Sigma-additive set function

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

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

One can prove by mathematical induction that an additive function satisfies $${\displaystyle \mu \left(\bigcup _{n=1}^{N}A_{n}\right)=\sum _{n=1}^{N}\mu \left(A_{n}\right)}$${\displaystyle \mu \left(\bigcup _{n=1}^{N}A_{n}\right)=\sum _{n=1}^{N}\mu \left(A_{n}\right)} for any ${\displaystyle A_{1},A_{2},\ldots ,A_{N}}${\displaystyle A_{1},A_{2},\ldots ,A_{N}} disjoint sets in ${\textstyle {\mathcal {A}}.}${\textstyle {\mathcal {A}}.}

Suppose that ${\displaystyle \scriptstyle {\mathcal {A}}}${\displaystyle \scriptstyle {\mathcal {A}}} is a σ-algebra. If for every sequence ${\displaystyle A_{1},A_{2},\ldots ,A_{n},\ldots }${\displaystyle A_{1},A_{2},\ldots ,A_{n},\ldots } of pairwise disjoint sets in ${\displaystyle \scriptstyle {\mathcal {A}},}${\displaystyle \scriptstyle {\mathcal {A}},} $${\displaystyle \mu \left(\bigcup _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n}),}$${\displaystyle \mu \left(\bigcup _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n}),} holds then ${\displaystyle \mu }${\displaystyle \mu } is said to be countably additive or σ-additive. Every σ-additive function is additive but not vice versa, as shown below.

Suppose that in addition to a sigma algebra ${\textstyle {\mathcal {A}},}${\textstyle {\mathcal {A}},} we have a topology ${\displaystyle \tau .}${\displaystyle \tau .} If for every directed family of measurable open sets ${\textstyle {\mathcal {G}}\subseteq {\mathcal {A}}\cap \tau ,}${\textstyle {\mathcal {G}}\subseteq {\mathcal {A}}\cap \tau ,} $${\displaystyle \mu \left(\bigcup {\mathcal {G}}\right)=\sup _{G\in {\mathcal {G}}}\mu (G),}$${\displaystyle \mu \left(\bigcup {\mathcal {G}}\right)=\sup _{G\in {\mathcal {G}}}\mu (G),} we say that ${\displaystyle \mu }${\displaystyle \mu } is ${\displaystyle \tau }${\displaystyle \tau }-additive. In particular, if ${\displaystyle \mu }${\displaystyle \mu } is inner regular (with respect to compact sets) then it is ${\displaystyle \tau }${\displaystyle \tau }-additive.^[1]

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

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

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

A set function ${\displaystyle \mu }${\displaystyle \mu } on a family of sets ${\displaystyle {\mathcal {S}}}${\displaystyle {\mathcal {S}}} is called a modular set function and a valuation if whenever ${\displaystyle A,}${\displaystyle A,} ${\displaystyle B,}${\displaystyle B,} ${\displaystyle A\cup B,}${\displaystyle A\cup B,} and ${\displaystyle A\cap B}${\displaystyle A\cap B} are elements of ${\displaystyle {\mathcal {S}},}${\displaystyle {\mathcal {S}},} then $${\displaystyle \mu (A\cup B)+\mu (A\cap B)=\mu (A)+\mu (B)}$${\displaystyle \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 ${\displaystyle A}${\displaystyle A} and ${\displaystyle B,}${\displaystyle B,} ${\displaystyle \mu (A\cup B)+\mu (A\cap B)=\mu (A)+\mu (B).}${\displaystyle \mu (A\cup B)+\mu (A\cap B)=\mu (A)+\mu (B).} Proof: write ${\displaystyle A=(A\cap B)\cup (A\setminus B)}${\displaystyle A=(A\cap B)\cup (A\setminus B)} and ${\displaystyle B=(A\cap B)\cup (B\setminus A)}${\displaystyle B=(A\cap B)\cup (B\setminus A)} and ${\displaystyle A\cup B=(A\cap B)\cup (A\setminus B)\cup (B\setminus A),}${\displaystyle 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 ${\displaystyle \mu (A\setminus B)+\mu (B\setminus A)+2\mu (A\cap B).}${\displaystyle \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.

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

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

If ${\displaystyle A_{1},A_{2},\ldots ,A_{n},\ldots }${\displaystyle 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 $${\displaystyle \mu \left(\bigcup _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n})}$${\displaystyle \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 ${\displaystyle \varnothing }${\displaystyle \varnothing } to ${\displaystyle 0.}${\displaystyle 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 example of an additive function which is not σ-additive is obtained by considering ${\displaystyle \mu }${\displaystyle \mu }, defined over the Lebesgue sets of the real numbers ${\displaystyle \mathbb {R} }${\displaystyle \mathbb {R} } by the formula $${\displaystyle \mu (A)=\lim _{k\to \infty }{\frac {1}{k}}\cdot \lambda (A\cap (0,k)),}$${\displaystyle \mu (A)=\lim _{k\to \infty }{\frac {1}{k}}\cdot \lambda (A\cap (0,k)),} where ${\displaystyle \lambda }${\displaystyle \lambda } denotes the Lebesgue measure and ${\displaystyle \lim }${\displaystyle \lim } the Banach limit. It satisfies ${\displaystyle 0\leq \mu (A)\leq 1}${\displaystyle 0\leq \mu (A)\leq 1} and if ${\displaystyle \sup A<\infty }${\displaystyle \sup A<\infty } then ${\displaystyle \mu (A)=0.}${\displaystyle \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 $${\displaystyle A_{n}=[n,n+1)}$${\displaystyle A_{n}=[n,n+1)} for ${\displaystyle n=0,1,2,\ldots }${\displaystyle n=0,1,2,\ldots } The union of these sets is the positive reals, and ${\displaystyle \mu }${\displaystyle \mu } applied to the union is then one, while ${\displaystyle \mu }${\displaystyle \mu } applied to any of the individual sets is zero, so the sum of ${\displaystyle \mu (A_{n})}${\displaystyle \mu (A_{n})} is also zero, which proves the counterexample.

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.

• 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
• Subadditive set function
• τ-additivity
• ba space – The set of bounded charges on a given sigma-algebra

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• Measure theory
• Additive functions

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