In mathematics, especially measure theory, a set function is a function whose domain is a family of subsets of some given set and that (usually) takes its values in the extended real number line R ∪ { ± ∞ } , {\displaystyle \mathbb {R} \cup \{\pm \infty \},} which consists of the real numbers R {\displaystyle \mathbb {R} } and ± ∞ . {\displaystyle \pm \infty .}
A set function generally aims to measure subsets in some way. Measures are typical examples of "measuring" set functions. Therefore, the term "set function" is often used for avoiding confusion between the mathematical meaning of "measure" and its common language meaning.
Definitions If F {\displaystyle {\mathcal {F}}} is a family of sets over Ω {\displaystyle \Omega } (meaning that F ⊆ ℘ ( Ω ) {\displaystyle {\mathcal {F}}\subseteq \wp (\Omega )} where ℘ ( Ω ) {\displaystyle \wp (\Omega )} denotes the powerset) then a set function on F {\displaystyle {\mathcal {F}}} is a function μ {\displaystyle \mu } with domain F {\displaystyle {\mathcal {F}}} and codomain [ − ∞ , ∞ ] {\displaystyle [-\infty ,\infty ]} or, sometimes, the codomain is instead some vector space, as with vector measures, complex measures, and projection-valued measures. The domain of a set function may have any number properties; the commonly encountered properties and categories of families are listed in the table below.
In general, it is typically assumed that μ ( E ) + μ ( F ) {\displaystyle \mu (E)+\mu (F)} is always well-defined for all E , F ∈ F , {\displaystyle E,F\in {\mathcal {F}},} or equivalently, that μ {\displaystyle \mu } does not take on both − ∞ {\displaystyle -\infty } and + ∞ {\displaystyle +\infty } as values. This article will henceforth assume this; although alternatively, all definitions below could instead be qualified by statements such as "whenever the sum/series is defined". This is sometimes done with subtraction, such as with the following result, which holds whenever μ {\displaystyle \mu } is finitely additive:
Set difference formula: μ ( F ) − μ ( E ) = μ ( F ∖ E ) whenever μ ( F ) − μ ( E ) {\displaystyle \mu (F)-\mu (E)=\mu (F\setminus E){\text{ whenever }}\mu (F)-\mu (E)} is defined with E , F ∈ F {\displaystyle E,F\in {\mathcal {F}}} satisfying E ⊆ F {\displaystyle E\subseteq F} and F ∖ E ∈ F . {\displaystyle F\setminus E\in {\mathcal {F}}.}
Null sets A set F ∈ F {\displaystyle F\in {\mathcal {F}}} is called a null set (with respect to μ {\displaystyle \mu } ) or simply null if μ ( F ) = 0. {\displaystyle \mu (F)=0.} Whenever μ {\displaystyle \mu } is not identically equal to either − ∞ {\displaystyle -\infty } or + ∞ {\displaystyle +\infty } then it is typically also assumed that:
null empty set: μ ( ∅ ) = 0 {\displaystyle \mu (\varnothing )=0} if ∅ ∈ F . {\displaystyle \varnothing \in {\mathcal {F}}.}
Variation and mass The total variation of a set S {\displaystyle S} is
| μ | ( S ) = def sup { | μ ( F ) | : F ∈ F and F ⊆ S } {\displaystyle |\mu |(S)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\sup\{|\mu (F)|:F\in {\mathcal {F}}{\text{ and }}F\subseteq S\}}
where | ⋅ | {\displaystyle |\,\cdot \,|} denotes the absolute value (or more generally, it denotes the norm or seminorm if μ {\displaystyle \mu } is vector-valued in a (semi)normed space). Assuming that ∪ F = def ⋃ F ∈ F F ∈ F , {\displaystyle \cup {\mathcal {F}}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\textstyle \bigcup \limits _{F\in {\mathcal {F}}}F\in {\mathcal {F}},} then | μ | ( ∪ F ) {\displaystyle |\mu |\left(\cup {\mathcal {F}}\right)} is called the total variation of μ {\displaystyle \mu } and μ ( ∪ F ) {\displaystyle \mu \left(\cup {\mathcal {F}}\right)} is called the mass of μ . {\displaystyle \mu .} A set function is called finite if for every F ∈ F , {\displaystyle F\in {\mathcal {F}},} the value μ ( F ) {\displaystyle \mu (F)} is finite (which by definition means that μ ( F ) ≠ ∞ {\displaystyle \mu (F)\neq \infty } and μ ( F ) ≠ − ∞ {\displaystyle \mu (F)\neq -\infty } ; an infinite value is one that is equal to ∞ {\displaystyle \infty } or − ∞ {\displaystyle -\infty } ). Every finite set function must have a finite mass.
