Let $(X,\mathcal{A},μ)$ a measurable space and let $A_1,A_2,...∈\mathcal{A}$, assume that $\sum\limits_{j=1}^{\infty}=\mu (A_j)<\infty$
We have $E=\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{k=n}^{\infty}(A_k)$
I want to show that $\mu(E)=0$
what I've done so far is that for an element x in E, we have
$x\in A_k$ for atleast one $k\ge n$ but for all $n$, so x must be in $A_k$ for infinite k's.
I'm not sure if that's right, or where I can go from here.
Use monotonicity: $\mu(E) \le \displaystyle \mu\left(\bigcup_{k=n}^\infty A_k\right)$ for all $n \ge 1$.
Use subadditivity: $\displaystyle \mu\left(\bigcup_{k=n}^\infty A_k\right) \le \sum_{k=n}^\infty \mu(A_k)$.
Conclude: $\displaystyle \mu(E) \le \sum_{k=n}^\infty \mu(A_k)$ for all $n \ge 1$.
What happens as $n \to \infty$?