Trace Sigma-Algebra of Generated Sigma-Algebra

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

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

Then the following equality of $\sigma$-algebras on $A$ holds:


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

where


 * $\sigma$ denotes a generated $\sigma$-algebra
 * $A \cap \sigma \left({\mathcal G}\right)$ denotes the trace $\sigma$-algebra on $A$
 * $A \cap \mathcal G$ is a shorthand for $\left\{{A \cap G: G \in \mathcal G}\right\}$

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


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

whence from Intersection Preserves Subsets:


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

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


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

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


 * $\Sigma := \left\{{E \subseteq X: A \cap X \in \sigma \left({A \cap \mathcal G}\right)}\right\}$

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

Since $A \in \sigma \left({A \cap \mathcal G}\right)$, we have:


 * $A \cap X = A \in \sigma \left({A \cap \mathcal G}\right)$

and therefore $X \in \Sigma$.

Now, suppose that $E \in \Sigma$. Then by Intersection Distributes over Set Difference and Intersection with Subset is Subset:


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

Since $E \cap A \in \sigma \left({A \cap \mathcal G}\right)$ and this is a $\sigma$-algebra on $A$:


 * $A \setminus \left({E \cap A}\right) \in \sigma \left({A \cap \mathcal G}\right)$

Finally, let $\left({E_n}\right)_{n \in \N}$ be a sequence in $\Sigma$.

Then by Intersection Distributes over Union:


 * $\displaystyle \left({\bigcup_{n \mathop \in \N} E_n}\right) \cap A = \bigcup_{n \mathop \in \N} \left({E_n \cap A}\right)$

The latter expression is a countable union of elements of $\sigma \left({A \cap \mathcal G}\right)$, hence again in $\sigma \left({A \cap \mathcal G}\right)$.

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

It is also apparent that $\mathcal G \subseteq \Sigma$ since:


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

by definition of generated $\sigma$-algebra.

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


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

and therefore:


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

Hence the result, by definition of set equality.