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 ${\mathcal {F))$ is a family of sets over $\Omega$ (meaning that ${\mathcal {F))\subseteq \wp (\Omega )$ where $\wp (\Omega )$ denotes the powerset) then a set function on ${\mathcal {F))$ is a function $\mu$ with domain${\mathcal {F))$ and codomain$[-\infty ,\infty ]$ 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.
Additionally, a semiring is a π-system where every complement $B\setminus A$ is equal to a finite disjoint union of sets in ${\mathcal {F)).$
A semialgebra is a semiring where every complement $\Omega \setminus A$ is equal to a finite disjoint union of sets in ${\mathcal {F)).$ $A,B,A_{1},A_{2},\ldots$ are arbitrary elements of ${\mathcal {F))$ and it is assumed that ${\mathcal {F))\neq \varnothing .$
In general, it is typically assumed that $\mu (E)+\mu (F)$ is always well-defined for all $E,F\in {\mathcal {F)),$ or equivalently, that $\mu$ does not take on both $-\infty$ and $+\infty$ 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 $\mu$ is finitely additive:
Set difference formula: $\mu (F)-\mu (E)=\mu (F\setminus E){\text{ whenever ))\mu (F)-\mu (E)$ is defined with $E,F\in {\mathcal {F))$ satisfying $E\subseteq F$ and $F\setminus E\in {\mathcal {F)).$
Null sets
A set $F\in {\mathcal {F))$ is called a null set (with respect to $\mu$) or simply null if $\mu (F)=0.$
Whenever $\mu$ is not identically equal to either $-\infty$ or $+\infty$ then it is typically also assumed that:
$|\mu |(S)~{\stackrel {\scriptscriptstyle {\text{def))}{=))~\sup\{|\mu (F)|:F\in {\mathcal {F)){\text{ and ))F\subseteq S\))$
where $|\,\cdot \,|$ denotes the absolute value (or more generally, it denotes the norm or seminorm if $\mu$ is vector-valued in a (semi)normed space).
Assuming that $\cup {\mathcal {F))~{\stackrel {\scriptscriptstyle {\text{def))}{=))~\textstyle \bigcup \limits _{F\in {\mathcal {F))}F\in {\mathcal {F)),$ then $|\mu |\left(\cup {\mathcal {F))\right)$ is called the total variation of $\mu$ and $\mu \left(\cup {\mathcal {F))\right)$ is called the mass of $\mu .$
A set function is called finite if for every $F\in {\mathcal {F)),$ the value $\mu (F)$ is finite (which by definition means that $\mu (F)\neq \infty$ and $\mu (F)\neq -\infty$; an infinite value is one that is equal to $\infty$ or $-\infty$).
Every finite set function must have a finite mass.
Common properties of set functions
A set function $\mu$ on ${\mathcal {F))$ is said to be^{[1]}
non-negative if it is valued in $[0,\infty ].$
finitely additive if $\textstyle \sum \limits _{i=1}^{n}\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{n}F_{i}\right)$ for all pairwise disjoint finite sequences $F_{1},\ldots ,F_{n}\in {\mathcal {F))$ such that $\textstyle \bigcup \limits _{i=1}^{n}F_{i}\in {\mathcal {F)).$
If ${\mathcal {F))$ is closed under binary unions then $\mu$ is finitely additive if and only if $\mu (E\cup F)=\mu (E)+\mu (F)$ for all disjoint pairs $E,F\in {\mathcal {F)).$
If $\mu$ is finitely additive and if $\varnothing \in {\mathcal {F))$ then taking $E:=F:=\varnothing$ shows that $\mu (\varnothing )=\mu (\varnothing )+\mu (\varnothing )$ which is only possible if $\mu (\varnothing )=0$ or $\mu (\varnothing )=\pm \infty ,$ where in the latter case, $\mu (E)=\mu (E\cup \varnothing )=\mu (E)+\mu (\varnothing )=\mu (E)+(\pm \infty )=\pm \infty$ for every $E\in {\mathcal {F))$ (so only the case $\mu (\varnothing )=0$ is useful).
