Additive function

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See Halmos' measure theory book too

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Needs to include everything the old page did, link to propositions and lead to measures

Definition

$$\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }$$ $$\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}$$ $$\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }$$ A real valued set function on a class of sets, $\mathcal{A}$, $f:\mathcal{A}\rightarrow\mathbb{R}$ is called additive or finitely additive if:

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

Finitely additive

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

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

Claim 1: $f$ is finitely additive $\implies$ $f$ is additive

Countably additive

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

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

Immediate properties

[Expand]

Claim: if $\emptyset\in\mathcal{A}$ then $f(\emptyset)=0$

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Let $\emptyset,A\in\mathcal{A}$, then:

• $f(A)=f(A\udot\emptyset)=f(A)+f(\emptyset)$ by hypothesis.
• Thus $f(A)=f(A)+f(\emptyset)$
• This means $f(A)-f(A)=f(\emptyset)$

We see $f(\emptyset)=0$, as required

Proof of claims

[Expand]

Claim 1: $f$ is additive $\implies$ $f$ is finitely additive

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Additive function/Claim 1 - additive implies finitely additive

See also

• Subadditive function

References

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

The same goes for $(Y,+_Y:Y\times Y\rightarrow Y)$.

A function $f$ is additive if for $a,b\in X$

$$f(a+_Xb)=f(a)+_Yf(b)$$

Warning about structure

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

$$x=x+0\implies f(x)+0=f(x)=f(x+0)=f(x)+f(0)\implies f(0)=0$$ so one must be careful!

On set functions

A set function, $\mu$, is called additive if whenever:

• $A\in X$
• $B\in X$
• $A\cap B=\emptyset$

We have:

$$\mu(A\cup B)=\mu(A)+\mu(B)$$ for valued set functions (set functions that map to values)

A shorter notation:

$$\mu(A\uplus B)=\mu(A)+\mu(B)$$ , where $\uplus$ 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:

• $$f\Big(\sum^n_{i=1}A_i\Big)=\sum^n_{i=1}f(A_i)$$ for additive functions
• $$\mu\Big(\biguplus^n_{i=1}A_i\Big)=\sum^n_{i=1}\mu(A_i)$$ 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.

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

Countable additivity can imply additivity

If

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

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

Thus:

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