Sigma-additive set function

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Short description: Mapping function

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, [math]\displaystyle{ \mu(A \cup B) = \mu(A) + \mu(B). }[/math] 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, [math]\displaystyle{ \mu\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mu(A_n). }[/math]

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

One can prove by mathematical induction that an additive function satisfies [math]\displaystyle{ \mu\left(\bigcup_{n=1}^N A_n\right)=\sum_{n=1}^N \mu\left(A_n\right) }[/math] for any [math]\displaystyle{ A_1, A_2, \ldots, A_N }[/math] disjoint sets in [math]\displaystyle{ \mathcal{A}. }[/math]

σ-additive set functions

Suppose that [math]\displaystyle{ \scriptstyle\mathcal{A} }[/math] is a σ-algebra. If for every sequence [math]\displaystyle{ A_1, A_2, \ldots, A_n, \ldots }[/math] of pairwise disjoint sets in [math]\displaystyle{ \scriptstyle\mathcal{A}, }[/math] [math]\displaystyle{ \mu\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mu(A_n), }[/math] holds then [math]\displaystyle{ \mu }[/math] 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 [math]\displaystyle{ \mathcal{A}, }[/math] we have a topology [math]\displaystyle{ \tau. }[/math] If for every directed family of measurable open sets [math]\displaystyle{ \mathcal{G} \subseteq \mathcal{A} \cap \tau, }[/math] [math]\displaystyle{ \mu\left(\bigcup \mathcal{G} \right) = \sup_{G\in\mathcal{G}} \mu(G), }[/math] we say that [math]\displaystyle{ \mu }[/math] is [math]\displaystyle{ \tau }[/math]-additive. In particular, if [math]\displaystyle{ \mu }[/math] is inner regular (with respect to compact sets) then it is τ-additive.[1]

Properties

Useful properties of an additive set function [math]\displaystyle{ \mu }[/math] include the following.

Value of empty set

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

Monotonicity

If [math]\displaystyle{ \mu }[/math] is non-negative and [math]\displaystyle{ A \subseteq B }[/math] then [math]\displaystyle{ \mu(A) \leq \mu(B). }[/math] That is, [math]\displaystyle{ \mu }[/math] is a monotone set function. Similarly, If [math]\displaystyle{ \mu }[/math] is non-positive and [math]\displaystyle{ A \subseteq B }[/math] then [math]\displaystyle{ \mu(A) \geq \mu(B). }[/math]

Modularity

A set function [math]\displaystyle{ \mu }[/math] on a family of sets [math]\displaystyle{ \mathcal{S} }[/math] is called a modular set function and a valuation if whenever [math]\displaystyle{ A, }[/math] [math]\displaystyle{ B, }[/math] [math]\displaystyle{ A\cup B, }[/math] and [math]\displaystyle{ A\cap B }[/math] are elements of [math]\displaystyle{ \mathcal{S}, }[/math] then [math]\displaystyle{ \phi(A\cup B)+ \phi(A\cap B) = \phi(A) + \phi(B) }[/math] The above property is called modularity and the argument below proves that additivity implies modularity.

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

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 [math]\displaystyle{ A \subseteq B }[/math] and [math]\displaystyle{ \mu(B) - \mu(A) }[/math] is defined, then [math]\displaystyle{ \mu(B \setminus A) = \mu(B) - \mu(A). }[/math]

Examples

An example of a 𝜎-additive function is the function [math]\displaystyle{ \mu }[/math] defined over the power set of the real numbers, such that [math]\displaystyle{ \mu (A)= \begin{cases} 1 & \mbox{ if } 0 \in A \\ 0 & \mbox{ if } 0 \notin A. \end{cases} }[/math]

If [math]\displaystyle{ A_1, A_2, \ldots, A_n, \ldots }[/math] 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 [math]\displaystyle{ \mu\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mu(A_n) }[/math] 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 [math]\displaystyle{ \varnothing }[/math] to [math]\displaystyle{ 0. }[/math][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

An example of an additive function which is not σ-additive is obtained by considering [math]\displaystyle{ \mu }[/math], defined over the Lebesgue sets of the real numbers [math]\displaystyle{ \R }[/math] by the formula [math]\displaystyle{ \mu(A) = \lim_{k\to\infty} \frac{1}{k} \cdot \lambda(A \cap (0,k)), }[/math] where [math]\displaystyle{ \lambda }[/math] denotes the Lebesgue measure and [math]\displaystyle{ \lim }[/math] the Banach limit. It satisfies [math]\displaystyle{ 0 \leq \mu(A) \leq 1 }[/math] and if [math]\displaystyle{ \sup A \lt \infty }[/math] then [math]\displaystyle{ \mu(A) = 0. }[/math]

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 [math]\displaystyle{ A_n = [n,n + 1) }[/math] for [math]\displaystyle{ n = 0, 1, 2, \ldots }[/math] The union of these sets is the positive reals, and [math]\displaystyle{ \mu }[/math] applied to the union is then one, while [math]\displaystyle{ \mu }[/math] applied to any of the individual sets is zero, so the sum of [math]\displaystyle{ \mu(A_n) }[/math]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


References

  1. D. H. Fremlin Measure Theory, Volume 4, Torres Fremlin, 2003.
  2. Bhaskara Rao, K. P. S.; Bhaskara Rao, M. (1983). Theory of charges: a study of finitely additive measures. London: Academic Press. pp. 35. ISBN 0-12-095780-9. OCLC 21196971. https://www.worldcat.org/oclc/21196971.