s-finite measure

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Short description: Mathematical function in measure theory


In measure theory, a branch of mathematics that studies generalized notions of volumes, an s-finite measure is a special type of measure. An s-finite measure is more general than a finite measure, but allows one to generalize certain proofs for finite measures.

The s-finite measures should not be confused with the σ-finite (sigma-finite) measures.

Definition

Let [math]\displaystyle{ (X, \mathcal A ) }[/math] be a measurable space and [math]\displaystyle{ \mu }[/math] a measure on this measurable space. The measure [math]\displaystyle{ \mu }[/math] is called an s-finite measure, if it can be written as a countable sum of finite measures [math]\displaystyle{ \nu_n }[/math] ([math]\displaystyle{ n \in \N }[/math]),[1]

[math]\displaystyle{ \mu= \sum_{n=1}^\infty \nu_n. }[/math]

Example

The Lebesgue measure [math]\displaystyle{ \lambda }[/math] is an s-finite measure. For this, set

[math]\displaystyle{ B_n= (-n,-n+1] \cup [n-1,n) }[/math]

and define the measures [math]\displaystyle{ \nu_n }[/math] by

[math]\displaystyle{ \nu_n(A)= \lambda(A \cap B_n) }[/math]

for all measurable sets [math]\displaystyle{ A }[/math]. These measures are finite, since [math]\displaystyle{ \nu_n(A) \leq \nu_n(B_n)=2 }[/math] for all measurable sets [math]\displaystyle{ A }[/math], and by construction satisfy

[math]\displaystyle{ \lambda = \sum_{n=1}^{\infty} \nu_n. }[/math]

Therefore the Lebesgue measure is s-finite.

Properties

Relation to σ-finite measures

Every σ-finite measure is s-finite, but not every s-finite measure is also σ-finite.

To show that every σ-finite measure is s-finite, let [math]\displaystyle{ \mu }[/math] be σ-finite. Then there are measurable disjoint sets [math]\displaystyle{ B_1, B_2, \dots }[/math] with [math]\displaystyle{ \mu(B_n)\lt \infty }[/math] and

[math]\displaystyle{ \bigcup_{n=1}^\infty B_n=X }[/math]

Then the measures

[math]\displaystyle{ \nu_n(\cdot):= \mu(\cdot \cap B_n) }[/math]

are finite and their sum is [math]\displaystyle{ \mu }[/math]. This approach is just like in the example above.

An example for an s-finite measure that is not σ-finite can be constructed on the set [math]\displaystyle{ X=\{a\} }[/math] with the σ-algebra [math]\displaystyle{ \mathcal A= \{\{a\}, \emptyset\} }[/math]. For all [math]\displaystyle{ n \in \N }[/math], let [math]\displaystyle{ \nu_n }[/math] be the counting measure on this measurable space and define

[math]\displaystyle{ \mu:= \sum_{n=1}^\infty \nu_n. }[/math]

The measure [math]\displaystyle{ \mu }[/math] is by construction s-finite (since the counting measure is finite on a set with one element). But [math]\displaystyle{ \mu }[/math] is not σ-finite, since

[math]\displaystyle{ \mu(\{a\})= \sum_{n=1}^\infty \nu_n(\{a\})= \sum_{n=1}^\infty 1= \infty. }[/math]

So [math]\displaystyle{ \mu }[/math] cannot be σ-finite.

Equivalence to probability measures

For every s-finite measure [math]\displaystyle{ \mu =\sum_{n=1}^\infty \nu_n }[/math], there exists an equivalent probability measure [math]\displaystyle{ P }[/math], meaning that [math]\displaystyle{ \mu \sim P }[/math].[1] One possible equivalent probability measure is given by

[math]\displaystyle{ P= \sum_{n=1}^\infty 2^{-n} \frac{\nu_n}{\nu_n(X)}. }[/math]

References

  1. 1.0 1.1 Kallenberg, Olav (2017). Random Measures, Theory and Applications. Probability Theory and Stochastic Modelling. 77. Switzerland: Springer. p. 21. doi:10.1007/978-3-319-41598-7. ISBN 978-3-319-41596-3.