s-finite measure

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 $(X,{\mathcal {A}})$ be a measurable space and $\mu$ a measure on this measurable space. The measure $\mu$ is called an s-finite measure, if it can be written as a countable sum of finite measures $\nu _{n}$ ($n\in \mathbb {N}$),

$\mu =\sum _{n=1}^{\infty }\nu _{n}.$

Example

The Lebesgue measure $\lambda$ is an s-finite measure. For this, set

$B_{n}=(-n,-n+1]\cup [n-1,n)$

and define the measures $\nu _{n}$ by

$\nu _{n}(A)=\lambda (A\cap B_{n})$

for all measurable sets $A$. These measures are finite, since $\nu _{n}(A)\leq \nu _{n}(B_{n})=2$ for all measurable sets $A$, and by construction satisfy

$\lambda =\sum _{n=1}^{\infty }\nu _{n}.$

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 $\mu$ be σ-finite. Then there are measurable disjoint sets $B_{1},B_{2},\dots$ with $\mu (B_{n})<\infty$ and

$\bigcup _{n=1}^{\infty }B_{n}=X$

Then the measures

$\nu _{n}(\cdot ):=\mu (\cdot \cap B_{n})$

are finite and their sum is $\mu$. 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 $X=\{a\}$ with the σ-algebra ${\mathcal {A}}=\{\{a\},\emptyset \}$. For all $n\in \mathbb {N}$, let $\nu _{n}$ be the counting measure on this measurable space and define

$\mu :=\sum _{n=1}^{\infty }\nu _{n}.$

The measure $\mu$ is by construction s-finite (since the counting measure is finite on a set with one element). But $\mu$ is not σ-finite, since

$\mu (\{a\})=\sum _{n=1}^{\infty }\nu _{n}(\{a\})=\sum _{n=1}^{\infty }1=\infty .$

So $\mu$ cannot be σ-finite.

Equivalence to probability measures

For every s-finite measure $\mu =\sum _{n=1}^{\infty }\nu _{n}$, there exists an equivalent probability measure $P$, meaning that $\mu \sim P$. One possible equivalent probability measure is given by

$P=\sum _{n=1}^{\infty }2^{-n}{\frac {\nu _{n}}{\nu _{n}(X)}}.$

References

• Falkner, Neil (2009). "Reviews". American Mathematical Monthly. 116 (7): 657–664. doi:10.4169/193009709X458654. ISSN 0002-9890.
• Olav Kallenberg (12 April 2017). Random Measures, Theory and Applications. Springer. ISBN 978-3-319-41598-7.
• Günter Last; Mathew Penrose (26 October 2017). Lectures on the Poisson Process. Cambridge University Press. ISBN 978-1-107-08801-6.
• R.K. Getoor (6 December 2012). Excessive Measures. Springer Science & Business Media. ISBN 978-1-4612-3470-8.