# 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.

Next, let $\sequence {E_n}_{n \mathop \in \N} \uparrow E$ 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 $\sequence {E_n}_{n \mathop \in \N}$ is an increasing sequence.

By construction, have for all $k \in \N$ that:

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

and so:

- $\ds E = \bigcup_{n \mathop \in \N} F_n$

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

\(\ds \map \mu E\) | \(=\) | \(\ds \map \mu {\bigcup_{n \mathop \in \N} F_n}\) | by the above reasoning | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \map \mu {F_n}\) | $\mu$ is a Measure | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{k \mathop \to \infty} \sum_{n \mathop = 1}^k \map \mu {F_n}\) | Definition of Series | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{k \mathop \to \infty} \map \mu {\bigcup_{n \mathop = 1}^k F_n}\) | Measure is Finitely Additive Function, Finite Union of Sets in Additive Function | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{k \mathop \to \infty} \map \mu {E_k}\) | by the above reasoning |

This establishes property $(3)$ for measures.

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:

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

Also, by Additive Function is Strongly Additive:

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

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 $(3)$.

$\blacksquare$

## Sources

- 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $4.4$