Intersection Measure is Measure

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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:

\(\displaystyle \mu_F \left({E}\right)\) \(=\) \(\displaystyle \mu \left({E \cap F}\right)\) Definition of $\mu_F$
\(\displaystyle \) \(\ge\) \(\displaystyle 0\) $\mu$ is a measure

$\Box$


Axiom $(2)$

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:

\(\displaystyle \mu_F \left({\bigcup_{n \mathop \in \N} E_n}\right)\) \(=\) \(\displaystyle \mu \left({ \left({\bigcup_{n \mathop \in \N} E_n}\right) \cap F}\right)\) Definition of $\mu_F$
\(\displaystyle \) \(=\) \(\displaystyle \mu \left({ \bigcup_{n \mathop \in \N} \left({E_n \cap F}\right) }\right)\) Intersection Distributes over Union: General Result
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop \in \N} \mu \left({E_n \cap F}\right)\) $\mu$ is a measure
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop \in \N} \mu_F \left({E_n}\right)\) Definition of $\mu_F$

$\Box$


Axiom $(3')$

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

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


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

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

because $\mu$ is a measure.

$\Box$


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

$\blacksquare$


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