S-finite measure

Template:Lowercase title 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 $(X,𝒜)$ 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 ${\nu }_n$ ($n\in \mathbb{N}$),

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

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 ${\nu }_n(A)\le {\nu }_n(B_n)=2$ for all measurable sets $A$, and by construction satisfy

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

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 $\mu (B_n)<\mathrm{\infty }$ and

$\bigcup _{n=1}^{\mathrm{\infty }}{B}_n=X$

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 $𝒜=\{\{a\},\mathrm{\varnothing }\}$. For all $n\in \mathbb{N}$, let ${\nu }_n$ be the counting measure on this measurable space and define

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

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

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

So $\mu$ cannot be σ-finite.

Equivalence to probability measures

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

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

References

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