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:

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

Proof

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 Bound for Variation of Complex Measure in terms of Jordan Decomposition:

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

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 $A \in \Sigma$.

$\blacksquare$