Characterization of Measures

Theorem
Let $\left({X, \Sigma}\right)$ 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 :


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

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

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

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.

Next, let $\left({E_n}\right)_{n \mathop \in \N} \uparrow E$ in $\Sigma$ be an increasing sequence.

Define $F_1 = E_1$, and, for $n \in \N$:
 * $F_{n+1} = E_{n+1} \setminus E_n$

Then as $\Sigma$ is a $\sigma$-algebra:
 * $\forall n \in \N: F_n \in \Sigma$

Also, the $F_n$ are pairwise disjoint as $\left({E_n}\right)_{n \mathop \in \N}$ is an increasing sequence.

By construction, have for all $k \in \N$ that:
 * $\displaystyle E_k = \bigcup_{n \mathop = 1}^k F_n$

and so:
 * $\displaystyle E = \bigcup_{n \mathop \in \N} F_n$

Hence, as $\mu$ is a measure, compute:

This establishes property $(3)$ for measures.

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

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

Suppose that $\mu \left({E_1}\right) < +\infty$.

By Measure is Monotone, this implies:
 * $\forall n \in \N: \mu \left({E_n}\right) < +\infty$

and also:
 * $\mu \left({E}\right) < +\infty$

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:

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

It follows that:
 * $\displaystyle \mu \left({E}\right) = \lim_{n \mathop \to \infty} \mu \left({E_n}\right)$

as required.

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 $\left({E_n}\right)_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.

Define, for $n \in \N$:
 * $F_n = \displaystyle \bigcup_{k \mathop = 1}^n E_k$

Then:
 * $\forall n \in \N: F_n \subseteq F_{n+1}$

Also, by Additive Function is Strongly Additive:


 * $\displaystyle \forall n \in \N: \mu \left({F_n}\right) = \mu \left({\bigcup_{k \mathop = 1}^n E_k}\right) = \sum_{k \mathop = 1}^n \mu \left({E_k}\right)$

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

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 $(3)$.