Trace Sigma-Algebra is Sigma-Algebra
Theorem
Let $X$ be a set.
Let $\Sigma$ be a $\sigma$-algebra on $X$.
Let $E \subseteq X$ be a subset of $X$.
Then the trace $\sigma$-algebra $\Sigma_E$ is a $\sigma$-algebra on $E$.
Proof
Verifying the axioms for a $\sigma$-algebra in turn:
Axiom $(1)$
Have $E = X \cap E$ from Intersection with Subset is Subset.
Hence, as $X \in \Sigma$ by axiom $(1)$, $E \in \Sigma_E$ by definition of trace $\sigma$-algebra.
$\Box$
Axiom $(2)$
Suppose that $S \in \Sigma_E$.
Then there is some $A \in \Sigma$ such that $S = E \cap A$, by definition of trace $\sigma$-algebra.
Hence $X \setminus A \in \Sigma$ by axiom $(2)$ for $\sigma$-algebras.
So $E \cap \paren {X \setminus A}$ in $\Sigma_E$.
By Intersection with Set Difference is Set Difference with Intersection, this equals $\paren {X \cap E} \setminus A = E \setminus A$.
Now $E \setminus A = E \setminus S$ by Set Difference with Intersection is Difference.
Hence $E \setminus S \in \Sigma_E$.
$\Box$
Axiom $(3)$
Let $\sequence {S_i}_{i \mathop \in \N}$ be a sequence in $\Sigma_E$.
For each $i \in \N$, let $A_i \in \Sigma$ such that $S_i = E \cap A_i$.
Then observe:
\(\ds \bigcup_{i \mathop \in \N} S_i\) | \(=\) | \(\ds \bigcup_{i \mathop \in \N} E \cap A_i\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds E \cap \bigcup_{i \mathop \in \N} A_i\) | Intersection Distributes over Union: General Result |
As $\Sigma$ is a $\sigma$-algebra, $\ds \bigcup_{i \mathop \in \N} A_i \in \Sigma$.
Hence $\ds \bigcup_{i \mathop \in \N} S_i \in \Sigma_E$ by definition of trace $\sigma$-algebra.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $3.3 \ \text{(vi)}$, $\S 3$: Problem $2$