countably additive or σ-additive^{[2]} if in addition to being finitely additive, for all pairwise disjoint sequences $F_{1},F_{2},\ldots \,$ in ${\mathcal {F))$ such that $\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\in {\mathcal {F)),$ all of the following hold:
The series on the left hand side is defined in the usual way as the limit $\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)~{\stackrel {\scriptscriptstyle {\text{def))}{=))~ \lim _{n\to \infty ))\mu \left(F_{1}\right)+\cdots +\mu \left(F_{n}\right).$
As a consequence, if $\rho :\mathbb {N} \to \mathbb {N}$ is any permutation/bijection then $\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)=\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{\rho (i)}\right);$ this is because $\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}=\textstyle \bigcup \limits _{i=1}^{\infty }F_{\rho (i)))$ and applying this condition (a) twice guarantees that both $\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)$ and $\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{\rho (i)}\right)=\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{\rho (i)}\right)$ 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 $F_{1},F_{2},\ldots$ to the new order $F_{\rho (1)},F_{\rho (2)},\ldots$ does not affect the sum of their measures. This is desirable since just as the union $F~{\stackrel {\scriptscriptstyle {\text{def))}{=))~\textstyle \bigcup \limits _{i\in \mathbb {N} }F_{i))$ does not depend on the order of these sets, the same should be true of the sums $\mu (F)=\mu \left(F_{1}\right)+\mu \left(F_{2}\right)+\cdots$ and $\mu (F)=\mu \left(F_{\rho (1)}\right)+\mu \left(F_{\rho (2)}\right)+\cdots \,.$
if $\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)$ is not infinite then this series $\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)$ must also converge absolutely, which by definition means that $\textstyle \sum \limits _{i=1}^{\infty }\left|\mu \left(F_{i}\right)\right|$ must be finite. This is automatically true if $\mu$ 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 $\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)= \lim _{N\to \infty ))\mu \left(F_{1}\right)+\mu \left(F_{2}\right)+\cdots +\mu \left(F_{N}\right)$ 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 $\mu$ is valued in $[-\infty ,\infty ].$
if $\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)=\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)$ is infinite then it is also required that the value of at least one of the series $\textstyle \sum \limits _{\stackrel {i\in \mathbb {N} }{\mu \left(F_{i}\right)>0))\mu \left(F_{i}\right)\;{\text{ and ))\;\textstyle \sum \limits _{\stackrel {i\in \mathbb {N} }{\mu \left(F_{i}\right)<0))\mu \left(F_{i}\right)\;$ be finite (so that the sum of their values is well-defined). This is automatically true if $\mu$ is non-negative.
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 signed measure if it is countably additive, has a null empty set, and $\mu$ does not take on both $-\infty$ and $+\infty$ as values.
complete if every subset of every null set is null; explicitly, this means: whenever $F\in {\mathcal {F)){\text{ satisfies ))\mu (F)=0$ and $N\subseteq F$ is any subset of $F$ then $N\in {\mathcal {F))$ and $\mu (N)=0.$
Unlike many other properties, completeness places requirements on the set $\operatorname {domain} \mu ={\mathcal {F))$ (and not just on $\mu$'s values).
𝜎-finite if there exists a sequence $F_{1},F_{2},F_{3},\ldots \,$ in ${\mathcal {F))$ such that $\mu \left(F_{i}\right)$ is finite for every index $i,$ and also $\textstyle \bigcup \limits _{n=1}^{\infty }F_{n}=\textstyle \bigcup \limits _{F\in {\mathcal {F))}F.$
decomposable if there exists a subfamily ${\mathcal {P))\subseteq {\mathcal {F))$ of pairwise disjoint sets such that $\mu (P)$ is finite for every $P\in {\mathcal {P))$ and also $\textstyle \bigcup \limits _{P\in {\mathcal {P))}\,P=\textstyle \bigcup \limits _{F\in {\mathcal {F))}F$ (where ${\mathcal {F))=\operatorname {domain} \mu$).
Every 𝜎-finite set function is decomposable although not conversely. For example, the counting measure on $\mathbb {R}$ (whose domain is $\wp (\mathbb {R} )$) is decomposable but not 𝜎-finite.
