Signed Measure Finite iff Finite Total Variation

Theorem

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

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

Let $\size \mu$ be the variation of $\mu$.

Then $\mu$ is finite if and only if:

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

Proof

Sufficient Condition

Suppose that:

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

Then, from Absolute Value of Signed Measure Bounded Above by Variation, we have:

$\size {\map \mu X} \le \map {\size \mu} X$

so:

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

So $\mu$ is finite.

$\Box$

Necessary Condition

Suppose that $\mu$ is finite.

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

From Jordan Decomposition of Finite Signed Measure, $\mu^+$ and $\mu^-$ are finite measures.

So:

$\map {\mu^+} X < \infty$ and $\map {\mu^-} X < \infty$.

Then, we have:

\(\ds \map {\size \mu} X\)

\(=\)

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

Definition of Total Variation of Signed Measure

\(\ds \)

\(<\)

\(\ds \infty\)

$\blacksquare$