Set function

A set function ${\displaystyle \mu }$ on ${\displaystyle {\mathcal {F}}}$ is said to be[1]

• non-negative if it is valued in ${\displaystyle [0,\infty ].}$
• finitely additive if ${\displaystyle \textstyle \sum \limits _{i=1}^{n}\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{n}F_{i}\right)}$ for all pairwise disjoint finite sequences ${\displaystyle F_{1},\ldots ,F_{n}\in {\mathcal {F}}}$ such that ${\displaystyle \textstyle \bigcup \limits _{i=1}^{n}F_{i}\in {\mathcal {F}}.}$
  • If ${\displaystyle {\mathcal {F}}}$ is closed under binary unions then ${\displaystyle \mu }$ is finitely additive if and only if ${\displaystyle \mu (E\cup F)=\mu (E)+\mu (F)}$ for all disjoint pairs ${\displaystyle E,F\in {\mathcal {F}}.}$
  • If ${\displaystyle \mu }$ is finitely additive and if ${\displaystyle \varnothing \in {\mathcal {F}}}$ then taking ${\displaystyle E:=F:=\varnothing }$ shows that ${\displaystyle \mu (\varnothing )=\mu (\varnothing )+\mu (\varnothing )}$ which is only possible if ${\displaystyle \mu (\varnothing )=0}$ or ${\displaystyle \mu (\varnothing )=\pm \infty ,}$ where in the latter case, ${\displaystyle \mu (E)=\mu (E\cup \varnothing )=\mu (E)+\mu (\varnothing )=\mu (E)+(\pm \infty )=\pm \infty }$ for every ${\displaystyle E\in {\mathcal {F}}}$ (so only the case ${\displaystyle \mu (\varnothing )=0}$ is useful).
• countably additive or σ-additive[2] if in addition to being finitely additive, for all pairwise disjoint sequences ${\displaystyle F_{1},F_{2},\ldots \,}$ in ${\displaystyle {\mathcal {F}}}$ such that ${\displaystyle \textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\in {\mathcal {F}},}$ all of the following hold:
  1. ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)}$
     • The series on the left hand side is defined in the usual way as the limit ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\displaystyle \lim _{n\to \infty }}\mu \left(F_{1}\right)+\cdots +\mu \left(F_{n}\right).}$
     • As a consequence, if ${\displaystyle \rho$ :\mathbb {N} \to \mathbb {N} } is any permutation/bijection then ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)=\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{\rho (i)}\right);}$ this is because ${\displaystyle \textstyle \bigcup \limits _{i=1}^{\infty }F_{i}=\textstyle \bigcup \limits _{i=1}^{\infty }F_{\rho (i)}}$ and applying this condition (a) twice guarantees that both ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)}$ and ${\displaystyle \mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{\rho (i)}\right)=\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{\rho (i)}\right)}$ 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 ${\displaystyle F_{1},F_{2},\ldots }$ to the new order ${\displaystyle F_{\rho (1)},F_{\rho (2)},\ldots }$ does not affect the sum of their measures. This is desirable since just as the union ${\displaystyle F~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\textstyle \bigcup \limits _{i\in \mathbb {N} }F_{i}}$ does not depend on the order of these sets, the same should be true of the sums ${\displaystyle \mu (F)=\mu \left(F_{1}\right)+\mu \left(F_{2}\right)+\cdots }$ and ${\displaystyle \mu (F)=\mu \left(F_{\rho (1)}\right)+\mu \left(F_{\rho (2)}\right)+\cdots \,.}$
  2. if ${\displaystyle \mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)}$ is not infinite then this series ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)}$ must also converge absolutely, which by definition means that ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\left|\mu \left(F_{i}\right)\right|}$ must be finite. This is automatically true if ${\displaystyle \mu }$ 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 ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)={\displaystyle \lim _{N\to \infty }}\mu \left(F_{1}\right)+\mu \left(F_{2}\right)+\cdots +\mu \left(F_{N}\right)}$ 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 ${\displaystyle \mu }$ is valued in ${\displaystyle [-\infty ,\infty ].}$
