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 be σ-finite. Then there are measurable disjoint sets with and
Then the measures
are finite and their sum is . 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 with the σ-algebra . For all , let be the counting measure on this measurable space and define
- :=\sum _{n=1}^{\infty }\nu _{n}.}
The measure is by construction s-finite (since the counting measure is finite on a set with one element). But is not σ-finite, since
So cannot be σ-finite.