Jordan Decomposition of Finite Signed Measure

Theorem

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

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

Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.

Then $\mu^+$ and $\mu^-$ are finite measures.

Proof

From the definition of Jordan decomposition, we have:

$\mu = \mu^+ - \mu^-$

with at least one of $\mu^+$ and $\mu^-$ finite. From the definition of a finite signed measure, we have:

$\cmod {\map \mu X} < \infty$

We show that:

if exactly one of $\mu^+$ and $\mu^-$ is finite, then $\mu$ is not a finite signed measure

that is:

if exactly one of $\mu^+$ and $\mu^-$ is finite, then $\map \mu X = \infty$ or $\map \mu X = -\infty$.

At which point we will have a contradiction, giving:

$\mu^+$ and $\mu^-$ are both finite.

Aiming for a contradiction, suppose suppose exactly one of $\mu^+$ and $\mu^-$ is finite.

Consider first the case of $\mu^+$ non-finite.

Then $\mu^-$ is finite, so:

$\map {\mu^-} X < \infty$

Since $\mu^+$ is not finite, we have:

$\map {\mu^+} X = \infty$

From the definition of extended real subtraction, we then have:

$\map \mu X = \map {\mu^+} X - \map {\mu^-} X = \infty$

which contradicts:

$\cmod {\map \mu X} < \infty$

So we have a contradiction in the case of $\mu^+$ non-finite.

Now suppose that $\mu^-$ is non-finite.

Then $\mu^+$ is finite, so:

$\map {\mu^+} X < \infty$

Since $\mu^-$ is not finite, we have:

$\map {\mu^-} X = \infty$

From the definition of extended real subtraction, we then have:

$\map \mu X = \map {\mu^+} X - \map {\mu^-} X = -\infty$

which contradicts:

$\cmod {\map \mu X} < \infty$

So we have a contradiction in the case of $\mu^-$ non-finite.

So we must have that both $\mu^+$ and $\mu^-$ are finite, completing the proof.

$\blacksquare$