Characterization of Measures

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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 if and only if:

$(1):\quad \mu \left({\varnothing}\right) = 0$
$(2):\quad \mu$ 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.


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 $\left({E_n}\right)_{n \mathop \in \N} \uparrow E$ 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:

\(\displaystyle \mu \left({E}\right)\) \(=\) \(\displaystyle \mu \left({\bigcup_{n \mathop \in \N} F_n}\right)\) by the above reasoning
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \mu \left({F_n}\right)\) $\mu$ is a Measure
\(\displaystyle \) \(=\) \(\displaystyle \lim_{k \mathop \to \infty} \sum_{n \mathop = 1}^k \mu \left({F_n}\right)\) Definition of Series
\(\displaystyle \) \(=\) \(\displaystyle \lim_{k \mathop \to \infty} \mu \left({\bigcup_{n \mathop = 1}^k F_n}\right)\) Measure is Finitely Additive Function, Finite Union of Sets in Additive Function
\(\displaystyle \) \(=\) \(\displaystyle \lim_{k \mathop \to \infty} \mu \left({E_k}\right)\) 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 $\left({E_n}\right)_{n \mathop \in \N} \downarrow E$ 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:

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

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.

$\Box$


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 $\left({E_n}\right)_{n \mathop \in \N}$ 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:

\(\displaystyle \mu \left({\bigcup_{n \mathop \in \N} E_n}\right)\) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \mu \left({F_n}\right)\) Condition $(3)$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \mu \left({E_k}\right)\) by the reasoning above
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^\infty \mu \left({E_k}\right)\) 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 $(3)$.


$\blacksquare$


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