If $\mu$ is valued in a normed space$(X,\|\cdot \|)$ then it is countably additive if and only if for any pairwise disjoint sequence $F_{1},F_{2},\ldots \,$ in ${\mathcal {F)),$$\lim _{n\to \infty }\left\|\mu \left(F_{1}\right)+\cdots +\mu \left(F_{n}\right)-\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)\right\|=0.$ If $\mu$ is finitely additive and valued in a Banach space then it is countably additive if and only if for any pairwise disjoint sequence $F_{1},F_{2},\ldots \,$ in ${\mathcal {F)),$$\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 $\mu :{\mathcal {F))\to \mathbb {C}$ whose domain is a σ-algebra.
By definition, a complex measure never takes $\pm \infty$ as a value and so has a null empty set.
As described in this article's section on generalized series, for any family $\left(r_{i}\right)_{i\in I))$ of real numbers indexed by an arbitrary indexing set$I,$ it is possible to define their sum $\textstyle \sum \limits _{i\in I}r_{i))$ as the limit of the net of finite partial sums $F\in \operatorname {FiniteSubsets} (I)\mapsto \textstyle \sum \limits _{i\in F}r_{i))$ where the domain $\operatorname {FiniteSubsets} (I)$ is directed by $\,\subseteq .\,$
Whenever this net converges then its limit is denoted by the symbols $\textstyle \sum \limits _{i\in I}r_{i))$ while if this net instead diverges to $\pm \infty$ then this may be indicated by writing $\textstyle \sum \limits _{i\in I}r_{i}=\pm \infty .$
Any sum over the empty set is defined to be zero; that is, if $I=\varnothing$ then $\textstyle \sum \limits _{i\in \varnothing }r_{i}=0$ by definition.
For example, if $z_{i}=0$ for every $i\in I$ then $\textstyle \sum \limits _{i\in I}z_{i}=0.$
And it can be shown that $\textstyle \sum \limits _{i\in I}r_{i}=\textstyle \sum \limits _{\stackrel {i\in I,}{r_{i}=0))r_{i}+\textstyle \sum \limits _{\stackrel {i\in I,}{r_{i}\neq 0))r_{i}=0+\textstyle \sum \limits _{\stackrel {i\in I,}{r_{i}\neq 0))r_{i}=\textstyle \sum \limits _{\stackrel {i\in I,}{r_{i}\neq 0))r_{i}.$
If $I=\mathbb {N}$ then the generalized series $\textstyle \sum \limits _{i\in I}r_{i))$ converges in $\mathbb {R}$ if and only if $\textstyle \sum \limits _{i=1}^{\infty }r_{i))$converges unconditionally (or equivalently, converges absolutely) in the usual sense.
If a generalized series $\textstyle \sum \limits _{i\in I}r_{i))$ converges in $\mathbb {R}$ then both $\textstyle \sum \limits _{\stackrel {i\in I}{r_{i}>0))r_{i))$ and $\textstyle \sum \limits _{\stackrel {i\in I}{r_{i}<0))r_{i))$ also converge to elements of $\mathbb {R}$ and the set $\left\{i\in I:r_{i}\neq 0\right\))$ is necessarily countable (that is, either finite or countably infinite); this remains true if $\mathbb {R}$ is replaced with any normed space.^{[proof 1]}
It follows that in order for a generalized series $\textstyle \sum \limits _{i\in I}r_{i))$ to converge in $\mathbb {R}$ or $\mathbb {C} ,$ it is necessary that all but at most countably many $r_{i))$ will be equal to $0,$ which means that $\textstyle \sum \limits _{i\in I}r_{i}~=~\textstyle \sum \limits _{\stackrel {i\in I}{r_{i}\neq 0))r_{i))$ is a sum of at most countably many non-zero terms.
Said differently, if $\left\{i\in I:r_{i}\neq 0\right\))$ is uncountable then the generalized series $\textstyle \sum \limits _{i\in I}r_{i))$ does not converge.
In summary, due to the nature of the real numbers and its topology, every generalized series of real numbers (indexed by an arbitrary set) that converges can be reduced to an ordinary absolutely convergent series of countably many real numbers. So in the context of measure theory, there is little benefit gained by considering uncountably many sets and generalized series. In particular, this is why the definition of "countably additive" is rarely extended from countably many sets $F_{1},F_{2},\ldots \,$ in ${\mathcal {F))$ (and the usual countable series $\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)$) to arbitrarily many sets $\left(F_{i}\right)_{i\in I))$ (and the generalized series $\textstyle \sum \limits _{i\in I}\mu \left(F_{i}\right)$).
