Difference between revisions of "Additive function"

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(Warning about structure: correct chain of implications)
(Finitely additive: still less grandiose)
 
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{{Requires references|See Halmos' measure theory book too}}
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{{Stub page|Needs to include everything the old page did, link to propositions and lead to measures}}
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==Definition==
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<!--
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NOTE TO FUTURE EDITORS---------------------------------------------------
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    Right now this defines an additive SET function, if you add an additive function (which way have meaning in say algebra), be sure to update the SET FUNCTION redirects that point into this page
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-->{{Extra Maths}}A ''[[real valued function|real valued]]'' [[set function]] on a class of [[set|sets]], {{M|\mathcal{A} }}, {{M|f:\mathcal{A}\rightarrow\mathbb{R} }} is called ''additive'' or ''finitely additive'' if{{rMT1VIB}}:
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* For {{M|A,B\in\mathcal{A} }} with {{M|1=A\cap B=\emptyset}} ([[pairwise disjoint]]) and {{M|A\udot B\in\mathcal{A} }} we have:
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** {{M|1=f(A\udot B)=f(A)+f(B)}}
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===Finitely additive===
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With the same definition of {{M|f}}, we say that {{M|f}} is ''finitely additive'' if for a [[pairwise disjoint]] family of sets {{M|1=\{A_i\}_{i=1}^n\subseteq\mathcal{A} }} with {{M|1=\bigudot_{i=1}^nA_i\in\mathcal{A} }} we have<ref name="MT1VIB"/>:
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* {{MM|1=f\left(\mathop{\bigudot}_{i=1}^nA_i\right)=\sum^n_{i=1}f(A_i)}}.
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'''Claim 1: ''' {{M|f}} is finitely additive {{M|\implies}} {{M|f}} is additive<ref name="NoStrongerClaim1" group="Note">{{Todo|Example on [[Talk:Additive function|talk page]]}}</ref>
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===Countably additive===
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With the same definition of {{M|f}}, we say that {{M|f}} is ''countably additive'' if for a [[pairwise disjoint]] family of sets {{M|1=\{A_n\}_{n=1}^\infty\subseteq\mathcal{A} }} with {{M|1=\bigudot_{n=1}^\infty A_n\in\mathcal{A} }} we have<ref name="MT1VIB"/>:
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* {{MM|1=f\left(\mathop{\bigudot}_{n=1}^\infty A_n\right)=\sum^\infty_{n=1}f(A_n)}}.
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==Immediate properties==
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{{Begin Inline Theorem}}
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'''Claim: ''' if {{M|\emptyset\in\mathcal{A} }} then {{M|1=f(\emptyset)=0}}
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{{Begin Inline Proof}}
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Let {{M|\emptyset,A\in\mathcal{A} }}, then:
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* {{M|1=f(A)=f(A\udot\emptyset)=f(A)+f(\emptyset)}} by hypothesis.
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* Thus {{M|1=f(A)=f(A)+f(\emptyset)}}
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* This means {{M|1=f(A)-f(A)=f(\emptyset)}}
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We see {{M|1=f(\emptyset)=0}}, as required
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{{End Proof}}{{End Theorem}}
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==Proof of claims==
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{{Begin Inline Theorem}}
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'''[[Additive function/Claim 1 - additive implies finitely additive|Claim 1]]: ''' {{M|f}} is ''additive'' {{M|\implies}} {{M|f}} is ''finitely additive''<ref group="Note" name="NoStrongerClaim1"/>
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{{Begin Inline Proof}}
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{{:Additive function/Claim 1 - additive implies finitely additive}}
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{{End Proof}}{{End Theorem}}
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==See also==
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* [[Subadditive function]]
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==Notes==
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<references group="Note"/>
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==References==
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<references/>
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{{Measure theory navbox|plain}}
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{{Definition|Measure Theory}}
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{{Todo|Check algebra books for definition of additive, perhaps split into two cases, additive set function and additive function}}
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=OLD PAGE=
 
An additive function is a [[Homomorphism|homomorphism]] that preserves the operation of addition in place on the structure in question.
 
An additive function is a [[Homomorphism|homomorphism]] that preserves the operation of addition in place on the structure in question.
  
