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 ${\displaystyle \mu }$ 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

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

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