Absolute Value of Signed Measure Bounded Above by Variation
Jump to navigation
Jump to search
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 $A \in \Sigma$.
Proof
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then:
- $\mu = \mu^+ - \mu^-$
and:
- $\size \mu = \mu^+ + \mu^-$
We have:
\(\ds \size {\map \mu A}\) | \(=\) | \(\ds \size {\map {\mu^+} A - \map {\mu^-} A}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size {\map {\mu^+} A} + \size {\map {\mu^-} A}\) | Triangle Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\mu^+} A + \map {\mu^-} A\) | since $\mu^+ \ge 0$ and $\mu^- \ge 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\size \mu} A\) |
$\blacksquare$
Sources
- 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $4.1$