Elementary Lebesgue measure problem 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. 
 A: 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.
A: 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.
A: 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.
A: 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, 
$$
\Bbb P(\cap_{k=1}^n E_k)=\Bbb P(X=n)=\Bbb E[1_{\{X=n\}}]\ge\Bbb E[X-(n-1)]=\Bbb E[X]-(n-1)>0.
$$
