Uniqueness of Jordan Decomposition

Theorem
Let $\struct {X, \Sigma}$ be a measurable space.

Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.

Let $\tuple {P_1, N_1}$ and $\tuple {P_2, N_2}$ be Hahn decompositions of $\mu$.

Let $\tuple {\mu^+_1, \mu^-_1}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P_1, N_1}$.

Let $\tuple {\mu^+_2, \mu^-_2}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P_2, N_2}$.

Then:


 * $\mu^+_1 = \mu^+_2$

and:


 * $\mu^-_1 = \mu^-_2$

So:


 * the Jordan decomposition of a signed measure is unique.

Lemma
From the lemma, we have:


 * $\map {\mu^+_1} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$

and:


 * $\map {\mu^+_2} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$

for each $A \in \Sigma$.

This gives:


 * $\mu^+_1 = \mu^+_2$

We also obtain:


 * $\map {\mu^-_1} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$

and:


 * $\map {\mu^-_2} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$

for each $A \in \Sigma$, so:


 * $\mu^-_1 = \mu^-_2$

Hence the demand.