Sigma-additive set function

Let ${\displaystyle \mu }$ be a set function defined on an algebra of sets ${\displaystyle \scriptstyle {\mathcal {A}}}$ with values in ${\displaystyle [-\infty ,\infty ]}$ (see the extended real number line). The function ${\displaystyle \mu }$ is called additive or finitely additive, if whenever ${\displaystyle A}$ and ${\displaystyle B}$ are disjoint sets in ${\displaystyle \scriptstyle {\mathcal {A}},}$ then $${\displaystyle \mu (A\cup B)=\mu (A)+\mu (B).}$$ 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 $${\displaystyle \mu \left(\bigcup _{n=1}^{N}A_{n}\right)=\sum _{n=1}^{N}\mu \left(A_{n}\right)}$$ for any ${\displaystyle A_{1},A_{2},\ldots ,A_{N}}$ disjoint sets in ${\textstyle {\mathcal {A}}.}$

Suppose that ${\displaystyle \scriptstyle {\mathcal {A}}}$ is a σ-algebra. If for every sequence ${\displaystyle A_{1},A_{2},\ldots ,A_{n},\ldots }$ of pairwise disjoint sets in ${\displaystyle \scriptstyle {\mathcal {A}},}$ $${\displaystyle \mu \left(\bigcup _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n}),}$$ holds then ${\displaystyle \mu }$ 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 ${\textstyle {\mathcal {A}},}$ we have a topology ${\displaystyle \tau .}$ If for every directed family of measurable open sets ${\textstyle {\mathcal {G}}\subseteq {\mathcal {A}}\cap \tau ,}$ $${\displaystyle \mu \left(\bigcup {\mathcal {G}}\right)=\sup _{G\in {\mathcal {G}}}\mu (G),}$$ we say that ${\displaystyle \mu }$ is ${\displaystyle \tau }$-additive. In particular, if ${\displaystyle \mu }$ is inner regular (with respect to compact sets) then it is ${\displaystyle \tau }$-additive.^[1]

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

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

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

A set function ${\displaystyle \mu }$ on a family of sets ${\displaystyle {\mathcal {S}}}$ is called a modular set function and a valuation if whenever ${\displaystyle A,}$ ${\displaystyle B,}$ ${\displaystyle A\cup B,}$ and ${\displaystyle A\cap B}$ are elements of ${\displaystyle {\mathcal {S}},}$ then $${\displaystyle \mu (A\cup B)+\mu (A\cap B)=\mu (A)+\mu (B)}$$ The above property is called modularity and the argument below proves that additivity implies modularity.

Given ${\displaystyle A}$ and ${\displaystyle B,}$ ${\displaystyle \mu (A\cup B)+\mu (A\cap B)=\mu (A)+\mu (B).}$ Proof: write ${\displaystyle A=(A\cap B)\cup (A\setminus B)}$ and ${\displaystyle B=(A\cap B)\cup (B\setminus A)}$ and ${\displaystyle A\cup B=(A\cap B)\cup (A\setminus B)\cup (B\setminus A),}$ where all sets in the union are disjoint. Additivity implies that both sides of the equality equal ${\displaystyle \mu (A\setminus B)+\mu (B\setminus A)+2\mu (A\cap B).}$

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

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

If ${\displaystyle A_{1},A_{2},\ldots ,A_{n},\ldots }$ 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 $${\displaystyle \mu \left(\bigcup _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n})}$$ 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 ${\displaystyle 0.}$^[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 ${\displaystyle \mu }$, defined over the Lebesgue sets of the real numbers ${\displaystyle \mathbb {R} }$ by the formula $${\displaystyle \mu (A)=\lim _{k\to \infty }{\frac {1}{k}}\cdot \lambda (A\cap (0,k)),}$$ where ${\displaystyle \lambda }$ denotes the Lebesgue measure and ${\displaystyle \lim }$ the Banach limit. It satisfies ${\displaystyle 0\leq \mu (A)\leq 1}$ and if ${\displaystyle \sup A<\infty }$ then ${\displaystyle \mu (A)=0.}$

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 $${\displaystyle A_{n}=[n,n+1)}$$ for ${\displaystyle n=0,1,2,\ldots }$ The union of these sets is the positive reals, and ${\displaystyle \mu }$ applied to the union is then one, while ${\displaystyle \mu }$ applied to any of the individual sets is zero, so the sum of ${\displaystyle \mu (A_{n})}$ 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

This article incorporates material from additive on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.