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 &sigma;-additive set function is a function that has the additivity property even for countably infinite many sets, that is, $\mu\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mu(A_n).$

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 &sigma;-additivity; that is, &sigma;-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 $$\scriptstyle\mathcal{A}$$ with values in $$[-\infty, \infty]$$ (see the extended real number line). The function $$\mu$$ is called ' or ', 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 $\mathcal{A}.$

&sigma;-additive set functions
Suppose that $$\scriptstyle\mathcal{A}$$ is a &sigma;-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 or. Every 𝜎-additive function is additive but not vice versa, as shown below.

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

Properties
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 $$A,$$ $$\mu(A) = \mu(A \cup \varnothing) = \mu(A) + \mu( \varnothing).$$ 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 . 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  and a Valuation (geometry) 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  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 $$\mu (A)= \begin{cases} 1 & \mbox{ if } 0 \in A \\ 0 & \mbox{ if } 0 \notin A. \end{cases}$$

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 &sigma;-additive
An example of an additive function which is not &sigma;-additive is obtained by considering $$\mu$$, defined over the Lebesgue sets of the real numbers $$\R$$ 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 &sigma;-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
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.