Common properties of set functions A set function μ {\displaystyle \mu } on F {\displaystyle {\mathcal {F}}} is said to be
non-negative if it is valued in [ 0 , ∞ ] . {\displaystyle [0,\infty ].}
finitely additive if ∑ i = 1 n μ ( F i ) = μ ( ⋃ i = 1 n F i ) {\displaystyle \textstyle \sum \limits _{i=1}^{n}\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{n}F_{i}\right)} for all pairwise disjoint finite sequences F 1 , … , F n ∈ F {\displaystyle F_{1},\ldots ,F_{n}\in {\mathcal {F}}} such that ⋃ i = 1 n F i ∈ F . {\displaystyle \textstyle \bigcup \limits _{i=1}^{n}F_{i}\in {\mathcal {F}}.}
If F {\displaystyle {\mathcal {F}}} is closed under binary unions then μ {\displaystyle \mu } is finitely additive if and only if μ ( E ∪ F ) = μ ( E ) + μ ( F ) {\displaystyle \mu (E\cup F)=\mu (E)+\mu (F)} for all disjoint pairs E , F ∈ F . {\displaystyle E,F\in {\mathcal {F}}.}
If μ {\displaystyle \mu } is finitely additive and if ∅ ∈ F {\displaystyle \varnothing \in {\mathcal {F}}} then taking E := F := ∅ {\displaystyle E:=F:=\varnothing } shows that μ ( ∅ ) = μ ( ∅ ) + μ ( ∅ ) {\displaystyle \mu (\varnothing )=\mu (\varnothing )+\mu (\varnothing )} which is only possible if μ ( ∅ ) = 0 {\displaystyle \mu (\varnothing )=0} or μ ( ∅ ) = ± ∞ , {\displaystyle \mu (\varnothing )=\pm \infty ,} where in the latter case, μ ( E ) = μ ( E ∪ ∅ ) = μ ( E ) + μ ( ∅ ) = μ ( E ) + ( ± ∞ ) = ± ∞ {\displaystyle \mu (E)=\mu (E\cup \varnothing )=\mu (E)+\mu (\varnothing )=\mu (E)+(\pm \infty )=\pm \infty } for every E ∈ F {\displaystyle E\in {\mathcal {F}}} (so only the case μ ( ∅ ) = 0 {\displaystyle \mu (\varnothing )=0} is useful).
countably additive or σ-additive if in addition to being finitely additive, for all pairwise disjoint sequences F 1 , F 2 , … {\displaystyle F_{1},F_{2},\ldots \,} in F {\displaystyle {\mathcal {F}}} such that ⋃ i = 1 ∞ F i ∈ F , {\displaystyle \textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\in {\mathcal {F}},} all of the following hold:
∑ i = 1 ∞ μ ( F i ) = μ ( ⋃ i = 1 ∞ F i ) {\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)}
The series on the left hand side is defined in the usual way as the limit ∑ i = 1 ∞ μ ( F i ) = def lim n → ∞ μ ( F 1 ) + ⋯ + μ ( F n ) . {\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\displaystyle \lim _{n\to \infty }}\mu \left(F_{1}\right)+\cdots +\mu \left(F_{n}\right).}
As a consequence, if ρ : N → N {\displaystyle \rho :\mathbb {N} \to \mathbb {N} } is any permutation/bijection then ∑ i = 1 ∞ μ ( F i ) = ∑ i = 1 ∞ μ ( F ρ ( i ) ) ; {\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)=\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{\rho (i)}\right);} this is because ⋃ i = 1 ∞ F i = ⋃ i = 1 ∞ F ρ ( i ) {\displaystyle \textstyle \bigcup \limits _{i=1}^{\infty }F_{i}=\textstyle \bigcup \limits _{i=1}^{\infty }F_{\rho (i)}} and applying this condition (a) twice guarantees that both ∑ i = 1 ∞ μ ( F i ) = μ ( ⋃ i = 1 ∞ F i ) {\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)} and