Bound for Variation of Complex Measure in terms of Jordan Decomposition

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

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

Let $\cmod \mu$ be the variation of $\mu$.

Let $\tuple {\mu_1, \mu_2, \mu_3, \mu_4}$ be the Jordan decomposition of $\mu$.

Then:


 * $\map {\cmod \mu} A \le \map {\mu_1} A + \map {\mu_2} A + \map {\mu_3} A + \map {\mu_4} A$

for all $A \in \Sigma$.

Proof
Let $A \in \Sigma$.

Let $\map P A$ be the set of finite partitions of $A$ into $\Sigma$-measurable sets.

From the definition of variation, we have:


 * $\ds \map {\cmod \mu} A = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$

Let:


 * $\set {A_1, A_2, \ldots, A_n} \in \map P A$

Consider:


 * $\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} }$

We have, from the definition of Jordan decomposition:


 * $\mu = \mu_1 - \mu_2 + i \paren {\mu_3 - \mu_4}$

so that:

Since: $\set {A_1, A_2, \ldots, A_n}$ is a partition of $A$, we have:


 * $\set {A_1, A_2, \ldots, A_n}$ are pairwise disjoint with:


 * $\ds A = \bigcup_{i \mathop = 1}^n A_i$

So: