Theorem: Dynkin System with Generator Closed under Intersection is Sigma-Algebra

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 $E \in \delta_D$:
{{begin-eqn}}
{{eqn | l = \paren {X \setminus E} \cap D
      | r = \paren {\paren {X \setminus E} \cap D} \cup \paren {\paren {X \setminus D} \cap D}
      | c = Set Difference Intersection with Second Set is Empty Set, Union with Empty Set
}}
{{eqn | r = \paren {\paren {X \setminus E} \cup \paren {X \setminus D} } \cap D
      | c = Intersection Distributes over Union
}}
{{eqn | r = \paren {X \setminus \paren {E \cap D} } \cap D
      | c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \paren {X \setminus \paren {E \cap D} } \cap \paren {X \setminus \paren {X \setminus D} }
      | c = Set Difference with Set Difference
}}
{{eqn | r = X \setminus \paren {\paren {E \cap D} \cup \paren {X \setminus D} }
      | c = De Morgan's Laws: Difference with Union
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \paren {\bigcup_{n \mathop \in \N} E_n} \cap D
      | r = \bigcup_{n \mathop \in \N} \paren {E_n \cap D}
      | c = Intersection Distributes over Union
}}
{{end-eqn}}
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.
{{qed}}