S-finite measure

Remarks

Let $\lambda$ be Lebesgue measure on the Borel sets of $\R$. Let $\mu(A) = \sum_{n=1}^{\infty} \lambda(A \cap [0, 1])$, which is $0$ when $\lambda(A \cap [0, 1]) = 0$ and $\infty$ otherwise. It is s-finite, since each term is a finite measure. It is not σ-finite: a set of finite $\mu$-measure meets $[0, 1]$ in a $\lambda$-null set, and countably many such sets cannot cover $[0, 1]$. Every σ-finite measure is s-finite: write $X$ as a disjoint union of sets $B_n$ of finite measure, and let $\mu_n(A) = \mu(A \cap B_n)$. Counting measure on the Borel sets of $\R$ is not s-finite. A finite measure gives positive mass to at most countably many points. So countably many finite measures give positive mass to at most countably many points in total. Counting measure gives mass $1$ to every point of the uncountable set $\R$.