  3. if ${\displaystyle \mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)=\textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)}$ is infinite then it is also required that the value of at least one of the series ${\displaystyle \textstyle \sum \limits _{\stackrel {i\in \mathbb {N} }{\mu \left(F_{i}\right)>0}}\mu \left(F_{i}\right)\;{\text{ and }}\;\textstyle \sum \limits _{\stackrel {i\in \mathbb {N} }{\mu \left(F_{i}\right)<0}}\mu \left(F_{i}\right)\;}$ be finite (so that the sum of their values is well-defined). This is automatically true if ${\displaystyle \mu }$ 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 ${\displaystyle 1.}$
• an outer measure if it is non-negative, countably subadditive, has a null empty set, and has the power set ${\displaystyle \wp (\Omega )}$ 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 ${\displaystyle \mu }$ 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 ${\displaystyle F\in {\mathcal {F}}{\text{ satisfies }}\mu (F)=0}$ and ${\displaystyle N\subseteq F}$ is any subset of ${\displaystyle F}$ then ${\displaystyle N\in {\mathcal {F}}}$ and ${\displaystyle \mu (N)=0.}$
  • Unlike many other properties, completeness places requirements on the set ${\displaystyle \operatorname {domain} \mu ={\mathcal {F}}}$ (and not just on ${\displaystyle \mu }$'s values).
• 𝜎-finite if there exists a sequence ${\displaystyle F_{1},F_{2},F_{3},\ldots \,}$ in ${\displaystyle {\mathcal {F}}}$ such that ${\displaystyle \mu \left(F_{i}\right)}$ is finite for every index ${\displaystyle i,}$ and also ${\displaystyle \textstyle \bigcup \limits _{n=1}^{\infty }F_{n}=\textstyle \bigcup \limits _{F\in {\mathcal {F}}}F.}$
• decomposable if there exists a subfamily ${\displaystyle {\mathcal {P}}\subseteq {\mathcal {F}}}$ of pairwise disjoint sets such that ${\displaystyle \mu (P)}$ is finite for every ${\displaystyle P\in {\mathcal {P}}}$ and also ${\displaystyle \textstyle \bigcup \limits _{P\in {\mathcal {P}}}\,P=\textstyle \bigcup \limits _{F\in {\mathcal {F}}}F}$ (where ${\displaystyle {\mathcal {F}}=\operatorname {domain} \mu }$).
  • Every 𝜎-finite set function is decomposable although not conversely. For example, the counting measure on ${\displaystyle \mathbb {R} }$ (whose domain is ${\displaystyle \wp (\mathbb {R} )}$) is decomposable but not 𝜎-finite.
• a vector measure if it is a countably additive set function ${\displaystyle \mu$ :{\mathcal {F}}\to X} valued in a topological vector space ${\displaystyle X}$ (such as a normed space) whose domain is a σ-algebra.
  • If ${\displaystyle \mu }$ is valued in a normed space ${\displaystyle (X,\|\cdot \|)}$ then it is countably additive if and only if for any pairwise disjoint sequence ${\displaystyle F_{1},F_{2},\ldots \,}$ in ${\displaystyle {\mathcal {F}},}$ ${\displaystyle \lim _{n\to \infty }\left\|\mu \left(F_{1}\right)+\cdots +\mu \left(F_{n}\right)-\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)\right\|=0.}$ If ${\displaystyle \mu }$ is finitely additive and valued in a Banach space then it is countably additive if and only if for any pairwise disjoint sequence ${\displaystyle F_{1},F_{2},\ldots \,}$ in ${\displaystyle {\mathcal {F}},}$ ${\displaystyle \lim _{n\to \infty }\left\|\mu \left(F_{n}\cup F_{n+1}\cup F_{n+2}\cup \cdots \right)\right\|=0.}$
• a complex measure if it is a countably additive complex-valued set function ${\displaystyle \mu$ :{\mathcal {F}}\to \mathbb {C} } whose domain is a σ-algebra.
  • By definition, a complex measure never takes ${\displaystyle \pm \infty }$ as a value and so has a null empty set.
• a random measure if it is a measure-valued random element.

