Suppose $E_1,\cdots,E_n\subset [0,1]$ are Borel sets such that $\sum_{i=1}^n\mu(E_i)>n-1$, in which $\mu$ denotes Lebesgue measure. Prove that $\cap_{i=1}^nE_i$ is nonempty.

My attempts included using the famous equation, which is true as all sets in question are of finite lengths: $$\mu(\sum E_i)=\sum^1\mu(E_{i_1})+(-1)^1\sum^2\mu(E_{i_1}E_{i_2})+\cdots+(-1)^{n-2}\sum^{n-1}\mu(E_{i_1}\cdots E_{i_{n-1}})+(-1)^{n-1}\mu(E_1E_2\cdots E_n). $$ where $\sum^k$ denotes the $k$-th cyclic sum, and set addition and multiplication are used in place of union and intersection, for the sake of notational simplicity. Sadly, this came to no avail at all. Indeed I could do $n=2$ (which of course is too trivial to discuss here) but I couldn't even do $n=3$.

Whatever I think the ultimate goal is to show $\mu(\cap E_i)>0$, and some kind of elementary set operations must be involved. But now I'm at a loss of what to do.

  • $\begingroup$ I think you mean $\sum_i \mu(E_i)$ as $\sum_i E_i$ doesn't make sense in this context. $\endgroup$ – Math1000 Mar 14 '16 at 12:52
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    $\begingroup$ @Math1000 Here I use addition and multiplication in place of set union and intersection for notational simplicity. So $\sum E_i=\cup E_i$. $\endgroup$ – Vim Mar 14 '16 at 12:56

The same answer as John Dawkins, without the probabilistic formalism.

Set $A_i:=[0,1]\setminus E_i$. Then $$\mu\left(\bigcup_{i=1}^n A_i \right)\leq \sum_{i=1}^n \mu(A_i)=\sum_{i=1}^n(1-\mu(E_i))=n-\sum_{i=1}^n \mu(E_i)<1.$$

So $[0,1]\setminus\bigcup_{i=1}^n A_i\neq\emptyset$, which is what you want.

  • $\begingroup$ Thanks. It is the easiest one for me to understand. $\endgroup$ – Vim Mar 15 '16 at 4:25

Define $X:=\sum_{k=1}^n 1_{E_k}$, the number of $E_k$ that occur. Observe that $1_{\{X=n\}}\ge X-(n-1)$. Consequently, $$ \Bbb P(\cap_{k=1}^n E_k)=\Bbb P(X=n)=\Bbb E[1_{\{X=n\}}]\ge\Bbb E[X-(n-1)]=\Bbb E[X]-(n-1)>0. $$

  • $\begingroup$ So you are using probability theory? $\endgroup$ – Vim Mar 14 '16 at 16:44
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    $\begingroup$ Yes, I am. If it's a vice you do not share, just substitute $\mu$ for $\Bbb P$ and $\int \cdot\,d\mu$ for $\Bbb E$. $\endgroup$ – John Dawkins Mar 14 '16 at 17:00
  • $\begingroup$ Indeed, here the probability measure $\mathbb P$ is precisely Lebesgue measure on $[0,1]$. $\endgroup$ – Math1000 Mar 14 '16 at 18:03

Partial answer, for the cases $n=3$ and $n=4$:

$n=3$: Assume $E_1\cap E_2 \cap E_3=\phi$. Let $F_2=E_2\setminus E_3\cup E_3\setminus E_2$ and $F_3=E_2\cap E_3$. We can easily see:

  • $\mu(E_2)+\mu(E_3)=\mu(E_2\setminus E_3\cup E_2\cap E_3)+\mu(E_3\setminus E_2\cup E_2\cap E_3)=\mu(F_2)+2\mu(F_3)$
  • $E_1 \cap F_3=\phi\Rightarrow \mu(E_1)+\mu(F_3)\leq 1$
  • $F_2 \cap F_3=\phi\Rightarrow \mu(F_2)+\mu(F_3)\leq 1$

Using this we deduce \begin{eqnarray} \mu(E_1)+\mu(E_2)+\mu(E_3) & = & \mu(E_1)+\mu(F_2)+2\mu(F_3)\\ & = & \mu(E_1)+\mu(F_3)+\mu(F_2)+\mu(F_3)\\ & \leq & 2 \end{eqnarray}

which is a contradiction.

$n=4$: Assume $E_1\cap E_2 \cap E_3\cap E_4=\phi$. Let \begin{eqnarray} F_2 & = & (E_2\setminus (E_3\cup E_4))\cup (E_3\setminus (E_2\cup E_4))\cup (E_4\setminus (E_2\cup E_3))\\ F_3 & = & ((E_2\cap E_3)\setminus E_4)\cup ((E_2\cap E_4)\setminus E_3)\cup ((E_3\cap E_4)\setminus E_2)\\ F_4 & = & E_2\cap E_3\cap E_4 \end{eqnarray} We can see that:

  • $\mu(E_2)+\mu(E_3)+\mu(E_4)=\mu(F_2)+2\mu(F_3)+3\mu(F_4)$
  • $E_1 \cap F_4=\phi\Rightarrow \mu(E_1)+\mu(F_3)\leq 1$
  • $F_3 \cap F_4=\phi\Rightarrow \mu(F_3)+\mu(F_4)\leq 1$
  • $F_2 \cap F_3 \cap F_4=\phi\Rightarrow \mu(F_2)+\mu(F_3)+\mu(F_4)\leq 1$

Using this we deduce \begin{eqnarray} \mu(E_1)+\mu(E_2)+\mu(E_3)+\mu(E_4) & = & \mu(E_1)+\mu(F_2)+2\mu(F_3)+3\mu(F_4)\\ & = & (\mu(E_1)+\mu(F_4))+(\mu(F_2)+\mu(F_3)+\mu(F_4))+(\mu(F_3)+\mu(F_4))\\ & \leq & 3 \end{eqnarray}

which is a contradiction.


The most straightforwards way to do this is by induction; notice that if $A$ and $B$ are subsets of $[0,1]$ we can easily prove that $$m(A\cap B)\geq m(A) + m(B) - 1.$$ Then, define $E'_n=\bigcap_{i=1}^nE_i$. We can prove that $\mu(E'_n)\geq \left(\sum_{i=1}^n \mu(E_i)\right)-n+1$ by induction. Obviously $\mu(E'_1)=\mu(E_1)\geq \mu(E_1)$ and we have $$m(E'_{n+1})=m(E'_n\cap E_{n+1})\geq m(E'_n) + m(E_{n+1})-1 \geq \left(\sum_{i=1}^{n+1}m(E_n)\right)-n - 1 + 1$$ which is the desired inequality. This gives that $E'_n$ has positive measure, given the condition you have, and thus is non-empty.


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