# 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.

$\blacksquare$