Difference between revisions of "Measure"

From Maths
Jump to: navigation, search
m
m (Terminology)
Line 31: Line 31:
  
 
==Terminology==
 
==Terminology==
 +
These terms apply to [[Pre-measure|pre-measures]] to, rather {{M|\mathcal{A} }} you would use the ring the pre-measure is defined on.
 
===Complete measure===
 
===Complete measure===
 
A measure is complete if for {{M|A\in\mathcal{A} }} we have <math>[\mu(A)=0\wedge B\subset A]\implies B\in \mathcal{A}</math>
 
A measure is complete if for {{M|A\in\mathcal{A} }} we have <math>[\mu(A)=0\wedge B\subset A]\implies B\in \mathcal{A}</math>

Revision as of 14:10, 18 March 2015

\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }Not to be confused with Pre-measure


Definition

A \sigma-ring \mathcal{A} and a countably additive, extended real valued. non-negative set function \mu:\mathcal{A}\rightarrow[0,\infty] is a measure.

Contrast with pre-measure

Note: the family A_n must be pairwise disjoint

Property Measure Pre-measure
\mu:\mathcal{A}\rightarrow[0,\infty] \mu_0:R\rightarrow[0,\infty]
\mu(\emptyset)=0 \mu_0(\emptyset)=0
Finitely additive \mu(\bigudot^n_{i=1}A_i)=\sum^n_{i=1}\mu(A_i) \mu_0(\bigudot^n_{i=1}A_i)=\sum^n_{i=1}\mu_0(A_i)
Countably additive \mu(\bigudot^\infty_{n=1}A_n)=\sum^\infty_{n=1}\mu(A_n) If \bigudot^\infty_{n=1}A_n\in R then \mu_0(\bigudot^\infty_{n=1}A_n)=\sum^\infty_{n=1}\mu_0(A_n)

Terminology

These terms apply to pre-measures to, rather \mathcal{A} you would use the ring the pre-measure is defined on.

Complete measure

A measure is complete if for A\in\mathcal{A} we have [\mu(A)=0\wedge B\subset A]\implies B\in \mathcal{A}

Finite

A set A\in\mathcal{A} is finite if \mu(A)<\infty - we say "A has finite measure"

Finite measure

\mu is a finite measure if every set \in\mathcal{A} is finite.

Sigma-finite

A set A\in\mathcal{A} is \sigma-finite if \exists(A_n)_{n=1}^\infty:[A\subseteq\cup^\infty_{n=1}A_n\wedge(\forall A_n,\ \mu(A_n)<\infty)]

Sigma-finite measure

\mu is \sigma-finite if every set \in\mathcal{A} is \sigma-finite

Total

If \mathcal{A} is a \sigma-algebra rather than a ring (that is X\in\mathcal{A} where X is the space) then we use

Totally finite measure

If X is finite

Totally sigma-finite measure

If X is \sigma-finite

Examples

Trivial measures

Given the Measurable space (X,\mathcal{A}) we can define:

\mu:\mathcal{A}\rightarrow\{0,+\infty\} by \mu(A)=\left\{\begin{array}{lr} 0 & \text{if }A=\emptyset \\ +\infty & \text{otherwise} \end{array}\right.

Another trivial measure is:

v:\mathcal{A}\rightarrow\{0\} by v(A)=0 for all A\in\mathcal{A}

See also