I'm thinking about the following statement.
Let $(\Omega, F,\Bbb{P})$ be a discrete probability space. Is it true that if $$\Bbb{P}(\lim_{n\rightarrow \infty} sup~ A_n)=0$$then $\sum_{n=1}^\infty \Bbb{P}(A_n)<\infty$?
So I could show the other implication and now I'm thinking about a couterexample but I don't come up with one.
Could someone maybe help me where to start?
If $\Bbb P(\limsup_{n\to\infty} A_n)=0$ then $K:=\sum_n1_{A_n}<\infty$ a.s. Because the probability space is discrete (which I understand to mean that $\Omega$ is countable and each singleton is measurable and has strictly positive weight) you must have $K(\omega)<\infty$ for all $\omega\in\Omega$. Notice that $$ \sum_{n=1}^\infty\Bbb P(A_n) =\Bbb E(K). $$ And if $\Omega$ is finite then $\Bbb E(K)<\infty$. For a counterexample you therefore need $\Omega$ to be infinite. How about $\Omega=\{1,2,\ldots\}$ with $\Bbb P(\{k\})$ proportional to $k^{-2}$, and $A_n=\{n,n+1,\ldots,2n\}$.