In mathematics, additivity and sigma additivity (also called countable additivity) of a function defined on subsets of a given set are abstractions of the intuitive properties of size (length, area, volume) of a set.

Let ${\displaystyle \mu }$ be a function defined on an algebra of sets ${\displaystyle \scriptstyle {\mathcal {A}}}$ with values in [−∞, +∞] (see the extended real number line). The function ${\displaystyle \mu }$ is called additive, or finitely additive, if, whenever A and B are disjoint sets in ${\displaystyle \scriptstyle {\mathcal {A}}}$, one has

${\displaystyle \mu (A\cup B)=\mu (A)+\mu (B).\,}$

(A consequence of this is that an additive function cannot take both −∞ and +∞ as values, for the expression ∞ − ∞ is undefined.)

One can prove by mathematical induction that an additive function satisfies

${\displaystyle \mu \left(\bigcup _{n=1}^{N}A_{n}\right)=\sum _{n=1}^{N}\mu (A_{n})}$

for any ${\displaystyle A_{1},A_{2},\dots ,A_{N}}$ disjoint sets in ${\displaystyle \scriptstyle {\mathcal {A}}}$.

Suppose that ${\displaystyle \scriptstyle {\mathcal {A}}}$ is a σ-algebra. If for any sequence ${\displaystyle A_{1},A_{2},\dots ,A_{n},\dots }$ of pairwise disjoint sets in ${\displaystyle \scriptstyle {\mathcal {A}}}$, one has

${\displaystyle \mu \left(\bigcup _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n})}$,

Any σ-additive function is additive but not vice versa, as shown below.

Suppose that in addition to a sigma algebra ${\displaystyle \scriptstyle {\mathcal {A}}}$, we have a topology τ. If for any directed family of measurable open sets ${\displaystyle \scriptstyle {\mathcal {G}}}$${\displaystyle \scriptstyle {\mathcal {A}}}$∩τ,

${\displaystyle \mu \left(\bigcup {\mathcal {G}}\right)=\sup _{G\in {\mathcal {G}}}\mu (G)}$,

we say that μ is τ-additive. In particular, if μ is inner regular then it is τ-additive.[1]

Properties

Basic properties

Useful properties of an additive function μ include the following:

1. Either μ(∅) = 0, or μ assigns ∞ to all sets in its domain, or μ assigns −∞ to all sets in its domain.
2. If μ is non-negative and AB, then μ(A) ≤ μ(B).
3. If AB and μ(B) − μ(A) is defined, then μ(B \ A) = μ(B) − μ(A).
4. Given A and B, μ(AB) + μ(AB) = μ(A) + μ(B).

Examples

An example of a σ-additive function is the function μ 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 ${\displaystyle A_{1},A_{2},\dots ,A_{n},\dots }$ 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.

An example of an additive function which is not σ-additive is obtained by considering μ, defined over the Lebesgue sets of the real numbers by the formula

${\displaystyle \mu (A)=\lim _{k\to \infty }{\frac {1}{k}}\cdot \lambda \left(A\cap \left(0,k\right)\right),}$

where λ denotes the Lebesgue measure and lim the Banach limit.

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

${\displaystyle A_{n}=\left[n,n+1\right)}$

for n=0, 1, 2, ... The union of these sets is the positive reals, and μ applied to the union is then one, while μ applied to any of the individual sets is zero, so the sum of μ(An) is also zero, which proves the counterexample.

Generalizations

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.