TITLE: Elementary Lebesgue measure problem
QUESTION [9 upvotes]: 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.

REPLY [2 votes]: 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.