Definition:Measure (Measure Theory)

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Let $\struct {X, \Sigma}$ 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 $\Sigma$ if and only if $\mu$ has the following properties:

\((1)\)   $:$     \(\ds \forall E \in \Sigma:\)    \(\ds \map \mu E \)   \(\ds \ge \)   \(\ds 0 \)             
\((2)\)   $:$     \(\ds \forall \sequence {S_n}_{n \mathop \in \N} \subseteq \Sigma: \forall i, j \in \N: S_i \cap S_j = \O:\)    \(\ds \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} \)   \(\ds = \)   \(\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} \)             that is, $\mu$ is a countably additive function
\((3)\)   $:$     \(\ds \exists E \in \Sigma:\)    \(\ds \map \mu E \)   \(\ds \in \)   \(\ds \R \)             that is, there exists at least one $E \in \Sigma$ such that $\map \mu E$ is finite

Alternative Definition

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

$(3'):\quad \map \mu \O = 0$

Note that it is assured that $\O \in \Sigma$ 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 $\map \mu \O = 0$.

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

$\forall E, F \in \Sigma: E \cap F = \O \implies \map \mu {E \cup F} = \map \mu E + \map \mu F$

Also see