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<math>\mu(A\cup B)=\mu(A)+\mu(B)</math> for valued set functions (set functions that map to values)
 
<math>\mu(A\cup B)=\mu(A)+\mu(B)</math> for valued set functions (set functions that map to values)
  
An example would be a [[Measure|measure]]
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A shorter notation:
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<math>\mu(A\uplus B)=\mu(A)+\mu(B)</math>, where {{M|\uplus}} denotes "disjoint union" -- just the union when the sets are disjoint, otherwise undefined.
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An example would be a [[Measure|measure]].
  
 
==Variations==
 
==Variations==
 
===Finitely additive===
 
===Finitely additive===
 
This follows by induction on the additive property above. It states that:
 
This follows by induction on the additive property above. It states that:
* <math>f(\sum^n_{i=1}A_i)=\sum^n_{i=1}f(A_i)</math> for additive functions
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* <math>f\Big(\sum^n_{i=1}A_i\Big)=\sum^n_{i=1}f(A_i)</math> for additive functions
* <math>\mu(\bigcup^n_{i=1}A_i)=\sum^n_{i=1}\mu(A_i)</math> for valued set functions
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* <math>\mu\Big(\biguplus^n_{i=1}A_i\Big)=\sum^n_{i=1}\mu(A_i)</math> for valued set functions
  
 
===Countably additive===
 
===Countably additive===
This is a separate property, while given additivity we can get finite additivity we cannot get additivity, we cannot get countable additivity from just additivity.
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This is a separate property, while given additivity we can get finite additivity, but we cannot get countable additivity from just additivity.
  
* <math>f(\sum^\infty_{n=1}A_n)=\sum^\infty_{n=1}f(A_n)</math> for additive functions
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* <math>f\Big(\sum^\infty_{n=1}A_n\Big)=\sum^\infty_{n=1}f(A_n)</math> for additive functions
* <math>\mu(\bigcup^\infty_{n=1}A_n)=\sum^\infty_{n=1}\mu(A_n)</math> for valued set functions
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* <math>\mu\Big(\biguplus^\infty_{n=1}A_n\Big)=\sum^\infty_{n=1}\mu(A_n)</math> for valued set functions
  
 
====Countable additivity can imply additivity====
 
====Countable additivity can imply additivity====

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Definition

[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]A real valued set function on a class of sets, [ilmath]\mathcal{A} [/ilmath], [ilmath]f:\mathcal{A}\rightarrow\mathbb{R} [/ilmath] is called additive or finitely additive if[1]:

  • For [ilmath]A,B\in\mathcal{A} [/ilmath] with [ilmath]A\cap B=\emptyset[/ilmath] (pairwise disjoint) and [ilmath]A\udot B\in\mathcal{A} [/ilmath] we have:
    • [ilmath]f(A\udot B)=f(A)+f(B)[/ilmath]

Finitely additive

With the same definition of [ilmath]f[/ilmath], we say that [ilmath]f[/ilmath] is finitely additive if for a pairwise disjoint family of sets [ilmath]\{A_i\}_{i=1}^n\subseteq\mathcal{A}[/ilmath] with [ilmath]\bigudot_{i=1}^nA_i\in\mathcal{A}[/ilmath] we have[1]:

  • [math]f\left(\mathop{\bigudot}_{i=1}^nA_i\right)=\sum^n_{i=1}f(A_i)[/math].

Claim 1: [ilmath]f[/ilmath] is finitely additive [ilmath]\implies[/ilmath] [ilmath]f[/ilmath] is additive[Note 1]

Countably additive

With the same definition of [ilmath]f[/ilmath], we say that [ilmath]f[/ilmath] is countably additive if for a pairwise disjoint family of sets [ilmath]\{A_n\}_{n=1}^\infty\subseteq\mathcal{A}[/ilmath] with [ilmath]\bigudot_{n=1}^\infty A_n\in\mathcal{A}[/ilmath] we have[1]:

  • [math]f\left(\mathop{\bigudot}_{n=1}^\infty A_n\right)=\sum^\infty_{n=1}f(A_n)[/math].

