Show that the set $D_b$ is a Dynkin-System

elementary-set-theorymeasure-theory

let $X \neq \emptyset$, ${D_B} := \{Q \subseteq X: Q \cap B \in \sigma(\mathcal{E}) \}$, where $\sigma(\mathcal{E}):= \bigcup_{\mathcal{E} \subseteq \mathcal{D_{i}}} D_i$ and $D_i$ is a family of Dynkin systems $ \forall i \in I$.

That is, $\emptyset \neq X, D \subseteq \mathcal{P}(X)$ is a Dynkin system when:

  1. $\emptyset, X \in D$.
  2. $A \in D \to X\setminus A \in D$.
  3. For every sequence $(A_{n})_{n\in\mathbb{N}}$ of pairwise disjoint sets $A_i$ in $D$, the union $\bigcup_{n\in\mathbb{N}} A_n \in D$.

Moreover, $B \in \sigma(\mathcal{E})$ and $ \mathcal{E} \subseteq \mathcal{P}(X)$. What I want to prove is that $D_B$ is indeed a Dynkin System. I struggle with the second and third condition and would appreciate some help.

Best Answer

It seems that the Dynkin system generated by $\mathcal E$ is the intersection of all the Dynkin systems containing $\mathcal E$. Denoting by $\mathcal D$ the collection of the Dynkin systems containing $\mathcal E$, this means that $\sigma\left(\mathcal E\right)=\bigcap_{D\in\mathcal D}D$. With these notations, $$ D_B=\left\{Q\subseteq X: Q\cap B\in \bigcap_{D\in\mathcal D}D\right\} =\bigcap_{D\in\mathcal D} \left\{Q\subseteq X: Q\cap B\in D\right\}. $$ To conclude, we need the following two facts:

  1. An arbitrary intersection of Dynkin systems is a Dynkin system.
  2. For each Dynkin system $D$, the collection $\mathcal A_D=\left\{Q\subseteq X: Q\cap B\in D\right\}$ is a Dynkin system.

Let us show the second fact. Observe that for all $D\in\mathcal D$, $B\in\mathcal D$ hence $X\in \mathcal A_D$. Let $Q\in\mathcal A_D$. Write $$ \left(X\setminus Q\right)\cap B=B\cap Q^c=\left(B^c\cup \left(Q\cap B\right) \right)^c. $$ Since $D$ is Dynkin system, $B^c\in D$ and by assumption $Q\cap B\in D$ hence by the third point of the definition, the disjoint union $B^c\cup \left(Q\cap B\right) $ belongs to $D$. Using again the second point of the definition shows that $X\setminus Q\in \mathcal A_D$.

For the third condition, let $\left(Q_n\right)_{n\geqslant 1}$ be a pairwise disjoint sequence of elements of $\mathcal A_D$. We know that $Q_n\cap B\in D$ for all $n$; since the sequence $\left(Q_n\cap B\right)_{n\geqslant 1}$ is disjoint and consists of elements of $D$, we know that $\bigcup_{n\geqslant 1}\left(Q_n\cap B\right)= \left(\bigcup_{n\geqslant 1}Q_n\right)\cap B$ is an element of $D$ hence so is $\bigcup_{n\geqslant 1}Q_n$.

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