Intersection Measure is Measure

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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:

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


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.

$\Box$


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

$\blacksquare$


Sources

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