μ ( ⋃ i = 1 ∞ F ρ ( i ) ) = ∑ i = 1 ∞ μ ( F ρ ( i ) ) {\displaystyle \mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{\rho (i)}\right)=\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{\rho (i)}\right)} hold. By definition, a convergent series with this property is said to be unconditionally convergent. Stated in plain English, this means that rearranging/relabeling the sets F 1 , F 2 , … {\displaystyle F_{1},F_{2},\ldots } to the new order F ρ ( 1 ) , F ρ ( 2 ) , … {\displaystyle F_{\rho (1)},F_{\rho (2)},\ldots } does not affect the sum of their measures. This is desirable since just as the union F = def ⋃ i ∈ N F i {\displaystyle F~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\textstyle \bigcup \limits _{i\in \mathbb {N} }F_{i}} does not depend on the order of these sets, the same should be true of the sums μ ( F ) = μ ( F 1 ) + μ ( F 2 ) + ⋯ {\displaystyle \mu (F)=\mu \left(F_{1}\right)+\mu \left(F_{2}\right)+\cdots } and μ ( F ) = μ ( F ρ ( 1 ) ) + μ ( F ρ ( 2 ) ) + ⋯ . {\displaystyle \mu (F)=\mu \left(F_{\rho (1)}\right)+\mu \left(F_{\rho (2)}\right)+\cdots \,.}
if μ ( ⋃ i = 1 ∞ F i ) {\displaystyle \mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)} is not infinite then this series ∑ i = 1 ∞ μ ( F i ) {\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)} must also converge absolutely, which by definition means that ∑ i = 1 ∞ | μ ( F i ) | {\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\left|\mu \left(F_{i}\right)\right|} must be finite. This is automatically true if μ {\displaystyle \mu } is non-negative (or even just valued in the extended real numbers). As with any convergent series of real numbers, by the Riemann series theorem, the series ∑ i = 1 ∞ μ ( F i ) = lim N → ∞ μ ( F 1 ) + μ ( F 2 ) + ⋯ + μ ( F N ) {\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)={\displaystyle \lim _{N\to \infty }}\mu \left(F_{1}\right)+\mu \left(F_{2}\right)+\cdots +\mu \left(F_{N}\right)} converges absolutely if and only if its sum does not depend on the order of its terms (a property known as unconditional convergence). Since unconditional convergence is guaranteed by (a) above, this condition is automatically true if μ {\displaystyle \mu } is valued in [ − ∞ , ∞ ] . {\displaystyle [-\infty ,\infty ].}
if μ ( ⋃ i = 1 ∞ F i ) = ∑ i = 1 ∞ μ ( F i ) {\displaystyle \mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)=\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)} is infinite then it is also required that the value of at least one of the series ∑ μ ( F i ) > 0 i ∈ N μ ( F i ) and ∑ μ ( F i ) < 0 i ∈ N μ ( F i ) {\displaystyle \textstyle \sum \limits _{\stackrel {i\in \mathbb {N} }{\mu \left(F_{i}\right)>0}}\mu \left(F_{i}\right)\;{\text{ and }}\;\textstyle \sum \limits _{\stackrel {i\in \mathbb {N} }{\mu \left(F_{i}\right)<0}}\mu \left(F_{i}\right)\;} be finite (so that the sum of their values is well-defined). This is automatically true if μ {\displaystyle \mu } is non-negative.
