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 $$ \nu_n $$ ($$ n \in \N $$),
 * $$ \mu= \sum_{n=1}^\infty \nu_n. $$

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.

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 $$ n \in \N $$, 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
 * $$ P= \sum_{n=1}^\infty 2^{-n} \frac{\nu_n}{\nu_n(X)}. $$