Intersection Measure is Measure

Theorem

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

Let $F \in \Sigma$ be a measurable set.

Then the intersection measure $\mu_F$ is a measure on the measurable space $\struct {X, \Sigma}$.

Proof

Verify the axioms for a measure in turn for $\mu_F$:

Axiom $(1)$

The statement of axiom $(1)$ for $\mu_F$ is:

$\forall E \in \Sigma: \map {\mu_F} E \ge 0$

For every $E \in \Sigma$ have:

\(\ds \map {\mu_F} E\)

\(=\)

\(\ds \map \mu {E \cap F}\)

Definition of $\mu_F$

\(\ds \)

\(\ge\)

\(\ds 0\)

$\mu$ is a measure

$\Box$

Axiom $(2)$

Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.

The statement of axiom $(2)$ for $\mu_F$ is:

$\ds \map {\mu_F} {\bigcup_{n \mathop \in \N} E_n} = \sum_{n \mathop \in \N} \map {\mu_F} {E_n}$

This is verified by the following computation:

\(\ds \map {\mu_F} {\bigcup_{n \mathop \in \N} E_n}\)

\(=\)

\(\ds \map \mu {\paren {\bigcup_{n \mathop \in \N} E_n} \cap F}\)

Definition of $\mu_F$

\(\ds \)

\(=\)

\(\ds \map \mu {\bigcup_{n \mathop \in \N} \paren {E_n \cap F} }\)

Intersection Distributes over Union: General Result

\(\ds \)

\(=\)

\(\ds \sum_{n \mathop \in \N} \map \mu {E_n \cap F}\)

$\mu$ is a measure

\(\ds \)

\(=\)

\(\ds \sum_{n \mathop \in \N} \map {\mu_F} {E_n}\)

Definition of $\mu_F$

$\Box$

Axiom $(3')$

The statement of axiom $(3')$ for $\mu_F$ is:

$\map {\mu_F} \O = 0$

We have:

\(\ds \map {\mu_F} \O =\)

\(=\)

\(\ds \map \mu {\O \cap F}\)

Intersection with Empty Set

\(\ds \)

\(=\)

\(\ds \map \mu {\O}\)

Intersection with Empty Set

\(\ds \)

\(=\)

\(\ds 0\)

$\mu$ is a measure

$\Box$

Having verified a suitable set of axioms, it follows that $\mu_F$ is a measure.

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 4$: Problem $7$