Dynkin System with Generator Closed under Intersection is Sigma-Algebra

Theorem
Let $X$ be a set.

Let $\mathcal G \subseteq \mathcal P \left({X}\right)$ be a collection of subsets of $X$.

Suppose that $\mathcal G$ satisfies the following condition:


 * $(1):\quad \forall G, H \in \mathcal G: G \cap H \in \mathcal G$

That is, $\mathcal G$ is closed under intersection.

Then:


 * $\delta \left({\mathcal G}\right) = \sigma \left({\mathcal G}\right)$

where $\delta$ denotes generated Dynkin system, and $\sigma$ denotes generated $\sigma$-algebra.

Proof
From Sigma-Algebra is Dynkin System and the definition of generated Dynkin system, it follows that:


 * $\delta \left({\mathcal G}\right) \subseteq \sigma \left({\mathcal G}\right)$

Let $D \in \delta \left({\mathcal G}\right)$, and define:


 * $\delta_D := \left\{{E \subseteq X: E \cap D \in \delta \left({\mathcal G}\right)}\right\}$

Let us verify that these $\delta_D$ form Dynkin systems.

First of all, note $X \cap D = D$, hence $X \in \delta_D$.

Next, compute, for any $E \in \delta_D$:

Now from Intersection is Associative, Set Difference Intersection with Second Set is Empty Set and Intersection with Empty Set:


 * $\left({E \cap D}\right) \cap \left({X \setminus D}\right) = E \cap \left({D \cap \left({X \setminus D}\right)}\right) = E \cap \varnothing = \varnothing$

Thus, since $E \cap D, X \setminus D \in \delta \left({\mathcal G}\right)$, it follows that their disjoint union is as well.

Finally, combining the above, it follows that:


 * $\left({X \setminus E}\right) \cap D \in \delta \left({\mathcal G}\right)$

Thus:


 * $E \in \delta_D \implies X \setminus E \in \delta_D$

Finally, let $\left({E_n}\right)_{n \mathop \in \N}$ be a pairwise disjoint sequence of sets in $\delta_D$.

Then it is immediate that $\left({E_n \cap D}\right)_{n \mathop \in \N}$ is also pairwise disjoint.

Hence:

and since the $E_n \cap D$ are in $\delta \left({\mathcal G}\right)$ by assumption, this disjoint union is also in $\delta \left({\mathcal G}\right)$.

Therefore, $\delta_D$ is a Dynkin system.

Now by definition of generated Dynkin system, $\mathcal G \subseteq \delta \left({\mathcal G}\right)$.

By this observation and $(1)$, we immediately obtain, for any $G \in \mathcal G$:


 * $\mathcal G \subseteq \delta_G$

Therefore, by definition of generated Dynkin system, for all $G \in \mathcal G$:


 * $\delta \left({\mathcal G}\right) \subseteq \delta_G$

Hence, for any $D \in \delta \left({\mathcal G}\right)$ and $G \in \mathcal G$:


 * $D \cap G \in \delta \left({\mathcal G}\right)$

Thus we have established that, for all $D \in \delta \left({\mathcal G}\right)$:


 * $\mathcal G \subseteq \delta_D$

whence, by the definition of $\delta_D$ and generated Dynkin system:


 * $\delta \left({\mathcal G}\right) \subseteq \delta_D$

That is to say:


 * $\forall D, E \in \delta \left({\mathcal G}\right): D \cap E \in \delta \left({\mathcal G}\right)$

Hence, by Dynkin System Closed under Intersections is Sigma-Algebra, $\delta \left({\mathcal G}\right)$ is a $\sigma$-algebra.

Thus, it follows that $\sigma \left({\mathcal G}\right) \subseteq \delta \left({\mathcal G}\right)$.

Hence the result, by definition of set equality.