Difference between revisions of "Measure"

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Not to be confused with [[Pre-measure]]
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{{Stub page|Requires further expansion|grade=A}}{{Extra Maths}}{{:Measure/Infobox}}
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__TOC__
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==Definition==
 
==Definition==
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A [[Sigma-ring|{{sigma|ring}}]] {{M|\mathcal{A} }} and a countably [[Additive function|additive]], [[Extended real value|extended real valued]]. non-negative [[Set function|set]] [[Function|function]] <math>\mu:\mathcal{A}\rightarrow[0,\infty]</math> is a measure.
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A (positive) ''measure'', {{M|\mu}} is a [[set function]] from a [[sigma-ring|{{sigma|ring}}]], {{M|\mathcal{R} }}, to the positive [[extended real values]]<ref group="Note">Recall {{M|\bar{\mathbb{R} }_{\ge0} }} is {{M|\mathbb{R}_{\ge0}\cup\{+\infty\} }}</ref>, {{M|\bar{\mathbb{R} }_{\ge 0} }}{{rMTH}}{{rMIAMRLS}}{{rMT1VIB}}:
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* {{M|\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge0} }}
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===Contrast with pre-measure===
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Such that:
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'''Note:''' the family <math>A_n</math> must be pairwise disjoint
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* {{M|1=\forall(A_n)_{n=1}^\infty\subseteq\mathcal{R}\text{ pairwise disjoint }[\mu\left(\bigudot_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty\mu(A_n)]}} ({{M|\mu}} is a [[countably additive set function]])
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** Recall that "''pairwise disjoint''" means {{M|1=\forall i,j\in\mathbb{N}[i\ne j\implies A_i\cap A_j=\emptyset]}}
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Entirely in words a (positive) ''measure'', {{M|\mu}} is:
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* An ''[[extended real valued]]'' [[countably additive set function]] from a [[sigma-ring|{{sigma|ring}}]], {{M|\mathcal{R} }}; {{M|\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} } }}.
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{{Note|Remember that every [[sigma-algebra|{{sigma|algebra}}]] is a {{sigma|ring}}, so this definition can be applied directly (and should be in the reader's mind) to {{sigma|algebras}}}}
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==Terminology==
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===For a set===
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We may say a set {{M|A\in\mathcal{R} }} (for a [[sigma-ring|{{sigma|ring}}]] {{M|\mathcal{R} }}) is:
 
{| class="wikitable" border="1"
 
{| class="wikitable" border="1"
 
|-
 
|-
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! Property
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! Term
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! Measure
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! Meaning
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! Pre-measure
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! Example
 
|-
 
|-
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! Finite<ref name="MTH"/>
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| if {{M|\mu(A)<\infty }}
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|
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* {{M|A}} is ''finite''
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* {{M|A}} is of ''finite measure''
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|-
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! {{sigma|finite}}<ref name="MTH"/>
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| if {{M|1=\exists(A_n)_{n=1}^\infty\subseteq\mathcal{R}\forall i\in\mathbb{N}[A\subseteq\bigcup_{n=1}^\infty A_n\wedge \mu(A_i)<\infty]}}<br/>
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* In words: if there exists a sequence of sets in {{M|\mathcal{R} }} such that {{M|A}} is in their union and each set has finite measure.
 
|
 
|
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| <math>\mu:\mathcal{A}\rightarrow[0,\infty]</math>
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* {{M|A}} is ''{{sigma|finite}}''
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| <math>\mu_0:R\rightarrow[0,\infty]</math>
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* {{M|A}} is of ''{{sigma|finite}} measure''
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|}
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===Of a measure===
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We may say a measure, {{M|\mu}} is:
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{| class="wikitable" border="1"
 
