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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.

user146925
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    do you see intuitively that this should be true???? –  Sep 24 '14 at 18:13
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    Your deduction is not correct. – T.J. Gaffney Sep 24 '14 at 18:15
  • @Gaffney : Why do you think that this is not correct? –  Sep 24 '14 at 18:16
  • @PraphullaKoushik I think so. Since our family of sets are subsets of $[0,1]$, and the sum of their measures is so large, there must be quite a bit of overlap between each set. I'm just not sure how to show it – user146925 Sep 24 '14 at 18:17
  • @Gaffney Why is it incorrect? – user146925 Sep 24 '14 at 18:18
  • what is $\mu[0,\frac{1}{2014})$ –  Sep 24 '14 at 18:19
  • How does a subset of $[0,1]$ have measure greater than $2013$? Look at the direction of your inequalities. – T.J. Gaffney Sep 24 '14 at 18:20
  • @Gaffney : It is not subset of $[0,1]$ which has measure greater than $2013$ it is sum of measures of subsets of $[0,1]$ –  Sep 24 '14 at 18:21
  • @Gaffney I think you've read incorrectly. The sum of the measures is greater than 2013, which means that each set must have measure very close to 1, thus there must be overlap. – user146925 Sep 24 '14 at 18:22
  • @PraphullaKoushik we that the outer measure of an interval is equal to the length of the interval, so since lebesgue measurable sets are a special case of outer measure, $\mu([0,\frac{1}{2014})$ must be $\frac{1}{2014}$ – user146925 Sep 24 '14 at 18:24
  • Consider replacing that $2014$ by $2$ and $2013$ by $1$... Big numbers may disturb your attention unnecessarily... –  Sep 24 '14 at 18:24
  • @PraphullaKoushik It's not really the large numbers that are causing the trouble, it's the transition from what I've deduced. I will try this, though, and let you know how it goes. – user146925 Sep 24 '14 at 18:29
  • I'm not misreading. http://imgur.com/Q9qRj95 – T.J. Gaffney Sep 24 '14 at 18:33
  • Oh I see now. Yeah that was pretty silly huh... – user146925 Sep 24 '14 at 18:37
  • @user146925 With your edit you haven't used countable subadditivity, so you ought as well remove the line before it. – T.J. Gaffney Sep 24 '14 at 18:40

2 Answers2

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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 $$m(\bigcap_{k=1}^{2014} E_k)=1-m([0,1]\setminus\bigcap_{k=1}^{2014} E_k)>0.$$

Nirvanacs
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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.

Petr Naryshkin
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