Sigma-additive_set_function Knowpia

In mathematics, an additive set function is a function mapping sets to numbers, with the property that its value on a union of two disjoint sets equals the sum of its values on these sets, namely, ${\textstyle \mu (A\cup B)=\mu (A)+\mu (B).}$ If this additivity property holds for any two sets, then it also holds for any finite number of sets, namely, the function value on the union of k disjoint sets (where k is a finite number) equals the sum of its values on the sets. Therefore, an additive set function is also called a finitely additive set function (the terms are equivalent). However, a finitely additive set function might not have the additivity property for a union of an infinite number of sets. A σ-additive set function is a function that has the additivity property even for countably infinite many sets, that is, ${\textstyle \mu \left(\bigcup _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n}).}$

Additivity and sigma-additivity are particularly important properties of measures. They are abstractions of how intuitive properties of size (length, area, volume) of a set sum when considering multiple objects. Additivity is a weaker condition than σ-additivity; that is, σ-additivity implies additivity.

The term modular set function is equivalent to additive set function; see modularity below.

Additive (or finitely additive) set functions edit

Let $\mu$ be a set function defined on an algebra of sets $\scriptstyle {\mathcal {A}}$ with values in $[-\infty ,\infty ]$ (see the extended real number line). The function $\mu$ is called additive or finitely additive, if whenever $A$ and $B$ are disjoint sets in $\scriptstyle {\mathcal {A}},$ then

\mu (A\cup B)=\mu (A)+\mu (B).

A consequence of this is that an additive function cannot take both

-\infty

and

+\infty

as values, for the expression

\infty -\infty

is undefined.

One can prove by mathematical induction that an additive function satisfies

\mu \left(\bigcup _{n=1}^{N}A_{n}\right)=\sum _{n=1}^{N}\mu \left(A_{n}\right)

for any

A_{1},A_{2},\ldots ,A_{N}

disjoint sets in

{\textstyle {\mathcal {A}}.}

σ-additive set functions edit

Suppose that $\scriptstyle {\mathcal {A}}$ is a σ-algebra. If for every sequence $A_{1},A_{2},\ldots ,A_{n},\ldots$ of pairwise disjoint sets in $\scriptstyle {\mathcal {A}},$

\mu \left(\bigcup _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n}),

holds then

\mu

is said to be countably additive or 𝜎-additive. Every 𝜎-additive function is additive but not vice versa, as shown below.

τ-additive set functions edit

Suppose that in addition to a sigma algebra ${\textstyle {\mathcal {A}},}$ we have a topology $\tau .$ If for every directed family of measurable open sets ${\textstyle {\mathcal {G}}\subseteq {\mathcal {A}}\cap \tau ,}$

\mu \left(\bigcup {\mathcal {G}}\right)=\sup _{G\in {\mathcal {G}}}\mu (G),

we say that

\mu

is

\tau

-additive. In particular, if

\mu

is inner regular (with respect to compact sets) then it is τ-additive.^[1]

Properties edit

Useful properties of an additive set function $\mu$ include the following.

Value of empty set edit

Either $\mu (\varnothing )=0,$ or $\mu$ assigns $\infty$ to all sets in its domain, or $\mu$ assigns $-\infty$ to all sets in its domain. Proof: additivity implies that for every set $A,$ $\mu (A)=\mu (A\cup \varnothing )=\mu (A)+\mu (\varnothing ).$ If $\mu (\varnothing )\neq 0,$ then this equality can be satisfied only by plus or minus infinity.

Monotonicity edit

If $\mu$ is non-negative and $A\subseteq B$ then $\mu (A)\leq \mu (B).$ That is, $\mu$ is a monotone set function. Similarly, If $\mu$ is non-positive and $A\subseteq B$ then $\mu (A)\geq \mu (B).$

Modularity edit

A set function $\mu$ on a family of sets ${\mathcal {S}}$ is called a modular set function and a valuation if whenever $A,$ $B,$ $A\cup B,$ and $A\cap B$ are elements of ${\mathcal {S}},$ then

\phi (A\cup B)+\phi (A\cap B)=\phi (A)+\phi (B)

The above property is called modularity and the argument below proves that additivity implies modularity.

Given $A$ and $B,$ $\mu (A\cup B)+\mu (A\cap B)=\mu (A)+\mu (B).$ Proof: write $A=(A\cap B)\cup (A\setminus B)$ and $B=(A\cap B)\cup (B\setminus A)$ and $A\cup B=(A\cap B)\cup (A\setminus B)\cup (B\setminus A),$ where all sets in the union are disjoint. Additivity implies that both sides of the equality equal $\mu (A\setminus B)+\mu (B\setminus A)+2\mu (A\cap B).$

However, the related properties of submodularity and subadditivity are not equivalent to each other.

Note that modularity has a different and unrelated meaning in the context of complex functions; see modular form.

Set difference edit

If $A\subseteq B$ and $\mu (B)-\mu (A)$ is defined, then $\mu (B\setminus A)=\mu (B)-\mu (A).$

Examples edit

An example of a 𝜎-additive function is the function $\mu$ defined over the power set of the real numbers, such that

\mu (A)={\begin{cases}1&{\mbox{ if }}0\in A\\0&{\mbox{ if }}0\notin A.\end{cases}}

If $A_{1},A_{2},\ldots ,A_{n},\ldots$ is a sequence of disjoint sets of real numbers, then either none of the sets contains 0, or precisely one of them does. In either case, the equality

\mu \left(\bigcup _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n})

holds.

See measure and signed measure for more examples of 𝜎-additive functions.

A charge is defined to be a finitely additive set function that maps $\varnothing$ to $0.$ ^[2] (Cf. ba space for information about bounded charges, where we say a charge is bounded to mean its range is a bounded subset of R.)

An additive function which is not σ-additive edit

An example of an additive function which is not σ-additive is obtained by considering $\mu$ , defined over the Lebesgue sets of the real numbers $\mathbb {R}$ by the formula

\mu (A)=\lim _{k\to \infty }{\frac {1}{k}}\cdot \lambda (A\cap (0,k)),

where

\lambda

denotes the Lebesgue measure and

\lim

the Banach limit. It satisfies

0\leq \mu (A)\leq 1

and if

\sup A<\infty

then

\mu (A)=0.

One can check that this function is additive by using the linearity of the limit. That this function is not σ-additive follows by considering the sequence of disjoint sets

A_{n}=[n,n+1)

for

n=0,1,2,\ldots

The union of these sets is the positive reals, and

\mu

applied to the union is then one, while

\mu

applied to any of the individual sets is zero, so the sum of

\mu (A_{n})

is also zero, which proves the counterexample.

Generalizations edit

One may define additive functions with values in any additive monoid (for example any group or more commonly a vector space). For sigma-additivity, one needs in addition that the concept of limit of a sequence be defined on that set. For example, spectral measures are sigma-additive functions with values in a Banach algebra. Another example, also from quantum mechanics, is the positive operator-valued measure.

References edit

^ D. H. Fremlin Measure Theory, Volume 4, Torres Fremlin, 2003.
^ Bhaskara Rao, K. P. S.; Bhaskara Rao, M. (1983). Theory of charges: a study of finitely additive measures. London: Academic Press. p. 35. ISBN 0-12-095780-9. OCLC 21196971.