|-
 
|-
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! Term
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! Meaning
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! Example
 +
|-
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! Finite<ref name="MTH"/>
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| If every set in the {{sigma|ring}} the measure is defined on ''is of finite measure''
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* Symbolically, if: {{M|1=\forall A\in\mathcal{R}[\mu(A)<\infty]}}
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|
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*{{M|\mu}} is a finite measure
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|-
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! {{sigma|finite}}<ref name="MTH"/>
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| If every set in the {{sigma|ring}} the measure is defined on ''is of {{sigma|finite}} measure''
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* Symbolically, if: {{M|1=\forall A\in\mathcal{R}\exists(A_n)_{n=1}^\infty\subseteq\mathcal{R}\forall i\in\mathbb{N}[A\subseteq\bigcup_{n=1}^\infty A_n\wedge \mu(A_i)<\infty]}}
 
|
 
|
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| <math>\mu(\emptyset)=0</math>
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* {{M|\mu}} is a {{sigma|finite}} measure
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| <math>\mu_0(\emptyset)=0</math>
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|-
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! Complete
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| if {{M|1=\forall A\in\mathcal{R}\forall B\in\mathcal{P}(A)[(\mu(A)=0)\implies(B\in\mathcal{R})]}}
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* In words: for every set of measure 0 in {{M|\mathcal{R} }} every subset of that set is also in {{M|\mathcal{R} }}
 +
|
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*{{M|\mu}} is a complete measure
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|}
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====Of a measure on a {{sigma|algebra}}====
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If {{M|\mu:\mathcal{A}\rightarrow\bar{\mathbb{R} }_{\ge0} }} for a [[sigma-algebra|{{sigma|algebra}}]] {{M|\mathcal{A} }}<ref group="Note">Remember a sigma-algebra is just a sigma-ring containing the entire space.</ref> then we can define:
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{| class="wikitable" border="1"
 
|-
 
|-
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| Finitely additive
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! Term
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| <math>\mu(\bigcup^n_{i=1}A_i)=\sum^n_{i=1}\mu(A_i)</math>
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! Meaning
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| <math>\mu_0(\bigcup^n_{i=1}A_i)=\sum^n_{i=1}\mu_0(A_i)</math>
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! Example
 
|-
 
|-
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| Countably additive
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! Totally finite<ref name="MTH"/>
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| <math>\mu(\bigcup^\infty_{n=1}A_n)=\sum^\infty_{n=1}\mu(A_n)</math>
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| if the measure of {{M|X}} is finite
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| If <math>\bigcup^\infty_{n=1}A_n\in R</math> then <math>\mu_0(\bigcup^\infty_{n=1}A_n)=\sum^\infty_{n=1}\mu_0(A_n)</math>
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* Symbolically, if {{M|\mu(X)<\infty}}
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|
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* {{M|\mu}} is totally finite
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|-
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! Totally {{sigma|finite}}<ref name="MTH"/>
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| if {{M|X}} is of {{sigma|finite}} measure
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* Symbolically, if: {{M|1=\exists(A_n)_{n=1}^\infty\subseteq\mathcal{R}\forall i\in\mathbb{N}[X=\bigcup_{n=1}^\infty A_n\wedge \mu(A_i)<\infty]}}
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|
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* {{M|\mu}} is totally {{sigma|finite}}
 
