Dynkin System with Generator Closed under Intersection is Sigma-Algebra

Theorem
Let $X$ be a set.

Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.

Suppose that $\GG$ satisfies the following condition:


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

That is, $\GG$ is closed under intersection.

Then:


 * $\map \delta \GG = \map \sigma \GG$

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:


 * $\map \delta \GG \subseteq \map \sigma \GG$

Let $D \in \map \delta \GG$, and define:


 * $\delta_D := set {E \subseteq X: E \cap D \in \map \delta \GG}$

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:


 * $\paren {E \cap D} \cap \paren {X \setminus D} = E \cap \paren {D \cap \paren {X \setminus D} } = E \cap \O = \O$

Thus, since $E \cap D, X \setminus D \in \map \delta \GG$, it follows that their disjoint union is as well.

Finally, combining the above, it follows that:


 * $\paren {X \setminus E} \cap D \in \map \delta \GG$

Thus:


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

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

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

Hence:

and since the $E_n \cap D$ are in $\map \delta \GG$ by assumption, this disjoint union is also in $\map \delta \GG$.

Therefore, $\delta_D$ is a Dynkin system.

Now by definition of generated Dynkin system, $\GG \subseteq \map \delta \GG$.

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


 * $\GG \subseteq \delta_G$

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


 * $\map \delta \GG \subseteq \delta_G$

Hence, for any $D \in \map \delta \GG$ and $G \in \GG$:


 * $D \cap G \in \map \delta \GG$

Thus we have established that, for all $D \in \map \delta \GG$:


 * $\GG \subseteq \delta_D$

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


 * $\map \delta \GG \subseteq \delta_D$

That is to say:


 * $\forall D, E \in \map \delta \GG: D \cap E \in \map \delta \GG$

Hence, by Dynkin System Closed under Intersections is Sigma-Algebra, $\map \delta \GG$ is a $\sigma$-algebra.

Thus, it follows that $\map \sigma \GG \subseteq \map \delta \GG$.

Hence the result, by definition of set equality.