Arbitrary sums

As described in this article's section on generalized series, for any family ${\displaystyle \left(r_{i}\right)_{i\in I}}$ of real numbers indexed by an arbitrary indexing set ${\displaystyle I,}$ it is possible to define their sum ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}}$ as the limit of the net of finite partial sums ${\displaystyle F\in \operatorname {FiniteSubsets} (I)\mapsto \textstyle \sum \limits _{i\in F}r_{i}}$ where the domain ${\displaystyle \operatorname {FiniteSubsets} (I)}$ is directed by ${\displaystyle \,\subseteq .\,}$ Whenever this net converges then its limit is denoted by the symbols ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}}$ while if this net instead diverges to ${\displaystyle \pm \infty }$ then this may be indicated by writing ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}=\pm \infty .}$ Any sum over the empty set is defined to be zero; that is, if ${\displaystyle I=\varnothing }$ then ${\displaystyle \textstyle \sum \limits _{i\in \varnothing }r_{i}=0}$ by definition.

For example, if ${\displaystyle z_{i}=0}$ for every ${\displaystyle i\in I}$ then ${\displaystyle \textstyle \sum \limits _{i\in I}z_{i}=0.}$ And it can be shown that ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}=\textstyle \sum \limits _{\stackrel {i\in I,}{r_{i}=0}}r_{i}+\textstyle \sum \limits _{\stackrel {i\in I,}{r_{i}\neq 0}}r_{i}=0+\textstyle \sum \limits _{\stackrel {i\in I,}{r_{i}\neq 0}}r_{i}=\textstyle \sum \limits _{\stackrel {i\in I,}{r_{i}\neq 0}}r_{i}.}$ If ${\displaystyle I=\mathbb {N} }$ then the generalized series ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}}$ converges in ${\displaystyle \mathbb {R} }$ if and only if ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }r_{i}}$ converges unconditionally (or equivalently, converges absolutely) in the usual sense. If a generalized series ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}}$ converges in ${\displaystyle \mathbb {R} }$ then both ${\displaystyle \textstyle \sum \limits _{\stackrel {i\in I}{r_{i}>0}}r_{i}}$ and ${\displaystyle \textstyle \sum \limits _{\stackrel {i\in I}{r_{i}<0}}r_{i}}$ also converge to elements of ${\displaystyle \mathbb {R} }$ and the set ${\displaystyle \left\{i\in I:r_{i}\neq 0\right\}}$ is necessarily countable (that is, either finite or countably infinite); this remains true if ${\displaystyle \mathbb {R} }$ is replaced with any normed space.[proof 1] It follows that in order for a generalized series ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}}$ to converge in ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} ,}$ it is necessary that all but at most countably many ${\displaystyle r_{i}}$ will be equal to ${\displaystyle 0,}$ which means that ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}~=~\textstyle \sum \limits _{\stackrel {i\in I}{r_{i}\neq 0}}r_{i}}$ is a sum of at most countably many non-zero terms. Said differently, if ${\displaystyle \left\{i\in I:r_{i}\neq 0\right\}}$ is uncountable then the generalized series ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}}$ 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 ${\displaystyle F_{1},F_{2},\ldots \,}$ in ${\displaystyle {\mathcal {F}}}$ (and the usual countable series ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)}$) to arbitrarily many sets ${\displaystyle \left(F_{i}\right)_{i\in I}}$ (and the generalized series ${\displaystyle \textstyle \sum \limits _{i\in I}\mu \left(F_{i}\right)}$).

