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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.

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

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

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

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

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

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

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

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

Therefore the Lebesgue measure is s-finite.

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

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

Then the measures

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

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

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

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

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

So ${\displaystyle \mu }$ cannot be σ-finite.

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

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

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• 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.