X=[0,1], consider the measure space (X,$F$,m), if there is no point in infinitely many Borel sets $E_n$, show that $m(E_n)=0$ for some n. Let me express the problem in the counter way: Consider Lebesgue space ([0,1],$F(0,1)$,m), $E_n$ be Borel sets. If there exists $c$>0 s.t. $m(E_n) \geq c$ for all n, show that there exists at least one point that in infinite many sets $E_n$. 
I try to use the Borel-Cantelli Lemma to find the n in tail term of a series, but is doesn't work. Could someone give me some help? Thanks.

Define $$A_k := \{x \in E_k : \forall n > k~~x \notin E_n\}$$
It is easy to see $A_k$ are disjoint and
$$\bigcup\limits_n E_n = \bigcup\limits_k A_k$$
From disjointness (and some work to see $A_k$ are measurable) you can conclude, that given $\varepsilon > 0$ you can find $N$, s.t.
$$m\left(\bigcup\limits_{k=N}^{\infty} A_k\right) < \varepsilon$$
but any $E_n \subseteq \bigcup\limits_{k=N}^{\infty} A_k$ for $n > N$.
contradiction to the fact $m(E_n) \geq c > 0$.
