Definition:Measure (Measure Theory)

From ProofWiki
Jump to: navigation, search

Definition

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

Let $\mu: \Sigma \to \overline{\R}$ be a mapping, where $\overline{\R}$ denotes the set of extended real numbers.

Then $\mu$ is called a measure on $\Sigma$ iff it has the following properties:


$(1): \quad$ For every $E \in \Sigma$:
$\mu \left({E}\right) \ge 0$


$(2): \quad$ For every sequence of pairwise disjoint sets $\left\{{S_{n}}\right\} \subseteq \Sigma$:
$\displaystyle \mu \left({\bigcup_{n \mathop = 1}^{\infty} S_n}\right) = \sum_{n \mathop = 1}^{\infty} \mu \left({S_{n}}\right)$
(that is, $\mu\ $ is a countably additive function).


$(3): \quad$ There exists at least one $E \in \Sigma$ such that $\mu \left({E}\right)$ is finite.


Alternative Definition

Alternatively, condition $(3)$ may be replaced by:

$(3'):\quad \mu \left({\varnothing}\right) = 0$

Note that it is assured that $\varnothing \in \Sigma$ by Sigma-Algebra Contains Empty Set.

That the two definitions are equivalent is shown on Equivalence of Definitions of Measure (Measure Theory).


Elementary Consequences

It follows from Measure of Empty Set is Zero that $\mu \left({\varnothing}\right) = 0$.


It then follows from Measure is Finitely Additive Function that $\mu$ is also finitely additive, i.e.:

$\forall E, F \in \Sigma: E \cap F = \varnothing \implies \mu \left({E \cup F}\right) = \mu \left({E}\right) + \mu \left({F}\right)$


Also see


Sources