Trace Sigma-Algebra is Sigma-Algebra

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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

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