a pre-measure if it is non-negative, countably additive (including finitely additive), and has a null empty set. a measure if it is a pre-measure whose domain is a σ-algebra. That is to say, a measure is a non-negative countably additive set function on a σ-algebra that has a null empty set. a probability measure if it is a measure that has a mass of 1. {\displaystyle 1.}
an outer measure if it is non-negative, countably subadditive, has a null empty set, and has the power set ℘ ( Ω ) {\displaystyle \wp (\Omega )} as its domain. Outer measures appear in the Carathéodory's extension theorem and they are often restricted to Carathéodory measurable subsets a signed measure if it is countably additive, has a null empty set, and μ {\displaystyle \mu } does not take on both − ∞ {\displaystyle -\infty } and + ∞ {\displaystyle +\infty } as values. complete if every subset of every null set is null; explicitly, this means: whenever F ∈ F satisfies μ ( F ) = 0 {\displaystyle F\in {\mathcal {F}}{\text{ satisfies }}\mu (F)=0} and N ⊆ F {\displaystyle N\subseteq F} is any subset of F {\displaystyle F} then N ∈ F {\displaystyle N\in {\mathcal {F}}} and μ ( N ) = 0. {\displaystyle \mu (N)=0.}
Unlike many other properties, completeness places requirements on the set domain μ = F {\displaystyle \operatorname {domain} \mu ={\mathcal {F}}} (and not just on μ {\displaystyle \mu } 's values). 𝜎-finite if there exists a sequence F 1 , F 2 , F 3 , … {\displaystyle F_{1},F_{2},F_{3},\ldots \,} in F {\displaystyle {\mathcal {F}}} such that μ ( F i ) {\displaystyle \mu \left(F_{i}\right)} is finite for every index i , {\displaystyle i,} and also ⋃ n = 1 ∞ F n = ⋃ F ∈ F F . {\displaystyle \textstyle \bigcup \limits _{n=1}^{\infty }F_{n}=\textstyle \bigcup \limits _{F\in {\mathcal {F}}}F.}
decomposable if there exists a subfamily P ⊆ F {\displaystyle {\mathcal {P}}\subseteq {\mathcal {F}}} of pairwise disjoint sets such that μ ( P ) {\displaystyle \mu (P)} is finite for every P ∈ P {\displaystyle P\in {\mathcal {P}}} and also ⋃ P ∈ P P = ⋃ F ∈ F F {\displaystyle \textstyle \bigcup \limits _{P\in {\mathcal {P}}}\,P=\textstyle \bigcup \limits _{F\in {\mathcal {F}}}F} (where F = domain μ {\displaystyle {\mathcal {F}}=\operatorname {domain} \mu } ). Every 𝜎-finite set function is decomposable although not conversely. For example, the counting measure on R {\displaystyle \mathbb {R} } (whose domain is ℘ ( R ) {\displaystyle \wp (\mathbb {R} )} ) is decomposable but not 𝜎-finite. a vector measure if it is a countably additive set function μ : F → X {\displaystyle \mu :{\mathcal {F}}\to X} valued in a topological vector space X {\displaystyle X} (such as a normed space) whose domain is a σ-algebra. If μ {\displaystyle \mu } is valued in a normed space ( X , ‖ ⋅ ‖ ) {\displaystyle (X,\|\cdot \|)} then it is countably additive if and only if for any pairwise disjoint sequence F 1 , F 2 , … {\displaystyle F_{1},F_{2},\ldots \,} in F , {\displaystyle {\mathcal {F}},} lim n → ∞ ‖ μ ( F 1 ) + ⋯ + μ ( F n ) − μ ( ⋃ i = 1 ∞ F i ) ‖ = 0. {\displaystyle \lim _{n\to \infty }\left\|\mu \left(F_{1}\right)+\cdots +\mu \left(F_{n}\right)-\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)\right\|=0.} If μ {\displaystyle \mu } is finitely additive and valued in a Banach space then it is countably additive if and only if for any pairwise disjoint sequence F 1 , F 2 , … {\displaystyle F_{1},F_{2},\ldots \,} in F , {\displaystyle {\mathcal {F}},} lim n → ∞ ‖ μ ( F n ∪ F n + 1 ∪ F n + 2 ∪ ⋯ ) ‖ = 0. {\displaystyle \lim _{n\to \infty }\left\|\mu \left(F_{n}\cup F_{n+1}\cup F_{n+2}\cup \cdots \right)\right\|=0.}
a complex measure if it is a countably additive complex-valued set function μ : F → C {\displaystyle \mu :{\mathcal {F}}\to \mathbb {C} } whose domain is a σ-algebra. By definition, a complex measure never takes