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:

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:

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


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

By Intersection with Empty Set, $\O \cap F = \O$. Hence:


 * $\map {\mu_F} \O = \map \mu {\O \cap F} = 0$

because $\mu$ is a measure.

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