Trace Sigma-Algebra of Generated Sigma-Algebra

Theorem
Let $X$ be a Set, and let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.

Let $A \subseteq X$ be a subset of $X$.

Then the following equality holds:


 * $A \cap \map \sigma \GG = \map \sigma {A \cap \GG}$

where


 * $\map \sigma \GG$ denotes the smallest $\sigma$-algebra on $X$ that contains $\GG$
 * $\map \sigma {A \cap \GG}$ denotes the smallest $\sigma$-algebra on $A$ that contains ${A \cap \GG}$
 * $A \cap \map \sigma \GG$ denotes the trace $\sigma$-algebra on $A$
 * $A \cap \GG$ is a shorthand for $\set {A \cap G: G \in \GG}$

Proof
By definition of generated $\sigma$-algebra:


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

whence from Set Intersection Preserves Subsets:


 * $A \cap \GG \subseteq A \cap \map \sigma \GG$

and therefore, by definition of generated $\sigma$-algebra:


 * $\map \sigma {A \cap \GG} \subseteq A \cap \map \sigma \GG$

For the reverse inclusion, define $\Sigma$ by:


 * $\Sigma := \set {E \subseteq X: A \cap E \in \map \sigma {A \cap \GG} }$

We will show that $\Sigma$ is a $\sigma$-algebra on $X$.

Since $A \in \map \sigma {A \cap \GG}$:


 * $A \cap X = A \in \map \sigma {A \cap \GG}$

and therefore $X \in \Sigma$.

Suppose that $E \in \Sigma$.

Then by Set Intersection Distributes over Set Difference and Intersection with Subset is Subset:


 * $\paren {X \setminus E} \cap A = \paren {X \cap A} \setminus \paren {E \cap A} = A \setminus \paren {E \cap A}$

Since $E \cap A \in \map \sigma {A \cap \GG}$ and this is a $\sigma$-algebra on $A$:


 * $A \setminus \paren {E \cap A} \in \map \sigma {A \cap \GG}$

Finally, let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\Sigma$.

Then by Intersection Distributes over Union:


 * $\displaystyle \paren {\bigcup_{n \mathop \in \N} E_n} \cap A = \bigcup_{n \mathop \in \N} \paren {E_n \cap A}$

The latter expression is a countable union of elements of $\map \sigma {A \cap \GG}$, hence again in $\map \sigma {A \cap \GG}$.

Therefore, $\Sigma$ is a $\sigma$-algebra.

It is also apparent that $\GG \subseteq \Sigma$ since:


 * $A \cap \GG \subseteq \map \sigma {A \cap \GG}$

by definition of generated $\sigma$-algebra.

Thus, as $\Sigma$ is a $\sigma$-algebra:


 * $\map \sigma \GG \subseteq \Sigma$

and therefore:


 * $A \cap \map \sigma \GG \subseteq \map \sigma {A \cap \GG}$

Hence the result, by definition of set equality.