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.

Proof

We first prove two useful identities.

Lemma

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:

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

$\Box$

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.

$\blacksquare$

Sources

• 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $4.1$