Finite Signed Measure is Complex Measure
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Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a finite signed measure on $\struct {X, \Sigma}$.
Then $\mu$ is a complex measure on $\struct {X, \Sigma}$.
Proof
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then:
- $\mu = \mu^+ - \mu^-$
for measures $\mu^+$ and $\mu^-$.
From Jordan Decomposition of Finite Signed Measure, $\mu^+$ and $\mu^-$ are finite measures.
Then, for each $A \in \Sigma$, we have:
\(\ds \cmod {\map \mu A}\) | \(\le\) | \(\ds \map {\mu^+} A + \map {\mu^-} A\) | Absolute Value of Signed Measure Bounded Above by Variation | |||||||||||
\(\ds \) | \(\le\) | \(\ds \map {\mu^+} X + \map {\mu^-} X\) | Measure is Monotone | |||||||||||
\(\ds \) | \(<\) | \(\ds \infty\) | Definition of Finite Measure |
So $\mu$ takes only real values.
That is, $\mu$ takes values in $\C$.
Since $\mu$ is a signed measure, we have:
- $\map \mu \O = 0$
and:
- for each sequence $\sequence {S_n}_{n \mathop \in \N}$ of pairwise disjoint $\Sigma$-measurable sets we have:
- $\ds \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = \sum_{n \mathop = 1}^\infty \map \mu {S_n}$
So $\mu$ is a complex measure.
$\blacksquare$