Characterization of Measures

Theorem
Let $\left({X, \mathcal A}\right)$ be a measurable space.

Denote $\overline{\R}_{\ge0}$ for the set of positive extended real numbers.

A mapping $\mu: \mathcal A \to \overline{\R}_{\ge0}$ is a measure iff:


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

where $A_n \uparrow A$ denotes limit of increasing sequence of sets.

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


 * $(3'):\quad$ For every decreasing sequence $\left({A_n}\right)_{n \in \N}$ in $\mathcal A$ for which $\mu \left({A_1}\right)$ is finite, if $A_n \downarrow A$, then:
 * $\mu \left({A}\right) = \displaystyle \lim_{n \to \infty} \mu \left({A_n}\right)$
 * $(3''):\quad$ For every decreasing sequence $\left({A_n}\right)_{n \in \N}$ in $\mathcal A$ for which $\mu \left({A_1}\right)$ is finite, if $A_n \downarrow \varnothing$, then:
 * $\displaystyle \lim_{n \to \infty} \mu \left({A_n}\right) = 0$

where $A_n \downarrow A$ 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 $\left({A_n}\right)_{n \in \N} \uparrow A$ in $\mathcal A$ be an increasing sequence.

Define $B_1 = A_1$, and, for $n \in \N$, $B_{n+1} = A_{n+1} \setminus A_n$.

Then $B_n \in \mathcal A$ for all $n \in \N$ as $\mathcal A$ is a $\sigma$-algebra.

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

By construction, have for all $k \in \N$ that $A_k = \displaystyle \bigcup_{n=1}^k B_n$, and so $A = \displaystyle \bigcup_{n \in \N} B_n$.

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 $\left({A_n}\right)_{n \in \N} \downarrow A$ be a decreasing sequence in $\mathcal A$.

Suppose that $\mu \left({A_1}\right) < +\infty$.

By Measure is Monotone, this implies $\mu \left({A_n}\right) < +\infty$ for all $n \in \N$, and also $\mu \left({A}\right) < +\infty$.

Now define, for all $n \in \N$, $B_n := A_1 \setminus A_n$.

Then $B_n \uparrow A_1 \setminus A$.

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

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

It follows that $\mu \left({A}\right) = \displaystyle \lim_{n \to \infty} \mu \left({A_n}\right)$ 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 $\left({S_n}\right)_{n \in \N}$ be a sequence of pairwise disjoint sets in $\mathcal A$.

Define, for $n \in \N$, $T_n = \displaystyle \bigcup_{k=1}^n S_k$.

Then for all $n \in \N$, $T_n \subseteq T_{n+1}$.

Also, note that for all $n \in \N$:


 * $\displaystyle \mu \left({T_n}\right) = \mu \left({\bigcup_{k=1}^n S_k}\right) = \sum_{k=1}^n \mu \left({S_k}\right)$

by Additive Function on Union of Sets.

Hence, using condition $(3)$ on the $T_n$, 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 $(3)$.