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Sigma-additive set function

Mapping function

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Source updated January 3, 2026

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, ${\textstyle \mu (A\cup B)=\mu (A)+\mu (B).$ 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.

01Additive (or finitely additive) set functions

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

02σ-additive set functions

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.

03τ-additive set functions

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 $\tau$ -additive.

04Properties

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

Value of empty set

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 $A,$ $\mu (A)=\mu (A\cup \varnothing )=\mu (A)+\mu (\varnothing )$ (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 $\mu (\varnothing )\neq 0,$ then this equality can be satisfied only by plus or minus infinity.

Monotonicity

If $\mu$ is non-negative and $A\subseteq B$ then $\mu (A)\leq \mu (B).$ 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

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 $\mu (A\cup B)+\mu (A\cap B)=\mu (A)+\mu (B)$ 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 $\mu (B\setminus A)=\mu (B)-\mu (A).$

05Examples

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 $0.$ (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 $\mathbb {R}$ 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.

06Generalizations

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.

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Sources and credits

This article is adapted from the Wikipedia article “Sigma-additive set function”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

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