# 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 $\mu$ , there exists an equivalent probability measure $P$ , meaning that $\mu \sim P$ . One possible equivalent probability measure is given by

$P=\sum _{n=1}^{\infty }2^{-n}{\frac {\nu _{n}}{\nu _{n}(X)}}.$ Here, the $\nu _{n}$ are finite measures that sum up to $\mu$ like in the definition.