Suppose $A_1, A_2, A_3...$ are measurable sets and that
$$\sum_{k=1}^{\infty}m(A_k) \lt \infty$$
Prove that $m(\lim \sup A_k)=0$
My attempt : Let $B_n= \bigcup_{k=n}^{\infty} A_k$ then clearly $B_n \supset B_{n+1}$ and let $B=\bigcap_{j=1}^{\infty}B_j$ then $B=\bigcap_{j=1}^{\infty}\bigcup_{k=j}^{\infty}A_k=\lim \sup A_k$
by hypothesis and countable additivity
$$m(b_1)=m(\bigcup_{k=1}^{\infty} A_k) \le\sum_{k=1}^{\infty}m(A_k) \lt \infty$$
So $$m(\lim \sup A_k)=m(B)=\lim_{n\to \infty}m(B_n)$$
At that point, I didn't show how to converge these sequence to zero. please help me