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 a 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:

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:

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.