Intersection Measure is Measure

Theorem
Let $\left({X, \Sigma, \mu}\right)$ 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 $\left({X, \Sigma}\right)$.

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

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


 * $\forall E \in \Sigma: \mu_F \left({E}\right) \ge 0$

For every $E \in \Sigma$ have:

$(2)$
Let $\left({E_n}\right)_{n \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.

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


 * $\displaystyle \mu_F \left({\bigcup_{n \mathop \in \N} E_n}\right) = \sum_{n \mathop \in \N} \mu_F \left({E_n}\right)$

This is verified by the following computation:

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


 * $\mu_F \left({\varnothing}\right) = 0$

By Intersection with Null, $\varnothing \cap F = \varnothing$. Hence:


 * $\mu_F \left({\varnothing}\right) = \mu \left({\varnothing \cap F}\right) = 0$

because $\mu$ is a measure.

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