Definition:Measure (Measure Theory)
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This page is about measures in the context of Measure Theory. For other uses, see Measure.
Definition
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.
Definition 1
$\mu$ is called a measure on $\Sigma$ if and only if $\mu$ fulfils the following axioms:
\((1)\) | $:$ | \(\ds \forall E \in \Sigma:\) | \(\ds \map \mu E \) | \(\ds \ge \) | \(\ds 0 \) | ||||
\((2)\) | $:$ | \(\ds \forall \sequence {S_n}_{n \mathop \in \N} \subseteq \Sigma: \forall i, j \in \N: S_i \cap S_j = \O:\) | \(\ds \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} \) | \(\ds = \) | \(\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} \) | 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 |
Definition 2
$\mu$ is called a measure on $\Sigma$ if and only if $\mu$ fulfils the following axioms:
\((1')\) | $:$ | \(\ds \forall E \in \Sigma:\) | \(\ds \map \mu E \) | \(\ds \ge \) | \(\ds 0 \) | ||||
\((2')\) | $:$ | \(\ds \forall \sequence {S_n}_{n \mathop \in \N} \subseteq \Sigma: \forall i, j \in \N: S_i \cap S_j = \O:\) | \(\ds \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} \) | \(\ds = \) | \(\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} \) | that is, $\mu$ is a countably additive function | |||
\((3')\) | $:$ | \(\ds \map \mu \O \) | \(\ds = \) | \(\ds 0 \) |
Also see
- Measure of Empty Set is Zero: $\map \mu \O = 0$.
- Measure is Finitely Additive Function
- Results about measures can be found here.
Sources
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- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $4.1$