Variation of Complex Measure is Finite Measure

== 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 $\mu$. 
Then $\cmod \mu$ is a finite measure on $\struct {X, \Sigma}$.
== Proof ==
We first show that $\map {\cmod \mu} A \ge 0$ for each $A \in \Sigma$. 
Let $A \in \Sigma$.
Let $\map P A$ be the set of finite partitions of $A$ into $\Sigma$-measurable sets.
Then, for each $A \in \Sigma$, we have: 
:$\ds \map {\cmod \mu} A = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
We clearly have: 
:$\set A \in \map P A$ 
so:
:$\ds \size {\map \mu A} \in \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
From the definition of supremum, we therefore have: 
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} \ge \size {\map \mu A} \ge 0$
giving:
:$\map {\cmod \mu} A \ge 0$
We will now verify the three conditions in Characterization of Measures. 
=== Proof of $(1)$ ===
We show that: 
:$\map {\cmod \mu} \O = 0$ 
We have: 
:$\map P \O = \set \O$
since the empty set does not have any subsets aside from itself. 
So:
:$\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P \O} = \set {\cmod {\map \mu \O} } = \set 0$ 
From the definition of supremum, we then have: 
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P \O} = 0$
so:
:$\map {\cmod \mu} \O = 0$
Let $\tuple {\mu_1, \mu_2, \mu_3, \mu_4}$ be the Jordan decomposition of $\mu$.
We obtain the following lemma:
=== Lemma ===
=== Proof of $(2)$ ===
Let $B_1$ and $B_2$ be disjoint $\Sigma$-measurable sets.
We show that: 
:$\map {\cmod \mu} {B_1 \cup B_2} = \map {\cmod \mu} {B_1} + \map {\cmod \mu} {B_2}$
at which point we will have finite additivity. 
We will do this by showing that: 
:$\map {\cmod \mu} {B_1 \cup B_2} \le \map {\cmod \mu} {B_1} + \map {\cmod \mu} {B_2}$
and then: 
:$\map {\cmod \mu} {B_1} + \map {\cmod \mu} {B_2} \le \map {\cmod \mu} {B_1 \cup B_2}$
We have: 
:$\ds \map {\cmod \mu} {B_1 \cup B_2} = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_1 \cup B_2} }$
Let: 
:$\set {A_1, A_2, \ldots, A_n} \in \map P {B_1 \cup B_2}$
Then from Sigma-Algebra Closed under Countable Intersection, we have: 
:$A_i \cap B_1 \in \Sigma$ for each $i$
and:
:$A_i \cap B_2 \in \Sigma$ for each $i$. 
We then have: 
:$\set {A_1 \cap B_1, A_2 \cap B_1, \ldots, A_n \cap B_1}$ is pairwise disjoint
and: 
so that: 
:$\set {A_1 \cap B_1, A_2 \cap B_1, \ldots, A_n \cap B_1} \in \map P {B_1}$
We therefore have, from the definition of supremum: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i \cap B_1} } \le \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_1} }$
That is: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i \cap B_1} } \le \map {\cmod \mu} {B_1}$
We also have: 
:$\set {A_1 \cap B_2, A_2 \cap B_2, \ldots, A_n \cap B_2}$ is pairwise disjoint
and: 
so that: 
:$\set {A_1 \cap B_2, A_2 \cap B_2, \ldots, A_n \cap B_2} \in \map P {B_2}$
We therefore have, from the definition of supremum: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i \cap B_2} } \le \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_2} }$
That is: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i \cap B_2} } \le \map {\cmod \mu} {B_2}$
Since from Intersection is Subset, we have:
:$A_i \cap B_1 \subseteq B_1$
and:
:$A_i \cap B_2 \subseteq B_2$
and $B_1$, $B_2$ are disjoint, we have that: 
:for each $i$, $A_i \cap B_1$ and $A_i \cap B_2$ are disjoint.
We now have: 
Since:
:$\sequence {A_1, A_2, \ldots, A_n}$ was an arbitrary element of $\map P {B_1 \cup B_2}$
We have that: 
:$S \le \map {\cmod \mu} {B_1} + \map {\cmod \mu} {B_2}$
for each:
:$\ds S \in \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_1 \cup B_2} }$
So, by the definition of supremum, we have: 
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_1 \cup B_2} } \le \map {\cmod \mu} {B_1} + \map {\cmod \mu} {B_2}$
giving: 
:$\map {\cmod \mu} {B_1 \cup B_2} \le \map {\cmod \mu} {B_1} + \map {\cmod \mu} {B_2}$
We now show that: 
:$\map {\cmod \mu} {B_1} + \map {\cmod \mu} {B_2} \le \map {\cmod \mu} {B_1 \cup B_2}$
Let: 
:$\set {E_1, E_2, \ldots, E_n} \in \map P {B_1}$
and:
:$\set {F_1, F_2, \ldots, F_m} \in \map P {B_2}$ 
Since: 
:$E_i \subseteq B_1$
and:
:$F_j \subseteq B_2$
for each $i, j$ and $B_1$, $B_2$ are disjoint, we have that: 
:$\set {E_1, E_2, \ldots, E_n, F_1, F_2, \ldots, F_m}$ is pairwise disjoint.
