mathcal{L}_B = \{A\subset X: A\cap B\in \mathcal{D}(\mathcal{E}) \},
$$
(which is already verified to be a Dynkin system), how can I show that $\mathcal{E}\subset \mathcal{L}_B$? Any of your help will be highly appreciated!
 A: Well, I mean... you can't prove this because it's not true. Dynkin systems aren't in general closed under finite intersection. So pick $\mathcal{E}$ to be any Dynkin Class that is not closed under finite intersection. For instance
$$
\mathcal{E}=\{\emptyset, (-\infty,1],(0,\infty), (1,\infty),(-\infty,0], \mathbb{R}\}
$$
It's easy to check that $\mathcal{E}$ is a Dynkin System. Hence, $\mathcal{D}(\mathcal{E})=\mathcal{E}$. It's obviously not closed under finite intersection. In particular, it's not true that $\mathcal{E}\subseteq \mathcal{L}_{(0,\infty)}.$
You are, presumably, trying to prove the uniqueness theorem for probability measures, in which case, you need the assumption that $\mathcal{E}$ is closed under finite intersection, in which case $\mathcal{E}\subseteq \mathcal{L}_B$ trivially.
A: Let $A \in \mathcal E$. $\{B  \in \mathcal D (\mathcal E): A\cap B \in \mathcal D (\mathcal E)\}$ is Dynkin system containing $\mathcal E$ (because $\mathcal E$ is closed under finite intersections). Hence it contains $\mathcal D (\mathcal E)$  and this is what we want to prove.
