The Problem
Suppose I have the measurable set $$E = \bigcup_{n = 1}^\infty\bigcap_{k = n}^\infty E_k$$ where $E_k \subseteq \mathbb{R}^d$ is measurable for all $k$. I want to prove that if $\sum_{k = 1}^\infty m(E_k) < \infty$ then $m(E) = 0$.
What I tried
I think it can be shown that $$E = \bigcup_{n = 1}^\infty\bigcap_{k = n}^\infty E_k \subseteq \bigcap_{n = 1}^\infty\bigcup_{k = n}^\infty E_k$$ Next let $B_N = \bigcup_{k = N}^\infty E_k$. Clearly since the sets may not be disjoint, $$m(B_N) \leq \sum_{k = N}^\infty m(E_k)$$
and since $\sum_{k = 1}^\infty m(E_k) < \infty$ then for all $\epsilon > 0$ there exists $N$ such that $$m(B_N) \leq \sum_{k = N}^\infty m(E_k) < \epsilon$$ Furthermore, it's not hard to see that $\bigcap_{n = 1}^\infty B_n \subseteq B_N$, so $$m(E) \leq m\left(\bigcap_{n = 1}^\infty\bigcup_{k = n}^\infty E_k\right) \leq m(B_N) < \epsilon$$ Since $\epsilon > 0$, then $m(E) = 0$.
Have I missed anything? If so, how can I correct it?
Thanks in advance!
We have another simple proof. We have $$\bigcap_{k=n}^{\infty}E_k\subset E_j$$ for every $j\ge n$, and from the convergence condition we have $m(E_j)\to 0$. Hence $$0\le m\left(\bigcap_{k=n}^{\infty}E_k\right)\le m(E_j)\to 0,$$ forcing the measure to be $0$. Hence we have, by countable sub-additivity, $$0\le m\left(\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}E_k\right)\le\sum_{n=1}^{\infty}m\left(\bigcap_{k=n}^{\infty}E_k\right) = \sum_{n=1}^{\infty}0=0$$ and this completes the proof.