Sigma-additive set function

In mathematics, additivity and sigma 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.

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

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

(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

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

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 μ 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.

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 power set of the real numbers by the slightly modified formula

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

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

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

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 infinity, 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.

Another counterexample can be obtained similarly, defining μ again over the power set of the real numbers by

$\mu (A)={\begin{cases}1&{\mbox{ if }}\exists a>0\ s.t.\ (0,a)\subset A\\0&{\mbox{ if }}\not \exists a>0\ s.t.\ (0,a)\subset A\end{cases}}$

At first sight, this is the same as the previous example, with the exception of negative sets. However rather than requiring the closure to include 0, it is required that the set include an interval next to 0. This will mean that any two sets that have measure 1 must overlap for some interval (0, a), this makes the measure additive (easily) but not sigma additive, using the same example of sets as above.

One can prove that each set has measure 0 by taking a to be half of 1/(n+1), do the sum will still be 0, but the union will once again be (0, 1) which clearly satisfies the condition required to have measure 1.

(A friend pointed out that this example does not work, since μ is not pairwise additive. Let S be the union of the intervals (1/2,1], (1/4,1/3], (1/6,1/5], (1/8,1/7], etc., and let S' be the union of the intervals (1/3,1/2], (1/5,1/4], (1/7,1/6], (1/9,1/8], etc. Neither S nor S' includes an interval of the form (0,a), so each has measure 0. Yet the union of S and S' contains (0,1), and hence has measure 1.

The example should be replaced with a correct one.)

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.

See also

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

additive at PlanetMath.