|}
 
|}
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==Immediate properties==
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{{Requires proof|Trivial|easy=Yes|grade=C}}
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{{Begin Inline Theorem}}
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'''Claim: ''' {{M|1=\mu(\emptyset)=0}}
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{{Begin Inline Proof}}
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PUT PROOF HERE
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{{End Proof}}{{End Theorem}}
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==Properties==
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{{Todo|Countable subadditivity and so forth}}
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===In common with a [[pre-measure]]===
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{{:Pre-measure/Properties in common with measure}}
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==Related theorems==
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* [[A function is a measure iff it measures the empty set as 0, disjoint sets add, and it is continuous from below (with equiv. conditions)]]
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==Examples==
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* [[Dirac measure]]
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* [[Counting measure]]
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* [[Discrete probability measure]]
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* [[Lebesgue measure]]
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===Trivial measures===
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Here {{M|\mathcal{R} }} is a [[sigma-ring|{{sigma|ring}}]]<ref group="Note">Remember every {{sigma|algebra}} is a {{sigma|ring}}, so {{M|\mathcal{R} }} could just as well be a {{sigma|algebra}}</ref>
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# <math>\mu:\mathcal{R}\rightarrow\{0,+\infty\}</math> by <math>\mu(A)=\left\{\begin{array}{lr}
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0 & \text{if }A=\emptyset \\
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+\infty & \text{otherwise}
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\end{array}\right.</math>
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#* Note that if we'd chosen a finite and non-zero value instead of {{M|+\infty}} it ''would not'' be a measure<ref group="Note">Unless {{M|\mathcal{R} }} was a ''trivial {{sigma|algebra}}'' consisting of the empty set and another set. </ref>, as take any non-empty {{M|A,B\in\mathcal{R} }} with {{M|1=A\cap B=\emptyset}}, for a measure we would have:
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#** {{M|1=\mu(A\cup B)=\mu(A)+\mu(B) }}, which will yield {{M|1=v=2v\implies v=0}} contradicting that {{M|\mu}} maps non-empty sets to finite non-zero values
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# <math>\mu:\mathcal{R}\rightarrow\{0\}</math> by <math>\mu:A\mapsto 0</math> is ''the'' trivial measure.
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{{Requires references|That this is the trivial measure}}
  
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==See also==
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* [[Pre-measure]]
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* [[Outer-measure]]
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* [[Constructing a measure from a pre-measure]]
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* [[Measurable space]]
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* [[Measure space]]
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==Notes==
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<references group="Note"/>
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==References==
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'''Note: ''' Inline with the [[Measure theory terminology doctrine]] the references do not define a ''measure'' exactly as such, only an object that fits the place we have named ''measure''. This sounds like a huge discrepancy but as is detailed on that page, it isn't.
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<references/>
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{{Measure theory navbox|plain}}
 
{{Definition|Measure Theory}}
 
{{Definition|Measure Theory}}

Latest revision as of 14:39, 16 August 2016

Stub grade: A
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Requires further expansion
[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]
(Positive) Measure
[ilmath]\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge0} [/ilmath]
For a [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath]
Properties
[ilmath]\forall\overbrace{(A_n)_{n=1}^\infty }^{\begin{array}{c}\text{pairwise}\\\text{disjoint}\end{array} }\subseteq\mathcal{R}[\mu\left(\bigudot_{n=1}^\infty A_n\right)=\sum^\infty_{n=1}\mu(A_n)][/ilmath]

Definition

A (positive) measure, [ilmath]\mu[/ilmath] is a set function from a [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath], to the positive extended real values[Note 1], [ilmath]\bar{\mathbb{R} }_{\ge 0} [/ilmath][1][2][3]:

  • [ilmath]\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge0} [/ilmath]

Such that:

  • [ilmath]\forall(A_n)_{n=1}^\infty\subseteq\mathcal{R}\text{ pairwise disjoint }[\mu\left(\bigudot_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty\mu(A_n)][/ilmath] ([ilmath]\mu[/ilmath] is a countably additive set function)
    • Recall that "pairwise disjoint" means [ilmath]\forall i,j\in\mathbb{N}[i\ne j\implies A_i\cap A_j=\emptyset][/ilmath]

Entirely in words a (positive) measure, [ilmath]\mu[/ilmath] is:

Remember that every [ilmath]\sigma[/ilmath]-algebra is a [ilmath]\sigma[/ilmath]-ring, so this definition can be applied directly (and should be in the reader's mind) to [ilmath]\sigma[/ilmath]-algebras

Terminology

For a set

We may say a set [ilmath]A\in\mathcal{R} [/ilmath] (for a [ilmath]\sigma[/ilmath]-ring [ilmath]\mathcal{R} [/ilmath]) is:

