In mathematics, an additive set function is a function μ {\textstyle \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, μ ( A ∪ B ) = μ ( A ) + μ ( 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, μ ( ⋃ n = 1 ∞ A n ) = ∑ n = 1 ∞ μ ( 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 μ {\displaystyle \mu } be a set function defined on an algebra of sets A {\displaystyle \scriptstyle {\mathcal {A}}} with values in [ − ∞ , ∞ ] {\displaystyle [-\infty ,\infty ]} (see the extended real number line). The function μ {\displaystyle \mu } is called additive or finitely additive, if whenever A {\displaystyle A} and B {\displaystyle B} are disjoint sets in A , {\displaystyle \scriptstyle {\mathcal {A}},} then μ ( A ∪ B ) = μ ( A ) + μ ( B ) . {\displaystyle \mu (A\cup B)=\mu (A)+\mu (B).} A consequence of this is that an additive function cannot take both − ∞ {\displaystyle -\infty } and + ∞ {\displaystyle +\infty } as values, for the expression ∞ − ∞ {\displaystyle \infty -\infty } is undefined.
One can prove by mathematical induction that an additive function satisfies μ ( ⋃ n = 1 N A n ) = ∑ n = 1 N μ ( A n ) {\displaystyle \mu \left(\bigcup _{n=1}^{N}A_{n}\right)=\sum _{n=1}^{N}\mu \left(A_{n}\right)} for any A 1 , A 2 , … , A N {\displaystyle A_{1},A_{2},\ldots ,A_{N}} disjoint sets in A . {\textstyle {\mathcal {A}}.}
Suppose that A {\displaystyle \scriptstyle {\mathcal {A}}} is a σ-algebra. If for every sequence A 1 , A 2 , … , A n , … {\displaystyle A_{1},A_{2},\ldots ,A_{n},\ldots } of pairwise disjoint sets in A , {\displaystyle \scriptstyle {\mathcal {A}},} μ ( ⋃ n = 1 ∞ A n ) = ∑ n = 1 ∞ μ ( A n ) , {\displaystyle \mu \left(\bigcup _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n}),} holds then μ {\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 A , {\textstyle {\mathcal {A}},} we have a topology τ . {\displaystyle \tau .} If for every directed family of measurable open sets G ⊆ A ∩ τ , {\textstyle {\mathcal {G}}\subseteq {\mathcal {A}}\cap \tau ,} μ ( ⋃ G ) = sup G ∈ G μ ( G ) , {\displaystyle \mu \left(\bigcup {\mathcal {G}}\right)=\sup _{G\in {\mathcal {G}}}\mu (G),} we say that μ {\displaystyle \mu } is τ {\displaystyle \tau } -additive. In particular, if μ {\displaystyle \mu } is inner regular (with respect to compact sets) then it is τ {\displaystyle \tau } -additive.[1]
Useful properties of an additive set function μ {\displaystyle \mu } include the following.
Either μ ( ∅ ) = 0 , {\displaystyle \mu (\varnothing )=0,} or μ {\displaystyle \mu } assigns ∞ {\displaystyle \infty } to all sets in its domain, or μ {\displaystyle \mu } assigns − ∞ {\displaystyle -\infty } to all sets in its domain. Proof: additivity implies that for every set A , {\displaystyle A,} μ ( A ) = μ ( A ∪ ∅ ) = μ ( A ) + μ ( ∅ ) {\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 A {\displaystyle A} is the empty set itself, but that still works). If μ ( ∅ ) ≠ 0 , {\displaystyle \mu (\varnothing )\neq 0,} then this equality can be satisfied only by plus or minus infinity.
If μ {\displaystyle \mu } is non-negative and A ⊆ B {\displaystyle A\subseteq B} then μ ( A ) ≤ μ ( B ) . {\displaystyle \mu (A)\leq \mu (B).} That is, μ {\displaystyle \mu } is a monotone set function. Similarly, If μ {\displaystyle \mu } is non-positive and A ⊆ B {\displaystyle A\subseteq B} then μ ( A ) ≥ μ ( B ) . {\displaystyle \mu (A)\geq \mu (B).}
A set function μ {\displaystyle \mu } on a family of sets S {\displaystyle {\mathcal {S}}} is called a modular set function and a valuation if whenever A , {\displaystyle A,} B , {\displaystyle B,} A ∪ B , {\displaystyle A\cup B,} and A ∩ B {\displaystyle A\cap B} are elements of S , {\displaystyle {\mathcal {S}},} then ϕ ( A ∪ B ) + ϕ ( A ∩ B ) = ϕ ( A ) + ϕ ( B ) {\displaystyle \phi (A\cup B)+\phi (A\cap B)=\phi (A)+\phi (B)} The above property is called modularity and the argument below proves that additivity implies modularity.
Given A {\displaystyle A} and B , {\displaystyle B,} μ ( A ∪ B ) + μ ( A ∩ B ) = μ ( A ) + μ ( B ) . {\displaystyle \mu (A\cup B)+\mu (A\cap B)=\mu (A)+\mu (B).} Proof: write A = ( A ∩ B ) ∪ ( A ∖ B ) {\displaystyle A=(A\cap B)\cup (A\setminus B)} and B = ( A ∩ B ) ∪ ( B ∖ A ) {\displaystyle B=(A\cap B)\cup (B\setminus A)} and A ∪ B = ( A ∩ B ) ∪ ( A ∖ B ) ∪ ( B ∖ 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 μ ( A ∖ B ) + μ ( B ∖ A ) + 2 μ ( A ∩ 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 A ⊆ B {\displaystyle A\subseteq B} and μ ( B ) − μ ( A ) {\displaystyle \mu (B)-\mu (A)} is defined, then μ ( B ∖ A ) = μ ( B ) − μ ( A ) . {\displaystyle \mu (B\setminus A)=\mu (B)-\mu (A).}
An example of a 𝜎-additive function is the function μ {\displaystyle \mu } defined over the power set of the real numbers, such that μ ( A ) = { 1 if 0 ∈ A 0 if 0 ∉ A . {\displaystyle \mu (A)={\begin{cases}1&{\mbox{ if }}0\in A\\0&{\mbox{ if }}0\notin A.\end{cases}}}
If A 1 , A 2 , … , A n , … {\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 μ ( ⋃ n = 1 ∞ A n ) = ∑ n = 1 ∞ μ ( 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 } to 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 μ {\displaystyle \mu } , defined over the Lebesgue sets of the real numbers R {\displaystyle \mathbb {R} } by the formula μ ( A ) = lim k → ∞ 1 k ⋅ λ ( A ∩ ( 0 , k ) ) , {\displaystyle \mu (A)=\lim _{k\to \infty }{\frac {1}{k}}\cdot \lambda (A\cap (0,k)),} where λ {\displaystyle \lambda } denotes the Lebesgue measure and lim {\displaystyle \lim } the Banach limit. It satisfies 0 ≤ μ ( A ) ≤ 1 {\displaystyle 0\leq \mu (A)\leq 1} and if sup A < ∞ {\displaystyle \sup A<\infty } then μ ( 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 A n = [ n , n + 1 ) {\displaystyle A_{n}=[n,n+1)} for n = 0 , 1 , 2 , … {\displaystyle n=0,1,2,\ldots } The union of these sets is the positive reals, and μ {\displaystyle \mu } applied to the union is then one, while μ {\displaystyle \mu } applied to any of the individual sets is zero, so the sum of μ ( 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.
This article incorporates material from additive on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.