Trace Sigma-Algebra is Sigma-Algebra

Theorem
Let $X$ be a set, and 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.

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 $E \cap \left({X \setminus A}\right)$ in $\Sigma_E$.

By Intersection with Set Difference is Set Difference with Intersection, this equals $\left({X \cap E}\right) \setminus A = E \setminus A$.

Now $E \setminus A = E \setminus S$ by Set Difference with Intersection is Difference.

Hence $E \setminus S \in \Sigma_E$.

Axiom $(3)$
Let $\left({S_i}\right)_{i \in \N}$ 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:

As $\Sigma$ is a $\sigma$-algebra, $\displaystyle \bigcup_{i \mathop \in \N} A_i \in \Sigma$.

Hence $\displaystyle \bigcup_{i \mathop \in \N} S_i \in \Sigma_E$ by definition of trace $\sigma$-algebra.