Difference between revisions of "Measure"
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| − | Not to be confused with [[Pre-measure]] | + | {{Extra Maths}}Not to be confused with [[Pre-measure]] |
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| Finitely additive | | Finitely additive | ||
| − | | <math>\mu(\ | + | | <math>\mu(\bigudot^n_{i=1}A_i)=\sum^n_{i=1}\mu(A_i)</math> |
| − | | <math>\mu_0(\ | + | | <math>\mu_0(\bigudot^n_{i=1}A_i)=\sum^n_{i=1}\mu_0(A_i)</math> |
|- | |- | ||
| Countably additive | | Countably additive | ||
| − | | <math>\mu(\ | + | | <math>\mu(\bigudot^\infty_{n=1}A_n)=\sum^\infty_{n=1}\mu(A_n)</math> |
| − | | If <math>\ | + | | If <math>\bigudot^\infty_{n=1}A_n\in R</math> then <math>\mu_0(\bigudot^\infty_{n=1}A_n)=\sum^\infty_{n=1}\mu_0(A_n)</math> |
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{{Definition|Measure Theory}} | {{Definition|Measure Theory}} | ||
Revision as of 22:32, 13 March 2015
[math]\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}} }[/math][math]\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}[/math][math]\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }[/math]Not to be confused with Pre-measure
Definition
A [ilmath]\sigma[/ilmath]-ring [ilmath]\mathcal{A} [/ilmath] and a countably additive, extended real valued. non-negative set function [math]\mu:\mathcal{A}\rightarrow[0,\infty][/math] is a measure.
Contrast with pre-measure
Note: the family [math]A_n[/math] must be pairwise disjoint
| Property | Measure | Pre-measure |
|---|---|---|
| [math]\mu:\mathcal{A}\rightarrow[0,\infty][/math] | [math]\mu_0:R\rightarrow[0,\infty][/math] | |
| [math]\mu(\emptyset)=0[/math] | [math]\mu_0(\emptyset)=0[/math] | |
| Finitely additive | [math]\mu(\bigudot^n_{i=1}A_i)=\sum^n_{i=1}\mu(A_i)[/math] | [math]\mu_0(\bigudot^n_{i=1}A_i)=\sum^n_{i=1}\mu_0(A_i)[/math] |
| Countably additive | [math]\mu(\bigudot^\infty_{n=1}A_n)=\sum^\infty_{n=1}\mu(A_n)[/math] | If [math]\bigudot^\infty_{n=1}A_n\in R[/math] then [math]\mu_0(\bigudot^\infty_{n=1}A_n)=\sum^\infty_{n=1}\mu_0(A_n)[/math] |
