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:

\(\ds \cmod {\map \mu A}\)

\(\le\)

\(\ds \map {\mu^+} A + \map {\mu^-} A\)

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.

$\blacksquare$