Uniqueness of Jordan Decomposition/Lemma

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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.

$\blacksquare$