Difference between revisions of "Additive function"

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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"/>:
 
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"/>:
 
* {{MM|1=f\left(\mathop{\bigudot}_{i=1}^nA_i\right)=\sum^n_{i=1}f(A_i)}}.
 
* {{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|\iff}} it is additive
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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===
 
===Countably additive===
 
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"/>:
 
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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Let {{M|\emptyset,A\in\mathcal{A} }}, then:
 
Let {{M|\emptyset,A\in\mathcal{A} }}, then:
 
* {{M|1=f(A)=f(A\udot\emptyset)=f(A)+f(\emptyset)}} by hypothesis.  
 
* {{M|1=f(A)=f(A\udot\emptyset)=f(A)+f(\emptyset)}} by hypothesis.  
−
* Thus {{M|1=f(A)=f(A)+f(\emptyset)}]
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* Thus {{M|1=f(A)=f(A)+f(\emptyset)}}
 
* This means {{M|1=f(A)-f(A)=f(\emptyset)}}
 
* This means {{M|1=f(A)-f(A)=f(\emptyset)}}
 
We see {{M|1=f(\emptyset)=0}}, as required
 
We see {{M|1=f(\emptyset)=0}}, as required
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==Proof of claims==
 
==Proof of claims==
 
{{Begin Inline Theorem}}
 
{{Begin Inline Theorem}}
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'''[[Additive function/Claim 1 - additive iff finitely additive|Claim 1]]: ''' {{M|f}} is ''additive'' {{iff}} {{M|f}} is ''finitely additive''
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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"/>
 
{{Begin Inline Proof}}
 
{{Begin Inline Proof}}
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{{:Additive function/Claim 1 - additive iff finitely additive}}
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{{:Additive function/Claim 1 - additive implies finitely additive}}
 
{{End Proof}}{{End Theorem}}
 
{{End Proof}}{{End Theorem}}
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==See also==
 
==See also==
 
* [[Subadditive function]]
 
* [[Subadditive function]]

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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]

Additive function/Claim 1 - additive implies finitely additive


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

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


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)
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