Clearly their union is $B_1 \cup B_2$.
So:
:$\set {E_1, E_2, \ldots, E_n, F_1, F_2, \ldots, F_m} \in \map P {B_1 \cup B_2}$
So, we have: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {E_i} } + \sum_{j \mathop = 1}^m \cmod {\map \mu {F_j} } \in \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_1 \cup B_2} }$
From the definition of supremum, this gives: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {E_i} } + \sum_{j \mathop = 1}^m \cmod {\map \mu {F_j} } \le \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_1 \cup B_2} }$
Since: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {E_i} }$ is an arbitrary element of $\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_1} }$
and:
:$\ds \sum_{j \mathop = 1}^m \cmod {\map \mu {F_j} }$ is an arbitrary element of $\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_2} }$
We obtain: 
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_1} } + \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_2} } \le \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P {B_1 \cup B_2} }$
so that: 
:$\map {\cmod \mu} {B_1} + \map {\cmod \mu} {B_2} \le \map {\cmod \mu} {B_1 \cup B_2}$
So we obtain: 
:$\map {\cmod \mu} {B_1 \cup B_2} = \map {\cmod \mu} {B_1} + \map {\cmod \mu} {B_2}$
so $\cmod \mu$ is finitely additive.
It remains to verify $(3)$.
=== Proof of $(3)$ ===
We will show that:
:for every decreasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ for which $\map {\cmod \mu} {E_1}$ is finite, if $E_n \downarrow \O$, we have:
::$\ds \lim_{n \mathop \to \infty} \map {\cmod \mu} {E_n} = 0$
Note that since $\map {\cmod \mu} A$ is finite for each $A \in \Sigma$ from $(2)$ in the Lemma, we can simply show: 
:for every decreasing sequence $\sequence {E_n}_{n \mathop \in \N}$, if $E_n \downarrow \O$ we have:
::$\ds \lim_{n \mathop \to \infty} \map {\cmod \mu} {E_n} = 0$
From Characterization of Measures again, since $\mu_1$, $\mu_2$, $\mu_3$ and $\mu_4$ are measures, we have: 
:for every decreasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ for which $\map {\mu_i} {E_1}$ is finite, if $E_n \downarrow \O$, we have:
::$\ds \lim_{n \mathop \to \infty} \map {\mu_i} {E_n} = 0$
for $i \in \set {1, 2, 3, 4}$.
From Measures in Jordan Decomposition of Complex Measure are Finite, we have that: 
:$\mu_1$, $\mu_2$, $\mu_3$ and $\mu_4$ are finite.
So, we have:
:for every decreasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$, if $E_n \downarrow \O$ we have:
::$\ds \lim_{n \mathop \to \infty} \map {\mu_i} {E_n} = 0$
for $i \in \set {1, 2, 3, 4}$.
Let $\sequence {E_n}_{n \mathop \in \N}$ be a decreasing sequence in $\Sigma$ with $E_n \downarrow \O$.
Then: 
:$\ds \lim_{n \mathop \to \infty} \map {\mu_i} {E_n} = 0 = 0$
for each $i \in \set {1, 2, 3, 4}$.
So:
:$\ds \lim_{n \mathop \to \infty} \paren {\map {\mu_1} {E_n} + \map {\mu_2} {E_n} + \map {\mu_3} {E_n} + \map {\mu_4} {E_n} } = 0$ 
From $(1)$ in the Lemma, we have: 
:$\map {\cmod \mu} {E_n} \le \map {\mu_1} {E_n} + \map {\mu_2} {E_n} + \map {\mu_3} {E_n} + \map {\mu_4} {E_n}$ 
From the Squeeze Theorem, we therefore obtain: 
:$\ds \lim_{n \mathop \to \infty} \map {\cmod \mu} {E_n} = 0$
verifying $(3)$. 
So, by Characterization of Measures, we have: 
:$\cmod \mu$ is a measure.
Since $\map {\cmod \mu} A$ is finite for each $A \in \Sigma$, we have that: 
:$\cmod \mu$ is a finite measure.
== Sources ==
* : $4.1$: Signed and Complex Measures: Proposition $4.1.7$
Category:Variation of Complex Measure is Finite Measure
Category:Variation of Complex Measure
Category:Finite Measures
Category:Complex Measures