Immediate properties

Claim: if [ilmath]\emptyset\in\mathcal{A} [/ilmath] then [ilmath]f(\emptyset)=0[/ilmath]


Let [ilmath]\emptyset,A\in\mathcal{A} [/ilmath], then:

  • [ilmath]f(A)=f(A\udot\emptyset)=f(A)+f(\emptyset)[/ilmath] by hypothesis.
  • Thus [ilmath]f(A)=f(A)+f(\emptyset)[/ilmath]
  • This means [ilmath]f(A)-f(A)=f(\emptyset)[/ilmath]

We see [ilmath]f(\emptyset)=0[/ilmath], as required

Proof of claims

Claim 1: [ilmath]f[/ilmath] is additive [ilmath]\implies[/ilmath] [ilmath]f[/ilmath] is finitely additive[Note 1]


See also

Notes

  1. 1.0 1.1

    TODO: Example on talk page


References

  1. 1.0 1.1 1.2 Measure Theory - Volume 1 - V. I. Bogachev



TODO: Check algebra books for definition of additive, perhaps split into two cases, additive set function and additive function



OLD PAGE

An additive function is a homomorphism that preserves the operation of addition in place on the structure in question.

In group theory (because there's only one operation) it is usually just called a "group homomorphism"

Definition

Here [ilmath](X,+_X:X\times X\rightarrow X)[/ilmath] (which we'll denote [ilmath]X[/ilmath] and [ilmath]+_X[/ilmath]) denotes a set endowed with a binary operation called addition.

The same goes for [ilmath](Y,+_Y:Y\times Y\rightarrow Y)[/ilmath].

A function [ilmath]f[/ilmath] is additive[1] if for [ilmath]a,b\in X[/ilmath]

[math]f(a+_Xb)=f(a)+_Yf(b)[/math]

Warning about structure

If the spaces X and Y have some sort of structure (example: Group) then some required properties follow, for example:

[math]x=x+0\implies f(x)+0=f(x)=f(x+0)=f(x)+f(0)\implies f(0)=0[/math] so one must be careful!

On set functions

A set function, [ilmath]\mu[/ilmath], is called additive if[2] whenever:

  • [ilmath]A\in X[/ilmath]
  • [ilmath]B\in X[/ilmath]
  • [ilmath]A\cap B=\emptyset[/ilmath]

We have:

[math]\mu(A\cup B)=\mu(A)+\mu(B)[/math] for valued set functions (set functions that map to values)

A shorter notation: [math]\mu(A\uplus B)=\mu(A)+\mu(B)[/math], where [ilmath]\uplus[/ilmath] denotes "disjoint union" -- just the union when the sets are disjoint, otherwise undefined.

An example would be a measure.

Variations

Finitely additive

This follows by induction on the additive property above. It states that:

  • [math]f\Big(\sum^n_{i=1}A_i\Big)=\sum^n_{i=1}f(A_i)[/math] for additive functions
  • [math]\mu\Big(\biguplus^n_{i=1}A_i\Big)=\sum^n_{i=1}\mu(A_i)[/math] for valued set functions

Countably additive

This is a separate property, while given additivity we can get finite additivity, but we cannot get countable additivity from just additivity.

  • [math]f\Big(\sum^\infty_{n=1}A_n\Big)=\sum^\infty_{n=1}f(A_n)[/math] for additive functions
  • [math]\mu\Big(\biguplus^\infty_{n=1}A_n\Big)=\sum^\infty_{n=1}\mu(A_n)[/math] for valued set functions

Countable additivity can imply additivity

If [math]f(0)=0[/math] or [math]\mu(\emptyset)=0[/math] then given a finite set [math]\{a_i\}_{i=1}^n[/math] we can define an infinite set [math]\{b_n\}_{n=1}^\infty[/math] by:

[math]b_i=\left\{\begin{array}a_i&\text{if }i\le n\\ 0\text{ or }\emptyset & \text{otherwise}\end{array}\right.[/math]

Thus:

  • [math]f(\sum^\infty_{n=1}b_n)= \begin{array}{lr} f(\sum^n_{i=1}a_i) \\ \sum^\infty_{n=1}f(b_n)=\sum^n_{i=1}f(a_i)+f(0)=\sum^n_{i=1}f(a_i) \end{array}[/math]
  • Or indeed [math]\mu(\sum^\infty_{n=1}b_n)= \begin{array}{lr} \mu(\sum^n_{i=1}a_i) \\ \sum^\infty_{n=1}\mu(b_n)=\sum^n_{i=1}\mu(a_i)+\mu(0)=\sum^n_{i=1}\mu(a_i) \end{array}[/math]

References

  1. http://en.wikipedia.org/w/index.php?title=Additive_function&oldid=630245379
  2. Halmos - p30 - Measure Theory - Springer - Graduate Texts in Mathematics (18)