# 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:

$\map {\cmod \mu} \O = 0$

$\Box$

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

We obtain the following lemma:

### Lemma

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

$\Box$

### 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:

 $\ds \bigcup_{i \mathop = 1}^n \paren {A_i \cap B_1}$ $=$ $\ds \paren {\bigcup_{i \mathop = 1}^n A_i} \cap B_1$ Intersection Distributes over Union $\ds$ $=$ $\ds \paren {B_1 \cup B_2} \cap B_1$ $\ds$ $=$ $\ds B_1$ Intersection with Subset is Subset

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:

 $\ds \bigcup_{i \mathop = 1}^n \paren {A_i \cap B_2}$ $=$ $\ds \paren {\bigcup_{i \mathop = 1}^n A_i} \cap B_2$ Intersection Distributes over Union $\ds$ $=$ $\ds \paren {B_1 \cup B_2} \cap B_2$ $\ds$ $=$ $\ds B_2$ Intersection with Subset is Subset

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:

 $\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} }$ $=$ $\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {\paren {B_1 \cup B_2} \cap A_i} }$ Intersection with Subset is Subset $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {\paren {B_1 \cap A_i} \cup \paren {B_2 \cap A_i} } }$ Intersection Distributes over Union $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {B_1 \cap A_i} + \map \mu {B_2 \cap A_i} }$ since $\mu$ is countably additive $\ds$ $\le$ $\ds \sum_{i \mathop = 1}^n \paren {\cmod {\map \mu {B_1 \cap A_i} } + \cmod {\map \mu {B_2 \cap A_i} } }$ Triangle Inequality: Complex Numbers $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {B_1 \cap A_i} } + \sum_{i \mathop = 1}^n \cmod {\map \mu {B_2 \cap A_i} }$ $\ds$ $\le$ $\ds \map {\cmod \mu} {B_1} + \map {\cmod \mu} {B_2}$

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.

$\Box$

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 $(3)$.

$\Box$

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.

$\blacksquare$