Finite Signed Measure is Complex Measure

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:

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.