Inner measures, outer measures, and other properties
A set function $\mu$ is said to be/satisfies^{[1]}
modular if it satisfies the following condition, known as modularity: $\mu (E\cup F)+\mu (E\cap F)=\mu (E)+\mu (F)$ for all $E,F\in {\mathcal {F))$ such that $E\cup F,E\cap F\in {\mathcal {F)).$
Every finitely additive function on a field of sets is modular.
submodular if $\mu (E\cup F)+\mu (E\cap F)\leq \mu (E)+\mu (F)$ for all $E,F\in {\mathcal {F))$ such that $E\cup F,E\cap F\in {\mathcal {F)).$
finitely subadditive if $|\mu (F)|\leq \textstyle \sum \limits _{i=1}^{n}\left|\mu \left(F_{i}\right)\right|$ for all finite sequences $F,F_{1},\ldots ,F_{n}\in {\mathcal {F))$ that satisfy $F\;\subseteq \;\textstyle \bigcup \limits _{i=1}^{n}F_{i}.$
countably subadditive or σ-subadditive if $|\mu (F)|\leq \textstyle \sum \limits _{i=1}^{\infty }\left|\mu \left(F_{i}\right)\right|$ for all sequences $F,F_{1},F_{2},F_{3},\ldots \,$ in ${\mathcal {F))$ that satisfy $F\;\subseteq \;\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}.$
If ${\mathcal {F))$ is closed under finite unions then this condition holds if and only if $|\mu (F\cup G)|\leq |\mu (F)|+|\mu (G)|$ for all $F,G\in {\mathcal {F)).$ If $\mu$ is non-negative then the absolute values may be removed.
If $\mu$ is a measure then this condition holds if and only if $\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)\leq \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)$ for all $F_{1},F_{2},F_{3},\ldots \,$ in ${\mathcal {F)).$^{[3]} If $\mu$ is a probability measure then this inequality is Boole's inequality.
If $\mu$ is countably subadditive and $\varnothing \in {\mathcal {F))$ with $\mu (\varnothing )=0$ then $\mu$ is finitely subadditive.
superadditive if $\mu (E)+\mu (F)\leq \mu (E\cup F)$ whenever $E,F\in {\mathcal {F))$ are disjoint with $E\cup F\in {\mathcal {F)).$
continuous from above if $\lim _{n\to \infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\right)$ for all non-increasing sequences of sets $F_{1}\supseteq F_{2}\supseteq F_{3}\cdots \,$ in ${\mathcal {F))$ such that $\textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\in {\mathcal {F))$ with $\mu \left(\textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\right)$ and all $\mu \left(F_{i}\right)$ finite.
Lebesgue measure $\lambda$ is continuous from above but it would not be if the assumption that all $\mu \left(F_{i}\right)$ are eventually finite was omitted from the definition, as this example shows: For every integer $i,$ let $F_{i))$ be the open interval $(i,\infty )$ so that $\lim _{n\to \infty }\lambda \left(F_{i}\right)=\lim _{n\to \infty }\infty =\infty \neq 0=\lambda (\varnothing )=\lambda \left(\textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\right)$ where $\textstyle \bigcap \limits _{i=1}^{\infty }F_{i}=\varnothing .$
continuous from below if $\lim _{n\to \infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)$ for all non-decreasing sequences of sets $F_{1}\subseteq F_{2}\subseteq F_{3}\cdots \,$ in ${\mathcal {F))$ such that $\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\in {\mathcal {F)).$
infinity is approached from below if whenever $F\in {\mathcal {F))$ satisfies $\mu (F)=\infty$ then for every real $r>0,$ there exists some $F_{r}\in {\mathcal {F))$ such that $F_{r}\subseteq F$ and $r\leq \mu \left(F_{r}\right)<\infty .$
atomic if every measurable set of positive measure contains an atom.
