Absolute Value of Signed Measure Bounded Above by 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:

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

for each $A \in \Sigma$.

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

Then:

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

and:

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

We have:

$\ds \size {\map \mu A}$

$=$

$\ds \size {\map {\mu^+} A - \map {\mu^-} A}$

$\ds $

$\le$

$\ds \size {\map {\mu^+} A} + \size {\map {\mu^-} A}$

$\ds $

$=$

$\ds \map {\mu^+} A + \map {\mu^-} A$

since $\mu^+ \ge 0$ and $\mu^- \ge 0$

$\ds $

$=$

$\ds \map {\size \mu} A$

$\blacksquare$