An exercise from Probability by Shiryaev:
Give an example to show that if a measure $\mu$ on the algebra $\mathcal{A}$ takes the value $+\infty$, countable additivity ($\sigma$-additivity) in general does not imply continuity at $\emptyset$.
Here, countable additivity for $\mu$ is given by, if, for all pairwise disjoint subsets $A_{1},A_{2},\ldots\in\mathcal{A}$ with $\bigcup_{n=1}^{\infty}A_{n}\in\mathcal{A}$, then
$$\mu\left(\bigcup_{n=1}^{\infty}A_{n}\right)=\sum_{n=1}^{\infty}\mu(A_{n}).$$
Here, continuity at $\emptyset$ means that if $\{A_{n}\}_{n=1}^{\infty}$ is a non-increasing sequence in the algebra $\mathcal{A}$ such that $\bigcap_{n=1}^{\infty}A_{n}=\emptyset$, then
$$\lim_{n\rightarrow\infty}\mu(A_{n})=0.$$
Best Answer
You could take for instance $\Omega = \Bbb Z$, $\mu$ to be counting measure, and the $\sigma$-algebra $\mathcal A$ to be the power set of $\Bbb Z$. Then $(\Omega,\mathcal A,\mu)$ is a $\sigma$-finite measure space, and the sets $A_n = [n,\infty)\cap\Bbb Z$ are non-increasing with $\bigcap_{n\ge 1} A_n = \emptyset$, but $\mu(A_n) = \infty$ for each $n$.