Term Meaning Example
Finite[1] if [ilmath]\mu(A)<\infty [/ilmath]
  • [ilmath]A[/ilmath] is finite
  • [ilmath]A[/ilmath] is of finite measure
[ilmath]\sigma[/ilmath]-finite[1] if [ilmath]\exists(A_n)_{n=1}^\infty\subseteq\mathcal{R}\forall i\in\mathbb{N}[A\subseteq\bigcup_{n=1}^\infty A_n\wedge \mu(A_i)<\infty][/ilmath]
  • In words: if there exists a sequence of sets in [ilmath]\mathcal{R} [/ilmath] such that [ilmath]A[/ilmath] is in their union and each set has finite measure.
  • [ilmath]A[/ilmath] is [ilmath]\sigma[/ilmath]-finite
  • [ilmath]A[/ilmath] is of [ilmath]\sigma[/ilmath]-finite measure

Of a measure

We may say a measure, [ilmath]\mu[/ilmath] is:

Term Meaning Example
Finite[1] If every set in the [ilmath]\sigma[/ilmath]-ring the measure is defined on is of finite measure
  • Symbolically, if: [ilmath]\forall A\in\mathcal{R}[\mu(A)<\infty][/ilmath]
  • [ilmath]\mu[/ilmath] is a finite measure
[ilmath]\sigma[/ilmath]-finite[1] If every set in the [ilmath]\sigma[/ilmath]-ring the measure is defined on is of [ilmath]\sigma[/ilmath]-finite measure
  • Symbolically, if: [ilmath]\forall A\in\mathcal{R}\exists(A_n)_{n=1}^\infty\subseteq\mathcal{R}\forall i\in\mathbb{N}[A\subseteq\bigcup_{n=1}^\infty A_n\wedge \mu(A_i)<\infty][/ilmath]
  • [ilmath]\mu[/ilmath] is a [ilmath]\sigma[/ilmath]-finite measure
Complete if [ilmath]\forall A\in\mathcal{R}\forall B\in\mathcal{P}(A)[(\mu(A)=0)\implies(B\in\mathcal{R})][/ilmath]
  • In words: for every set of measure 0 in [ilmath]\mathcal{R} [/ilmath] every subset of that set is also in [ilmath]\mathcal{R} [/ilmath]
  • [ilmath]\mu[/ilmath] is a complete measure

Of a measure on a [ilmath]\sigma[/ilmath]-algebra

If [ilmath]\mu:\mathcal{A}\rightarrow\bar{\mathbb{R} }_{\ge0} [/ilmath] for a [ilmath]\sigma[/ilmath]-algebra [ilmath]\mathcal{A} [/ilmath][Note 2] then we can define:

Term Meaning Example
Totally finite[1] if the measure of [ilmath]X[/ilmath] is finite
  • Symbolically, if [ilmath]\mu(X)<\infty[/ilmath]
  • [ilmath]\mu[/ilmath] is totally finite
Totally [ilmath]\sigma[/ilmath]-finite[1] if [ilmath]X[/ilmath] is of [ilmath]\sigma[/ilmath]-finite measure
  • Symbolically, if: [ilmath]\exists(A_n)_{n=1}^\infty\subseteq\mathcal{R}\forall i\in\mathbb{N}[X=\bigcup_{n=1}^\infty A_n\wedge \mu(A_i)<\infty][/ilmath]
  • [ilmath]\mu[/ilmath] is totally [ilmath]\sigma[/ilmath]-finite

Immediate properties

Grade: C
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
Trivial

This proof has been marked as an page requiring an easy proof

Claim: [ilmath]\mu(\emptyset)=0[/ilmath]

PUT PROOF HERE


Properties

TODO: Countable subadditivity and so forth

In common with a pre-measure

  • Finitely additive: if [ilmath]A\cap B=\emptyset[/ilmath] then [ilmath]\mu_0(A\udot B)=\mu_0(A)+\mu_0(B)[/ilmath]

Follows immediately from definition (property 2)

