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:

\(\ds \map {\paren {\mu \restriction_{\Sigma_A} } } E\)

\(=\)

\(\ds \map \mu E\)

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

\(\ds \)

\(\ge\)

\(\ds 0\)

$\Box$

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:

\(\ds \map {\paren {\mu \restriction_{\Sigma_A} } } {\bigcup_{n \mathop = 1}^\infty E_n}\)

\(=\)

\(\ds \map \mu {\bigcup_{n \mathop = 1}^\infty E_n}\)

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

\(\ds \)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \map \mu {E_n}\)

since $\mu$ is a measure

\(\ds \)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \map {\paren {\mu \restriction_{\Sigma_A} } } {E_n}\)

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

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:

\(\ds \map {\paren {\mu \restriction_{\Sigma_A} } } \O\)

\(=\)

\(\ds \map \mu \O\)

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

\(\ds \)

\(=\)

\(\ds 0\)

Definition of Null Set

$\blacksquare$