Sigma-additive set function

In mathematics, an additive set function is a function $\mu$ 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, $\mu (A\cup B)=\mu (A)+\mu (B).$ 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, $\mu \left({\bigcup }_{n=1}^{\mathrm{\infty }}{A}_n\right)={\sum }_{n=1}^{\mathrm{\infty }}\mu (A_n).$

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

Let $\mu$ be a set function defined on an algebra of sets $𝒜$ with values in $[-\mathrm{\infty },\mathrm{\infty }]$ (see the extended real number line). The function $\mu$ is called additive or finitely additive, if whenever $A$ and $B$ are disjoint sets in $𝒜,$ then $$\mu (A\cup B)=\mu (A)+\mu (B).$$ A consequence of this is that an additive function cannot take both $-\mathrm{\infty }$ and $+\mathrm{\infty }$ as values, for the expression $\mathrm{\infty }-\mathrm{\infty }$ 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 $A_1,A_2,\dots ,A_N$ disjoint sets in $𝒜.$

σ-additive set functions

Suppose that $𝒜$ is a σ-algebra. If for every sequence $A_1,A_2,\dots ,A_n,\dots$ of pairwise disjoint sets in $𝒜,$ $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mu (A_n),$$ 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

Suppose that in addition to a sigma algebra $𝒜,$ we have a topology $\tau .$ If for every directed family of measurable open sets $𝒢\subseteq 𝒜\cap \tau ,$ $$\mu (\bigcup 𝒢)=\underset{G\in 𝒢}{\sup }\mu (G),$$ we say that $\mu$ is $\tau$-additive. In particular, if $\mu$ is inner regular (with respect to compact sets) then it is $\tau$-additive.

Properties

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

Value of empty set

Either $\mu (\varnothing )=0,$ or $\mu$ assigns $\mathrm{\infty }$ to all sets in its domain, or $\mu$ assigns $-\mathrm{\infty }$ to all sets in its domain. Proof: additivity implies that for every set $A,$ $\mu (A)=\mu (A\cup \varnothing )=\mu (A)+\mu (\varnothing )$ (it's possible in the edge case of an empty domain that the only choice for $A$ is the empty set itself, but that still works). If $\mu (\varnothing )\ne 0,$ then this equality can be satisfied only by plus or minus infinity.

Monotonicity

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

Modularity

See also: Valuation (geometry)

See also: Valuation (measure theory)

A set function $\mu$ on a family of sets $𝒮$ is called a modular set function and a valuation if whenever $A,$ $B,$ $A\cup B,$ and $A\cap B$ are elements of $𝒮,$ then $$\mu (A\cup B)+\mu (A\cap B)=\mu (A)+\mu (B)$$ 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

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

Examples

An example of a 𝜎-additive function is the function $\mu$ defined over the power set of the real numbers, such that $$\mu (A)=\{\begin{array}{cc}1& \text{ if }0\in A\\ 0& \text{ if }0\notin A.\end{array}$$

If $A_1,A_2,\dots ,A_n,\dots$ 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 $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mu (A_n)$$ 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 $0.$ (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

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 $ℝ$ by the formula $$\mu (A)=\underset{k\to \mathrm{\infty }}{\lim }\frac{1}{k}\cdot \lambda (A\cap (0,k)),$$ where $\lambda$ denotes the Lebesgue measure and $\lim$ the Banach limit. It satisfies $0\le \mu (A)\le 1$ and if $\sup A<\mathrm{\infty }$ 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 $n=0,1,2,\dots$ 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

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.

See also

• Additive map – Z-module homomorphism
• Hahn–Kolmogorov theorem – Theorem extending pre-measures to measuresPages displaying short descriptions of redirect targets
• Measure (mathematics) – Generalization of mass, length, area and volume
• σ-finite measure – Concept in measure theory
• Signed measure – Generalized notion of measure in mathematics
• Submodular set function – Set-to-real map with diminishing returns
• ba space – The set of bounded charges on a given sigma-algebra

This article incorporates material from additive on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.