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.

Relation to σ-finite measures edit

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.

Equivalence to probability measures edit

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

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

• 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.