If a binary operation$\,+\,$ is defined, then a set function $\mu$ is said to be
translation invariant if $\mu (\omega +F)=\mu (F)$ for all $\omega \in \Omega$ and $F\in {\mathcal {F))$ such that $\omega +F\in {\mathcal {F)).$
Topology related definitions
If $\tau$ is a topology on $\Omega$ then a set function $\mu$ is said to be:
a Borel measure if it is a measure defined on the σ-algebra of all Borel sets, which is the smallest σ-algebra containing all open subsets (that is, containing $\tau$).
locally finite if for every point $\omega \in \Omega$ there exists some neighborhood $U\in {\mathcal {F))\cap \tau$ of this point such that $\mu (U)$ is finite.
If $\mu$ is a finitely additive, monotone, and locally finite then $\mu (K)$ is necessarily finite for every compact measurable subset $K.$
$\tau$-additive if $\mu \left({\textstyle \bigcup }\,{\mathcal {D))\right)=\sup _{D\in {\mathcal {D))}\mu (D)$ whenever ${\mathcal {D))\subseteq \tau \cap {\mathcal {F))$ is directed with respect to $\,\subseteq \,$ and satisfies ${\textstyle \bigcup }\,{\mathcal {D))~{\stackrel {\scriptscriptstyle {\text{def))}{=))~\textstyle \bigcup \limits _{D\in {\mathcal {D))}D\in {\mathcal {F)).$
${\mathcal {D))$ is directed with respect to $\,\subseteq \,$ if and only if it is not empty and for all $A,B\in {\mathcal {D))$ there exists some $C\in {\mathcal {D))$ such that $A\subseteq C$ and $B\subseteq C.$
inner regular or tight if for every $F\in {\mathcal {F)),$$\mu (F)=\sup\{\mu (K):F\supseteq K{\text{ with ))K\in {\mathcal {F)){\text{ a compact subset of ))(\Omega ,\tau )\}.$
outer regular if for every $F\in {\mathcal {F)),$$\mu (F)=\inf\{\mu (U):F\subseteq U{\text{ and ))U\in {\mathcal {F))\cap \tau \}.$
regular if it is both inner regular and outer regular.
If $\mu$ and $\nu$ are $\sigma$-finite measures on the same measurable space and if $\mu \ll \nu ,$ then the Radon–Nikodym derivative${\frac {d\mu }{d\nu ))$ exists and for every measurable $F,$
$\mu$ and $\nu$ are singular, written $\mu \perp \nu ,$ if there exist disjoint sets $M$ and $N$ in the domains of $\mu$ and $\nu$ such that $M\cup N=\Omega ,$$\mu (F)=0$ for all $F\subseteq M$ in the domain of $\mu ,$ and $\nu (F)=0$ for all $F\subseteq N$ in the domain of $\nu .$
assigning densities to sufficiently well-behaved subsets $A\subseteq \{1,2,3,\ldots \},$ is a set function.
A probability measure assigns a probability to each set in a σ-algebra. Specifically, the probability of the empty set is zero and the probability of the sample space is $1,$ with other sets given probabilities between $0$ and $1.$
A possibility measure assigns a number between zero and one to each set in the powerset of some given set. See possibility theory.
The Jordan measure on $\mathbb {R} ^{n))$ is a set function defined on the set of all Jordan measurable subsets of $\mathbb {R} ^{n};$ it sends a Jordan measurable set to its Jordan measure.
Lebesgue measure
The Lebesgue measure on $\mathbb {R}$ is a set function that assigns a non-negative real number to every set of real numbers that belongs to the Lebesgue $\sigma$-algebra.^{[5]}
Its definition begins with the set $\operatorname {Intervals} (\mathbb {R} )$ of all intervals of real numbers, which is a semialgebra on $\mathbb {R} .$
The function that assigns to every interval $I$ its $\operatorname {length} (I)$ is a finitely additive set function (explicitly, if $I$ has endpoints $a\leq b$ then $\operatorname {length} (I)=b-a$).
