Uniqueness of Jordan Decomposition/Lemma
Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {P, N}$ be a Hahn decomposition of $\mu$.
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P, N}$.
Then, for each $A \in \Sigma$, we have:
- $\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
and:
- $\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
Proof
Since $\tuple {\mu^+, \mu^-}$ is a Jordan decomposition of $\mu$, we have:
- $\mu = \mu^+ - \mu^-$
with $\mu^+$ and $\mu^-$ measures.
Let $A \in \Sigma$.
We first show:
- $\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
Let $B \in \Sigma$ have $B \subseteq A$.
We have:
\(\ds \map \mu B\) | \(=\) | \(\ds \map {\mu^+} B - \map {\mu^-} B\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \map {\mu^+} B\) | $\mu^+ \ge 0$ since $\mu^+$ is a measure | |||||||||||
\(\ds \) | \(\le\) | \(\ds \map {\mu^+} A\) | Measure is Monotone |
So:
- $\map {\mu^+} A$ is an upper bound for $\set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
Note that from Intersection is Subset, we have:
- $A \cap P \subseteq A$
From Sigma-Algebra Closed under Countable Intersection, we also have $A \cap P \in \Sigma$ so:
- $\map \mu {A \cap P} \in \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
By definition of the Jordan decomposition:
- $\map {\mu^+} A = \map \mu {A \cap P}$
so:
- $\map {\mu^+} A \in \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
So:
- $\map {\mu^+} A$ is the greatest element of $\set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
From Greatest Element is Supremum, we therefore have:
- $\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
We now show that:
- $\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
Let $B \in \Sigma$ have $B \subseteq A$.
We have:
\(\ds -\map \mu B\) | \(=\) | \(\ds \map {\mu^-} B - \map {\mu^+} B\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \map {\mu^-} B\) | $\mu^- \ge 0$ since $\mu^-$ is a measure | |||||||||||
\(\ds \) | \(\le\) | \(\ds \map {\mu^-} A\) | Measure is Monotone |
So:
- $\map {\mu^-} A$ is an upper bound for $\set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
Note from Intersection is Subset, we have:
- $A \cap N \subseteq A$
From Sigma-Algebra Closed under Countable Intersection, we also have $A \cap N \in \Sigma$ so:
- $-\map \mu {A \cap N} \in \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
By definition of the Jordan decomposition:
- $\map {\mu^-} A = -\map \mu {A \cap N}$
so:
- $\map {\mu^-} A \in \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
So:
- $\map {\mu^-} A$ is the greatest element of $\set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
From Greatest Element is Supremum, we have:
- $\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
hence the result.
$\blacksquare$