# Characterization of Measures

## Theorem

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

Denote $\overline \R_{\ge 0}$ for the set of positive extended real numbers.

A mapping $\mu: \Sigma \to \overline \R_{\ge 0}$ is a measure if and only if:

$(1):\quad \map \mu \O = 0$
$(2):\quad \mu$ is finitely additive
$(3):\quad$ For every increasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$, if $E_n \uparrow E$, then:
$\map \mu E = \ds \lim_{n \mathop \to \infty} \map \mu {E_n}$

where $E_n \uparrow E$ denotes limit of increasing sequence of sets.

Alternatively, and equivalently, $(3)$ may be replaced by either of:

$(3'):\quad$ For every decreasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ for which $\map \mu {E_1}$ is finite, if $E_n \downarrow E$, then:
$\map \mu E = \ds \lim_{n \mathop \to \infty} \map \mu {E_n}$
$(3''):\quad$ For every decreasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ for which $\map \mu {E_1}$ is finite, if $E_n \downarrow \O$, then:
$\ds \lim_{n \mathop \to \infty} \map \mu {E_n} = 0$

where $E_n \downarrow E$ denotes limit of decreasing sequence of sets.

## Proof

### Necessary Condition

To show is that a measure $\mu$ has the properties $(1)$, $(2)$, $(3)$, $(3')$ and $(3'')$.

Property $(1)$ is also part of the definition of measure, and hence is immediate.

Property $(2)$ is precisely the statement of Measure is Finitely Additive Function.

Property $(3)$ is precisely the statement of Measure of Limit of Increasing Sequence of Measurable Sets.

For $(3'')$, note that it is a special case of $(3')$.

For property $(3')$, let $\sequence {E_n}_{n \mathop \in \N} \downarrow E$ be a decreasing sequence in $\Sigma$.

Suppose that $\map \mu {E_1} < +\infty$.

By Measure is Monotone, this implies:

$\forall n \in \N: \map \mu {E_n} < +\infty$

and also:

$\map \mu E < +\infty$

Now define:

$\forall n \in \N: F_n := E_1 \setminus E_n$

Then:

$F_n \uparrow E_1 \setminus E$

Hence, property $(3)$ can be applied as follows:

 $\ds \map \mu {E_1} - \map \mu E$ $=$ $\ds \map \mu {E_1 \setminus E}$ Measure of Set Difference with Subset $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \map \mu {E_1 \setminus E_n}$ by property $(3)$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \paren {\map \mu {E_1} - \map \mu {E_n} }$ Measure of Set Difference with Subset $\ds$ $=$ $\ds \map \mu {E_1} - \lim_{n \mathop \to \infty} \map \mu {E_n}$

Here, all expressions involving subtraction are well-defined as $\mu$ takes finite values.

It follows that:

$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$

as required.

$\Box$

### Sufficient Condition

The mapping $\mu$ is already satisfying axiom $(1)$ for a measure by the imposition on its codomain.

Also, axiom $(3')$ is identical to assumption $(1)$.

It remains to check axiom $(2)$.

So let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.

Define, for $n \in \N$:

$F_n = \ds \bigcup_{k \mathop = 1}^n E_k$

Then:

$\forall n \in \N: F_n \subseteq F_{n + 1}$
$\ds \forall n \in \N: \map \mu {F_n} = \map \mu {\bigcup_{k \mathop = 1}^n E_k} = \sum_{k \mathop = 1}^n \map \mu {E_k}$

Hence, using condition $(3)$ on the $F_n$, obtain:

 $\ds \map \mu {\bigcup_{n \mathop \in \N} E_n}$ $=$ $\ds \lim_{n \mathop \to \infty} \map \mu {F_n}$ Condition $(3)$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \map \mu {E_k}$ by the reasoning above $\ds$ $=$ $\ds \sum_{k \mathop = 1}^\infty \map \mu {E_k}$ Definition of Series

This establishes that $\mu$ also satisfies axiom $(2)$ for a measure, and so it is a measure.

Now to show that $(3')$ and $(3'')$ can validly replace $(3)$.

As $(3')$ clearly implies $(3'')$ (which is a special case of the former), it will suffice to show that $(3'')$ implies $(3)$.

$\blacksquare$