Lebesgue measure: Show $m(\cap_{k=1}^{2014} E_k)>0$...

Question

Here's the full problem:

Let the collection of measurable sets $\{E_k\}_{k=1}^{2014} \subset [0,1]$ be s.t $\sum_{k=1}^{2014} m(E_k) >2013$, where $m$ denotes Lebesgue measure. Show $$m(\cap_{k=1}^{2014} E_k)>0$$

Here is what I've been thinking:

We can deduce that $$1\geq m(\cup_{k=1}^{2014} E_k)>0$$ since the sum of the measures are so large

From this information though, I've no idea really where to go. I mean, we know that since the value is so large from the sum of the measures of our E's, there must be some overlap when intersecting our sets, which in turn would mean the arbitrary intersection is non-zero, but formally showing this has been very difficult for me. If anyone could give some aid, I'd be very grateful.

Answer 1

Note that$$m([0,1]\setminus\bigcap_{k=1}^{2014} E_k)=m(\bigcup_{k=1}^{2014}([0,1]\setminus E_k))$$and $$\sum_{k=1}^{2014}m([0,1]\setminus E_k)=2014-\sum_{k=1}^{2014}m(E_k)<1,$$it follows that $$m([0,1]\setminus\bigcap_{k=1}^{2014} E_k)=m(\bigcup_{k=1}^{2014}([0,1]\setminus E_k))\leqslant \sum_{k=1}^{2014}m([0,1]\setminus E_k)<1.$$Hence $$m(\bigcap_{k=1}^{2014} E_k)=1-m([0,1]\setminus\bigcap_{k=1}^{2014} E_k)>0.$$

Answer 2

I'm not sure, but: Let $A_1, A_2, \ldots A_{2^{2014}}$ be the dijoint sets so that for any $i_1, i_2, \ldots, i_n$ there exists $j_1, j_2, \ldots, j_m$ so that $\bigcap_{k\le n}E_{i_k} = \bigcup_{l\le m}A_{j_l}$. Then if $A_i$ is in $t_i$ different sets $E$, $\sum_{i=1}^{2^{2014}}\mu(A_i)t_i = \sum_{i=1}^{2014}\mu(E_k)$ by subaddivity. $A$s are disjoint so the sum of their measures is the measure of their union, no more than the measure of $\mu([0, 1]) = 1$. Then, if every $t_i$ is no more than 2013 then $\sum_{i=1}^{2^{2014}}\mu(A_i)t_i \le 2013 \sum_{i=1}^{2^{2014}}\mu(A_i) \le 2013 \times 1 =2013$. Then there exists $A_i$ for which $t_i$ is 2014, it means $A_i$ is in all of $E$s, and it's measure must be >0 or there will be a contradiction I wrote higher.