Question on proof of Borel-Cantelli Lemma The Borel-Cantelli Lemma: Let $\{E_k\}_{k=1}^\infty$ be a countable collection of measurable sets for which $\sum_{k=1}^\infty m(E_k) < \infty$. Then almost all $x \in \mathbb R$ belong to at most finitely many of the $E_k$'s. 
Here is the proof from Royden's book. 
For each n, the the countable subadditivity of measure, $$m(\bigcup_{k=n}^\infty E_k) \leq \sum_{k=n}^\infty m(E_k) < \infty.$$
Hence, by the countinuity of measure, $$m(\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)) = \lim_{n \to \infty} m(\bigcup_{k=n}^\infty E_k) \leq \lim_{n \to \infty} \sum_{k=n}^\infty m(E_k) = 0.$$
Therefor almost all $x \in \mathbb R$ fail to belong to $\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)$ and therefore belong to at most finitely many $E_k$'s. 
I don't understand the last sentence. How can we conclude that if almost all $x \in \mathbb R$ fail to belong to $\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)$ then they belong to at most finitely many $E_k$'s. If $x \in \bigcap_{n=1}^\infty (\bigcup_{k=2n}^\infty E_k)$, it may fail to belong to $\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)$, but it still belongs to infinitely many $E_k$'s. 

Suppose that $x \notin \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k$. Applying de Morgan, we find that
$$ x \in \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty \overline{E_k}. $$
Therefore for some $n \ge 1, x \in \overline{E_k}$ for $k \ge n$ or
$$ x \in \bigcap_{k=n}^\infty \overline{E_k}. $$
Applying de Morgan again,
$$ x \notin \bigcup_{k=n}^\infty E_k. $$
So $x$ can only belong to $E_1,\ldots,E_{n-1}$, a finite number of sets.
