Sigma-additive set function
Let μ be a function defined on an algebra of sets with values in [−∞ +∞] (see the extended real line). We say that μ is additive if, whenever A and B are disjoint sets in , we have
One can prove by mathematical induction that an additive function satisfies
for any A1, A2, ..., An disjoint sets in .
Suppose is a σ-algebra. If for any sequence A1, A2, ..., An, ... of disjoint sets in one has
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 include the following:
- μ(∅) = 0.
- If μ is non-negative and A ⊆ B, then μ(A) ≤ μ(B).
- If A ⊆ B, then μ(B - A) = μ(B) - μ(A).
- Given A and B, μ(A ∪ B) + μ(A ∩ B) = μ(A) + μ(B).</math>
Examples
An example of a σ-additive function is the function μ defined over the power set of the real numbers, such that
If A1, A2, ..., An, ... 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
holds.
An example of an additive function which is not σ-additive is obtained by considering μ defined by the slightly modified formula
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
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(An) is also zero, which proves the counterexample.
See also
additive at PlanetMath.