Measure space $(X,\mathcal{B},\mu)$, $B_n\in \mathcal{B}$, $\mu(B_n) \le \infty, n=1,2,...,$.
(1) If $\sum\limits_{n=1}^{\infty} \mu(B_n) \le \infty $, then $\mu(\lim\sup_{n \to \infty} B_n)=0$.
(2) $A_n\in\mathcal{B}, A_n \supset A_{n+1},n=1,2,..., \mu(A_1)\le \infty$
(i) $B_n:= A_n\setminus A_{n+1}$, then $\sum\limits_{n=1}^{\infty} B_n$ converges.
(ii) $\forall n\in N, \cup_{k=n}^{\infty}(A_k\setminus A_{k+1})=A_n$, then show $\mu(\cap_{n=1}^{\infty} A_n)=0$.
(1) http://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma
(2) Apply (1) for (ii), along with the definition of $\limsup$ (for (i), use $\mu(B_n) = \mu (A_n) - \mu (A_{n+1})$).