Restriction of Measure to Trace Sigma-Algebra of Measurable Set is Measure

Theorem
Let $\struct {X, \Sigma}$ be a measurable space.

Let $\mu$ be a measure on $\struct {X, \Sigma}$.

Let $A \in \Sigma$.

Let $\Sigma_A$ be the trace $\sigma$-algebra of $A$ in $\Sigma$.

Let $\mu \restriction_{\Sigma_A}$ be the restriction of $\mu$ to $\Sigma_A$.

Then $\mu \restriction_{\Sigma_A}$ is a measure on $\struct {A, \Sigma_A}$.

Proof
We verify the three conditions required of a measure for $\mu \restriction_{\Sigma_A}$.

Note that from Trace Sigma-Algebra of Measurable Set, we have $\Sigma_A \subseteq \Sigma$.

Condition $(1)$
Let $E \in \Sigma_A$.

Then, we have $E \in \Sigma$ and:

Condition $(2)$
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint $\Sigma_A$-measurable sets.

Then $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint $\Sigma$-measurable sets, and we have:

Condition $(3)$
From Empty Set is Null Set, we have that $\O$ is $\mu$-null.

From Sigma-Algebra Contains Empty Set, we have $\O \in \Sigma_A$.

So, we have: