Sigma-additive set function

Let μ be a function defined on an algebra of sets ${\mathcal {A}}$ with values in [−∞ +∞] (see the extended real line). We say that μ is additive if, whenever A and B are disjoint sets in ${\mathcal {A}}$, we have

$\mu (A\cup B)=\mu (A)+\mu (B).$

One can prove by mathematical induction that an additive function satisfies

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

for any A₁, A₂, ..., A_n disjoint sets in ${\mathcal {A}}$.

Suppose ${\mathcal {A}}$ is a σ-algebra. If for any sequence A₁, A₂, ..., A_n, ... of disjoint sets in ${\mathcal {A}}$ one has

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

we say that μ is countably additive or σ-additive.

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

Useful properties of an additive function $\mu$ include the following:

1. μ(∅) = 0.
2. If μ is non-negative and A ⊆ B, then μ(A) ≤ μ(B).
3. If A ⊆ B, then μ(B - A) = μ(B) - μ(A).
4. Given A and B, μ(A ∪ B) + μ(A ∩ B) = 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

$\mu (A)={\begin{cases}1&{\mbox{ if }}0\in A\\0&{\mbox{ if }}0\notin A.\end{cases}}$

If A₁, A₂, ..., A_n, ... 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

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

holds.

An example of an additive function which is not σ-additive is obtained by considering μ defined by the slightly modified formula

$\mu (A)={\begin{cases}1&{\mbox{ if }}0\in {\bar {A}}\\0&{\mbox{ if }}0\notin {\bar {A}}\end{cases}}$

where the bar denotes the closure of a set.

One can check that this function is additive by using the property that the closure of a finite union of sets is the union of the closures of the sets, and looking at the cases when 0 is in the closure of any of those sets or not. That this function is not σ-additive follows by considering the sequence of disjoint sets

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

for n=1, 2, 3, ... The union of these sets is the interval (0, 1] whose closure is [0, 1] and μ applied to the union is then 1, while μ applied to any of the individual sets is zero, so the sum of μ(A_n) is also zero, which proves the counterexample.

See also

• measure (mathematics)
• signed measure
• additive function
• subadditive function
• Hahn-Kolmogorov theorem

additive at PlanetMath.