This set function can be extended to the Lebesgue outer measure on $\mathbb {R} ,$ which is the translation-invariant set function $\lambda ^{\!*\!}:\wp (\mathbb {R} )\to [0,\infty ]$ that sends a subset $E\subseteq \mathbb {R}$ to the infimum
$\lambda ^{\!*\!}(E)=\inf \left\{\sum _{k=1}^{\infty }\operatorname {length} (I_{k}):{(I_{k})_{k\in \mathbb {N} )){\text{ is a sequence of open intervals with ))E\subseteq \bigcup _{k=1}^{\infty }I_{k}\right\}.$
Lebesgue outer measure is not countably additive (and so is not a measure) although its restriction to the 𝜎-algebra of all subsets $M\subseteq \mathbb {R}$ that satisfy the Carathéodory criterion:
$\lambda ^{\!*\!}(M)=\lambda ^{\!*\!}(M\cap E)+\lambda ^{\!*\!}(M\cap E^{c})\quad {\text{ for every ))S\subseteq \mathbb {R}$
Finitely additive translation-invariant set functions
The only translation-invariant measure on $\Omega =\mathbb {R}$ with domain $\wp (\mathbb {R} )$ that is finite on every compact subset of $\mathbb {R}$ is the trivial set function $\wp (\mathbb {R} )\to [0,\infty ]$ that is identically equal to $0$ (that is, it sends every $S\subseteq \mathbb {R}$ to $0$)^{[6]}
However, if countable additivity is weakened to finite additivity then a non-trivial set function with these properties does exist and moreover, some are even valued in $[0,1].$ In fact, such non-trivial set functions will exist even if $\mathbb {R}$ is replaced by any other abeliangroup$G.$^{[7]}
Theorem^{[8]} — If $(G,+)$ is any abelian group then there exists a finitely additive and translation-invariant^{[note 1]} set function $\mu :\wp (G)\to [0,1]$ of mass $\mu (G)=1.$
Suppose that $\mu$ is a set function on a semialgebra${\mathcal {F))$ over $\Omega$ and let
$\operatorname {algebra} ({\mathcal {F))):=\left\{F_{1}\sqcup \cdots \sqcup F_{n}:n\in \mathbb {N} {\text{ and ))F_{1},\ldots ,F_{n}\in {\mathcal {F)){\text{ are pairwise disjoint ))\right\},$
which is the algebra on $\Omega$ generated by ${\mathcal {F)).$
The archetypal example of a semialgebra that is not also an algebra is the family
${\mathcal {S))_{d}:=\{\varnothing \}\cup \left\{\left(a_{1},b_{1}\right]\times \cdots \times \left(a_{1},b_{1}\right]~:~-\infty \leq a_{i}<b_{i}\leq \infty {\text{ for all ))i=1,\ldots ,d\right\))$
on $\Omega :=\mathbb {R} ^{d))$ where $(a,b]:=\{x\in \mathbb {R} :a<x\leq b\))$ for all $-\infty \leq a<b\leq \infty .$^{[9]} Importantly, the two non-strict inequalities $\,\leq \,$ in $-\infty \leq a_{i}<b_{i}\leq \infty$ cannot be replaced with strict inequalities $\,<\,$ since semialgebras must contain the whole underlying set $\mathbb {R} ^{d};$ that is, $\mathbb {R} ^{d}\in {\mathcal {S))_{d))$ is a requirement of semialgebras (as is $\varnothing \in {\mathcal {S))_{d))$).
If $\mu$ is finitely additive then it has a unique extension to a set function ${\overline {\mu ))$ on $\operatorname {algebra} ({\mathcal {F)))$ defined by sending $F_{1}\sqcup \cdots \sqcup F_{n}\in \operatorname {algebra} ({\mathcal {F)))$ (where $\,\sqcup \,$ indicates that these $F_{i}\in {\mathcal {F))$ are pairwise disjoint) to:^{[9]}
This extension ${\overline {\mu ))$ will also be finitely additive: for any pairwise disjoint $A_{1},\ldots ,A_{n}\in \operatorname {algebra} ({\mathcal {F))),$^{[9]}
If in addition $\mu$ is extended real-valued and monotone (which, in particular, will be the case if $\mu$ is non-negative) then ${\overline {\mu ))$ will be monotone and finitely subadditive: for any $A,A_{1},\ldots ,A_{n}\in \operatorname {algebra} ({\mathcal {F)))$ such that $A\subseteq A_{1}\cup \cdots \cup A_{n},$^{[9]}
If $\mu :{\mathcal {F))\to [0,\infty ]$ is a pre-measure on a ring of sets (such as an algebra of sets) ${\mathcal {F))$ over $\Omega$ then $\mu$ has an extension to a measure ${\overline {\mu )):\sigma ({\mathcal {F)))\to [0,\infty ]$ on the σ-algebra$\sigma ({\mathcal {F)))$ generated by ${\mathcal {F)).$ If $\mu$ is σ-finite then this extension is unique.
