Intersection Complex Measure is Complex Measure

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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

Let $F \in \Sigma$.

Let $\mu_F$ be the intersection complex measure of $\mu$ by $F$.


Then $\mu_F$ is a complex measure.


Proof

Since $\mu$ is a complex measure, we have:

$\map \mu E \in \C$

for each $E \in \Sigma$.

So, in particular:

$\map \mu {E \cap F} \in \C$

for all $E \in \Sigma$.

That is:

$\map {\mu_F} E \in \C$

for all $E \in \Sigma$.

We verify the two conditions required of a complex measure.

We have:

\(\ds \map {\mu_F} \O\) \(=\) \(\ds \map \mu {F \cap \O}\) Definition of Intersection Complex Measure
\(\ds \) \(=\) \(\ds \map \mu \O\) Intersection with Empty Set
\(\ds \) \(=\) \(\ds 0\) since $\mu$ is a complex measure

Now let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.

Then, we have:

\(\ds \map {\mu_F} {\bigcup_{n \mathop = 1}^\infty S_n}\) \(=\) \(\ds \map \mu {F \cap \bigcup_{n \mathop = 1}^\infty S_n}\) Definition of Intersection Complex Measure
\(\ds \) \(=\) \(\ds \map \mu {\bigcup_{n \mathop = 1}^\infty \paren {F \cap S_n} }\) Intersection Distributes over Union
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map \mu {F \cap S_n}\) using countable additivity of $\mu$
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map {\mu_F} {S_n}\) Definition of Intersection Complex Measure

So $\mu$ is a complex measure.

$\blacksquare$