  • Monotonic: [Note 3] if [ilmath]A\subseteq B[/ilmath] then [ilmath]\mu_0(A)\le\mu_0(B)[/ilmath]


TODO: Be bothered to write out

  • If [ilmath]A\subseteq B[/ilmath] and [ilmath]\mu_0(A)<\infty[/ilmath] then [ilmath]\mu_0(B-A)=\mu_0(B)-\mu(A)[/ilmath]


TODO: Be bothered, note the significance of the finite-ness of [ilmath]A[/ilmath] - see Extended real value

  • Strongly additive: [ilmath]\mu_0(A\cup B)=\mu_0(A)+\mu_0(B)-\mu_0(A\cap B)[/ilmath]


TODO: Be bothered

  • Subadditive: [ilmath]\mu_0(A\cup B)\le\mu_0(A)+\mu_0(B)[/ilmath]


TODO: Again - be bothered

Related theorems

Examples

Trivial measures

Here [ilmath]\mathcal{R} [/ilmath] is a [ilmath]\sigma[/ilmath]-ring[Note 4]

  1. [math]\mu:\mathcal{R}\rightarrow\{0,+\infty\}[/math] by [math]\mu(A)=\left\{\begin{array}{lr} 0 & \text{if }A=\emptyset \\ +\infty & \text{otherwise} \end{array}\right.[/math]
    • Note that if we'd chosen a finite and non-zero value instead of [ilmath]+\infty[/ilmath] it would not be a measure[Note 5], as take any non-empty [ilmath]A,B\in\mathcal{R} [/ilmath] with [ilmath]A\cap B=\emptyset[/ilmath], for a measure we would have:
      • [ilmath]\mu(A\cup B)=\mu(A)+\mu(B)[/ilmath], which will yield [ilmath]v=2v\implies v=0[/ilmath] contradicting that [ilmath]\mu[/ilmath] maps non-empty sets to finite non-zero values
  2. [math]\mu:\mathcal{R}\rightarrow\{0\}[/math] by [math]\mu:A\mapsto 0[/math] is the trivial measure.
(Unknown grade)
This page requires references, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
The message provided is:
That this is the trivial measure

See also

Notes

  1. ↑ Recall [ilmath]\bar{\mathbb{R} }_{\ge0} [/ilmath] is [ilmath]\mathbb{R}_{\ge0}\cup\{+\infty\} [/ilmath]
  2. ↑ Remember a sigma-algebra is just a sigma-ring containing the entire space.
  3. ↑ Sometimes stated as monotone (it is monotone in Measures, Integrals and Martingales in fact!)
  4. ↑ Remember every [ilmath]\sigma[/ilmath]-algebra is a [ilmath]\sigma[/ilmath]-ring, so [ilmath]\mathcal{R} [/ilmath] could just as well be a [ilmath]\sigma[/ilmath]-algebra
  5. ↑ Unless [ilmath]\mathcal{R} [/ilmath] was a trivial [ilmath]\sigma[/ilmath]-algebra consisting of the empty set and another set.

References

Note: Inline with the Measure theory terminology doctrine the references do not define a measure exactly as such, only an object that fits the place we have named measure. This sounds like a huge discrepancy but as is detailed on that page, it isn't.

  1. ↑ 1.0 1.1 1.2 1.3 1.4 1.5 1.6 Measure Theory - Paul R. Halmos
  2. ↑ Measures, Integrals and Martingales - René L. Schilling
  3. ↑ Measure Theory - Volume 1 - V. I. Bogachev

v • d • e
 
Measure Theory
Overview of the basic and important objects studied in measure theory
Set-structures
Measures
Pre-measure [ilmath]\leadsto[/ilmath] outer-measure [ilmath]\leadsto[/ilmath] measure (see Extending pre-measures to measures (via: Extending pre-measures to outer-measures [ilmath]\rightarrow[/ilmath] "Restricting outer-measures to measures" maybe?). Alternatively: inner-measure can be used.
measure space
Functions
Spaces
Integrals
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