Additive function

From Maths
Jump to: navigation, search
This page is currently being refactored (along with many others)
Please note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.
(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:
See Halmos' measure theory book too
(Unknown grade)
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:
Needs to include everything the old page did, link to propositions and lead to measures


[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


  1. 1.0 1.1

    TODO: Example on talk page


  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


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"


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]


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.


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]


  • [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]


  2. Halmos - p30 - Measure Theory - Springer - Graduate Texts in Mathematics (18)