To define this extension, first extend $\mu$ to an outer measure$\mu ^{*))$ on $2^{\Omega }=\wp (\Omega )$ by
and then restrict it to the set ${\mathcal {F))_{M))$ of $\mu ^{*))$-measurable sets (that is, Carathéodory-measurable sets), which is the set of all $M\subseteq \Omega$ such that
$\mu ^{*}(S)=\mu ^{*}(S\cap M)+\mu ^{*}(S\cap M^{\mathrm {c} })\quad {\text{ for every subset ))S\subseteq \Omega .$
It is a $\sigma$-algebra and $\mu ^{*))$ is sigma-additive on it, by Caratheodory lemma.
If $\mu ^{*}:\wp (\Omega )\to [0,\infty ]$ is an outer measure on a set $\Omega ,$ where (by definition) the domain is necessarily the power set$\wp (\Omega )$ of $\Omega ,$ then a subset $M\subseteq \Omega$ is called $\mu ^{*))$–measurable or Carathéodory-measurable if it satisfies the following Carathéodory's criterion:
$\mu ^{*}(S)=\mu ^{*}(S\cap M)+\mu ^{*}(S\cap M^{\mathrm {c} })\quad {\text{ for every subset ))S\subseteq \Omega ,$
where $M^{\mathrm {c} }:=\Omega \setminus M$ is the complement of $M.$
The family of all $\mu ^{*))$–measurable subsets is a σ-algebra and the restriction of the outer measure $\mu ^{*))$ to this family is a measure.
^The function $\mu$ being translation-invariant means that $\mu (S)=\mu (g+S)$ for every $g\in G$ and every subset $S\subseteq G.$
Proofs
^Suppose the net ${\textstyle \textstyle \sum \limits _{i\in I}r_{i}~{\stackrel {\scriptscriptstyle {\text{def))}{=))~{\textstyle \lim \limits _{A\in \operatorname {Finite} (I)))\ \textstyle \sum \limits _{i\in A}r_{i}=\lim \left\{\textstyle \sum \limits _{i\in A}r_{i}\,:A\subseteq I,A{\text{ finite ))\right\))$ converges to some point in a metrizable topological vector space$X$ (such as $\mathbb {R} ,$$\mathbb {C} ,$ or a normed space), where recall that this net's domain is the directed set$(\operatorname {Finite} (I),\subseteq ).$
Like every convergent net, this convergent net of partial sums $A\mapsto \textstyle \sum \limits _{i\in A}r_{i))$ is a Cauchy net, which for this particular net means (by definition) that for every neighborhood $W$ of the origin in $X,$ there exists a finite subset $A_{0))$ of $I$ such that
${\textstyle \textstyle \sum \limits _{i\in B}r_{i}-\textstyle \sum \limits _{i\in C}r_{i}\in W}$ for all finite supersets $B,C\supseteq A_{0};$
this implies that $r_{i}\in W$ for every $i\in I\setminus A_{0))$ (by taking $B:=A_{0}\cup \{i\))$ and $C:=A_{0))$).
Since $X$ is metrizable, it has a countable neighborhood basis $U_{1},U_{2},\ldots$ at the origin, whose intersection is necessarily $U_{1}\cap U_{2}\cap \cdots =\{0\))$ (since $X$ is a Hausdorff TVS).
For every positive integer $n\in \mathbb {N} ,$ pick a finite subset $A_{n}\subseteq I$ such that $r_{i}\in U_{n))$ for every $i\in I\setminus A_{n}.$
If $i$ belongs to $(I\setminus A_{1})\cap (I\setminus A_{2})\cap \cdots =I\setminus \left(A_{1}\cup A_{2}\cup \cdots \right)$ then $r_{i))$ belongs to $U_{1}\cap U_{2}\cap \cdots =\{0\}.$
Thus $r_{i}=0$ for every index $i\in I$ that does not belong to the countable set $A_{1}\cup A_{2}\cup \cdots .$$\blacksquare$