A set function ${\displaystyle \mu }$ is said to be/satisfies[1]

• monotone if ${\displaystyle \mu (E)\leq \mu (F)}$ whenever ${\displaystyle E,F\in {\mathcal {F}}}$ satisfy ${\displaystyle E\subseteq F.}$
• modular if it satisfies the following condition, known as modularity: ${\displaystyle \mu (E\cup F)+\mu (E\cap F)=\mu (E)+\mu (F)}$ for all ${\displaystyle E,F\in {\mathcal {F}}}$ such that ${\displaystyle E\cup F,E\cap F\in {\mathcal {F}}.}$
  • Every finitely additive function on a field of sets is modular.
  • In geometry, a set function valued in some abelian semigroup that possess this property is known as a valuation. This geometric definition of "valuation" should not be confused with the stronger non-equivalent measure theoretic definition of "valuation" that is given below.
• submodular if ${\displaystyle \mu (E\cup F)+\mu (E\cap F)\leq \mu (E)+\mu (F)}$ for all ${\displaystyle E,F\in {\mathcal {F}}}$ such that ${\displaystyle E\cup F,E\cap F\in {\mathcal {F}}.}$
• finitely subadditive if ${\displaystyle |\mu (F)|\leq \textstyle \sum \limits _{i=1}^{n}\left|\mu \left(F_{i}\right)\right|}$ for all finite sequences ${\displaystyle F,F_{1},\ldots ,F_{n}\in {\mathcal {F}}}$ that satisfy ${\displaystyle F\;\subseteq \;\textstyle \bigcup \limits _{i=1}^{n}F_{i}.}$
• countably subadditive or σ-subadditive if ${\displaystyle |\mu (F)|\leq \textstyle \sum \limits _{i=1}^{\infty }\left|\mu \left(F_{i}\right)\right|}$ for all sequences ${\displaystyle F,F_{1},F_{2},F_{3},\ldots \,}$ in ${\displaystyle {\mathcal {F}}}$ that satisfy ${\displaystyle F\;\subseteq \;\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}.}$
  • If ${\displaystyle {\mathcal {F}}}$ is closed under finite unions then this condition holds if and only if ${\displaystyle |\mu (F\cup G)|\leq |\mu (F)|+|\mu (G)|}$ for all ${\displaystyle F,G\in {\mathcal {F}}.}$ If ${\displaystyle \mu }$ is non-negative then the absolute values may be removed.
  • If ${\displaystyle \mu }$ is a measure then this condition holds if and only if ${\displaystyle \mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)\leq \textstyle \sum \limits _{i=1}^{\infty }\mu \left(F_{i}\right)}$ for all ${\displaystyle F_{1},F_{2},F_{3},\ldots \,}$ in ${\displaystyle {\mathcal {F}}.}$[3] If ${\displaystyle \mu }$ is a probability measure then this inequality is Boole's inequality.
  • If ${\displaystyle \mu }$ is countably subadditive and ${\displaystyle \varnothing \in {\mathcal {F}}}$ with ${\displaystyle \mu (\varnothing )=0}$ then ${\displaystyle \mu }$ is finitely subadditive.
• superadditive if ${\displaystyle \mu (E)+\mu (F)\leq \mu (E\cup F)}$ whenever ${\displaystyle E,F\in {\mathcal {F}}}$ are disjoint with ${\displaystyle E\cup F\in {\mathcal {F}}.}$
• continuous from above if ${\displaystyle \lim _{n\to \infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\right)}$ for all non-increasing sequences of sets ${\displaystyle F_{1}\supseteq F_{2}\supseteq F_{3}\cdots \,}$ in ${\displaystyle {\mathcal {F}}}$ such that ${\displaystyle \textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\in {\mathcal {F}}}$ with ${\displaystyle \mu \left(\textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\right)}$ and all ${\displaystyle \mu \left(F_{i}\right)}$ finite.
  • Lebesgue measure ${\displaystyle \lambda }$ is continuous from above but it would not be if the assumption that all ${\displaystyle \mu \left(F_{i}\right)}$ are eventually finite was omitted from the definition, as this example shows: For every integer ${\displaystyle i,}$ let ${\displaystyle F_{i}}$ be the open interval ${\displaystyle (i,\infty )}$ so that ${\displaystyle \lim _{n\to \infty }\lambda \left(F_{i}\right)=\lim _{n\to \infty }\infty =\infty \neq 0=\lambda (\varnothing )=\lambda \left(\textstyle \bigcap \limits _{i=1}^{\infty }F_{i}\right)}$ where ${\displaystyle \textstyle \bigcap \limits _{i=1}^{\infty }F_{i}=\varnothing .}$
• continuous from below if ${\displaystyle \lim _{n\to \infty }\mu \left(F_{i}\right)=\mu \left(\textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\right)}$ for all non-decreasing sequences of sets ${\displaystyle F_{1}\subseteq F_{2}\subseteq F_{3}\cdots \,}$ in ${\displaystyle {\mathcal {F}}}$ such that ${\displaystyle \textstyle \bigcup \limits _{i=1}^{\infty }F_{i}\in {\mathcal {F}}.}$
• infinity is approached from below if whenever ${\displaystyle F\in {\mathcal {F}}}$ satisfies ${\displaystyle \mu (F)=\infty }$ then for every real ${\displaystyle r>0,}$ there exists some ${\displaystyle F_{r}\in {\mathcal {F}}}$ such that ${\displaystyle F_{r}\subseteq F}$ and ${\displaystyle r\leq \mu \left(F_{r}\right)<\infty .}$
• an outer measure if ${\displaystyle \mu }$ is non-negative, countably subadditive, has a null empty set, and has the power set ${\displaystyle \wp (\Omega )}$ as its domain.
• an inner measure if ${\displaystyle \mu }$ is non-negative, superadditive, continuous from above, has a null empty set, has the power set ${\displaystyle \wp (\Omega )}$ as its domain, and ${\displaystyle +\infty }$ is approached from below.
• atomic if every measurable set of positive measure contains an atom.

