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"

Definition

Here $(X,+_X:X\times X\rightarrow Y)$ (which we'll denote $X$ and $+_X$) denotes a space with an additive function assigned.

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

A function 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:

$$x=x+0\implies f(x)=f(x+0)=f(x)+f(0)\implies f(x)=f(x)+0\implies f(0)=0$$ 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)

An example would be a measure

Variations

Finitely additive

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

• $$f(\sum^n_{i=1}A_i)=\sum^n_{i=1}f(A_i)$$ for additive functions
• $$\mu(\bigcup^n_{i=1}A_i)=\sum^n_{i=1}\mu(A_i)$$ for valued set functions

Countably additive

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

• $$f(\sum^\infty_{n=1}A_n)=\sum^\infty_{n=1}f(A_n)$$ for additive functions
• $$\mu(\bigcup^\infty_{n=1}A_n)=\sum^\infty_{n=1}\mu(A_n)$$ 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:

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

Note that a Mathjax bug was found here.