Variation of Complex Measure is Finite Measure/Lemma

Lemma
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:
 * $(1) \quad$ $\map {\cmod \mu} A \le \map {\mu_1} A + \map {\mu_2} A + \map {\mu_3} A + \map {\mu_4} A$ for each $A \in \Sigma$
 * $(2) \quad$ $\map {\cmod \mu} A < \infty$ for each $A \in \Sigma$.

Proof of $(1)$
Let $A \in \Sigma$ and:


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

Consider:


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

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

Then:


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

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:

showing $(1)$.

Proof of $(2)$
From Measures in Jordan Decomposition of Complex Measure are Finite, we have:


 * $\mu_1$, $\mu_2$, $\mu_3$, $\mu_4$ are finite.

So there exist real numbers $M_1, M_2, M_3, M_4 > 0$ such that:


 * $\map {\mu_i} A \le M_i$

for each $A \in \Sigma$.

Taking:


 * $M_1 + M_2 + M_3 + M_4 = M$

we have, from $(1)$:


 * $\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} } \le \map {\mu_1} A + \map {\mu_2} A + \map {\mu_3} A + \map {\mu_4} A \le M < \infty$

for each $A \in \Sigma$, with $M > 0$ a real number independent of $A$.

So, for each:


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

we have:


 * $S \le M$

for some $M \in \R$.

So, from the definition of supremum, we have:


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

that is:


 * $\map {\cmod \mu} A \le M < \infty$

So:


 * $\map {\cmod \mu} A < \infty$ for each $A \in \Sigma$

showing $(2)$.