If a binary operation ${\displaystyle \,+\,}$ is defined, then a set function ${\displaystyle \mu }$ is said to be

• translation invariant if ${\displaystyle \mu (\omega +F)=\mu (F)}$ for all ${\displaystyle \omega \in \Omega }$ and ${\displaystyle F\in {\mathcal {F}}}$ such that ${\displaystyle \omega +F\in {\mathcal {F}}.}$

If ${\displaystyle \tau }$ is a topology on ${\displaystyle \Omega }$ then a set function ${\displaystyle \mu }$ 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 ${\displaystyle \tau }$).
• a Baire measure if it is a measure defined on the σ-algebra of all Baire sets.
• locally finite if for every point ${\displaystyle \omega \in \Omega }$ there exists some neighborhood ${\displaystyle U\in {\mathcal {F}}\cap \tau }$ of this point such that ${\displaystyle \mu (U)}$ is finite.
  • If ${\displaystyle \mu }$ is a finitely additive, monotone, and locally finite then ${\displaystyle \mu (K)}$ is necessarily finite for every compact measurable subset ${\displaystyle K.}$
• ${\displaystyle \tau }$-additive if ${\displaystyle \mu \left({\textstyle \bigcup }\,{\mathcal {D}}\right)=\sup _{D\in {\mathcal {D}}}\mu (D)}$ whenever ${\displaystyle {\mathcal {D}}\subseteq \tau \cap {\mathcal {F}}}$ is directed with respect to ${\displaystyle \,\subseteq \,}$ and satisfies ${\displaystyle {\textstyle \bigcup }\,{\mathcal {D}}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\textstyle \bigcup \limits _{D\in {\mathcal {D}}}D\in {\mathcal {F}}.}$
  • ${\displaystyle {\mathcal {D}}}$ is directed with respect to ${\displaystyle \,\subseteq \,}$ if and only if it is not empty and for all ${\displaystyle A,B\in {\mathcal {D}}}$ there exists some ${\displaystyle C\in {\mathcal {D}}}$ such that ${\displaystyle A\subseteq C}$ and ${\displaystyle B\subseteq C.}$
• inner regular or tight if for every ${\displaystyle F\in {\mathcal {F}},}$ ${\displaystyle \mu (F)=\sup\{\mu (K):F\supseteq K{\text{ with }}K\in {\mathcal {F}}{\text{ a compact subset of }}(\Omega ,\tau )\}.}$
• outer regular if for every ${\displaystyle F\in {\mathcal {F}},}$ ${\displaystyle \mu (F)=\inf\{\mu (U):F\subseteq U{\text{ and }}U\in {\mathcal {F}}\cap \tau \}.}$
• regular if it is both inner regular and outer regular.
• a Borel regular measure if it is a Borel measure that is also regular.
• a Radon measure if it is a regular and locally finite measure.
• strictly positive if every non-empty open subset has (strictly) positive measure.
• a valuation if it is non-negative, monotone, modular, has a null empty set, and has domain ${\displaystyle \tau .}$

If ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are two set functions over ${\displaystyle \Omega ,}$ then:

• ${\displaystyle \mu }$ is said to be absolutely continuous with respect to ${\displaystyle \nu }$ or dominated by ${\displaystyle \nu }$, written ${\displaystyle \mu \ll \nu ,}$ if for every set ${\displaystyle F}$ that belongs to the domain of both ${\displaystyle \mu }$ and ${\displaystyle \nu ,}$ if ${\displaystyle \nu (F)=0}$ then ${\displaystyle \mu (F)=0.}$
  • If ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are ${\displaystyle \sigma }$-finite measures on the same measurable space and if ${\displaystyle \mu \ll \nu ,}$ then the Radon–Nikodym derivative ${\displaystyle {\frac {d\mu }{d\nu }}}$ exists and for every measurable ${\displaystyle F,}$
    $${\displaystyle \mu (F)=\int _{F}{\frac {d\mu }{d\nu }}d\nu .}$$

    
• ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are called equivalent if each one is absolutely continuous with respect to the other. ${\displaystyle \mu }$ is called a supporting measure of a measure ${\displaystyle \nu }$ if ${\displaystyle \mu }$ is ${\displaystyle \sigma }$-finite and they are equivalent.[4]
• ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are singular, written ${\displaystyle \mu \perp \nu ,}$ if there exist disjoint sets ${\displaystyle M}$ and ${\displaystyle N}$ in the domains of ${\displaystyle \mu }$ and ${\displaystyle \nu }$ such that ${\displaystyle M\cup N=\Omega ,}$ ${\displaystyle \mu (F)=0}$ for all ${\displaystyle F\subseteq M}$ in the domain of ${\displaystyle \mu ,}$ and ${\displaystyle \nu (F)=0}$ for all ${\displaystyle F\subseteq N}$ in the domain of ${\displaystyle \nu .}$

The Lebesgue measure on ${\displaystyle \mathbb {R} }$ is a set function that assigns a non-negative real number to every set of real numbers that belongs to the Lebesgue ${\displaystyle \sigma }$-algebra.[5]

Its definition begins with the set ${\displaystyle \operatorname {Intervals} (\mathbb {R} )}$ of all intervals of real numbers, which is a semialgebra on ${\displaystyle \mathbb {R} .}$ The function that assigns to every interval ${\displaystyle I}$ its ${\displaystyle \operatorname {length} (I)}$ is a finitely additive set function (explicitly, if ${\displaystyle I}$ has endpoints ${\displaystyle a\leq b}$ then ${\displaystyle \operatorname {length} (I)=b-a}$). This set function can be extended to the Lebesgue outer measure on ${\displaystyle \mathbb {R} ,}$ which is the translation-invariant set function ${\displaystyle \lambda ^{\!*\!}:\wp (\mathbb {R} )\to [0,\infty ]}$ that sends a subset ${\displaystyle E\subseteq \mathbb {R} }$ to the infimum

$${\displaystyle \lambda ^{\!*\!}(E)=\inf \left\{\sum _{k=1}^{\infty }\operatorname {length} (I_{k}):{(I_{k})_{k\in \mathbb {N} }}{\text{ is a sequence of open intervals with }}E\subseteq \bigcup _{k=1}^{\infty }I_{k}\right\}.}$$

Lebesgue outer measure is not countably additive (and so is not a measure) although its restriction to the 𝜎-algebra of all subsets ${\displaystyle M\subseteq \mathbb {R} }$ that satisfy the Carathéodory criterion:

$${\displaystyle \lambda ^{\!*\!}(M)=\lambda ^{\!*\!}(M\cap E)+\lambda ^{\!*\!}(M\cap E^{c})\quad {\text{ for every }}S\subseteq \mathbb {R} }$$

is a measure that called Lebesgue measure. Vitali sets are examples of non-measurable sets of real numbers.

As detailed in the article on infinite-dimensional Lebesgue measure, the only locally finite and translation-invariant Borel measure on an infinite-dimensional separable normed space is the trivial measure. However, it is possible to define Gaussian measures on infinite-dimensional topological vector spaces. The structure theorem for Gaussian measures shows that the abstract Wiener space construction is essentially the only way to obtain a strictly positive Gaussian measure on a separable Banach space.

The only translation-invariant measure on ${\displaystyle \Omega =\mathbb {R} }$ with domain ${\displaystyle \wp (\mathbb {R} )}$ that is finite on every compact subset of ${\displaystyle \mathbb {R} }$ is the trivial set function ${\displaystyle \wp (\mathbb {R} )\to [0,\infty ]}$ that is identically equal to ${\displaystyle 0}$ (that is, it sends every ${\displaystyle S\subseteq \mathbb {R} }$ to ${\displaystyle 0}$)[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 ${\displaystyle [0,1].}$ In fact, such non-trivial set functions will exist even if ${\displaystyle \mathbb {R} }$ is replaced by any other abelian group ${\displaystyle G.}$[7]

Suppose that ${\displaystyle \mu }$ is a set function on a semialgebra ${\displaystyle {\mathcal {F}}}$ over ${\displaystyle \Omega }$ and let

$${\displaystyle \operatorname {algebra} ({\mathcal {F}}):=\left\{F_{1}\sqcup \cdots \sqcup F_{n}:n\in \mathbb {N} {\text{ and }}F_{1},\ldots ,F_{n}\in {\mathcal {F}}{\text{ are pairwise disjoint }}\right\},}$$

which is the algebra on ${\displaystyle \Omega }$ generated by ${\displaystyle {\mathcal {F}}.}$ The archetypal example of a semialgebra that is not also an algebra is the family

$${\displaystyle {\mathcal {S}}_{d}:=\{\varnothing \}\cup \left\{\left(a_{1},b_{1}\right]\times \cdots \times \left(a_{1},b_{1}\right]~:~-\infty \leq a_{i}<b_{i}\leq \infty {\text{ for all }}i=1,\ldots ,d\right\}}$$

on ${\displaystyle \Omega$ :=\mathbb {R} ^{d}} where ${\displaystyle (a,b]:=\{x\in \mathbb {R} :a<x\leq b\}}$ for all ${\displaystyle -\infty \leq a<b\leq \infty .}$[9] Importantly, the two non-strict inequalities ${\displaystyle \,\leq \,}$ in ${\displaystyle -\infty \leq a_{i}<b_{i}\leq \infty }$ cannot be replaced with strict inequalities ${\displaystyle \,<\,}$ since semialgebras must contain the whole underlying set ${\displaystyle \mathbb {R} ^{d};}$ that is, ${\displaystyle \mathbb {R} ^{d}\in {\mathcal {S}}_{d}}$ is a requirement of semialgebras (as is ${\displaystyle \varnothing \in {\mathcal {S}}_{d}}$).

If ${\displaystyle \mu }$ is finitely additive then it has a unique extension to a set function ${\displaystyle {\overline {\mu }}}$ on ${\displaystyle \operatorname {algebra} ({\mathcal {F}})}$ defined by sending ${\displaystyle F_{1}\sqcup \cdots \sqcup F_{n}\in \operatorname {algebra} ({\mathcal {F}})}$ (where ${\displaystyle \,\sqcup \,}$ indicates that these ${\displaystyle F_{i}\in {\mathcal {F}}}$ are pairwise disjoint) to:[9]