I am trying to show if $E_k \subset (0, 1)$, $(k=1,...,n)$, and $\sum_{1 \le k \le n}{\mu(E_k)} > n -1$, then $\mu(\bigcap_{1 \le k \le n}{E_k}) > 0$.
Intuitively this seems like such an obvious statement but I have been having a hard time showing this. Following is some analysis I was able to perform regarding this problem.
Using Demorgan's law we know $\bigcap_{1 \le k \le n}{E_k} = (\bigcup_{1 \le k \le n}{{E_k}^\mathsf{c}})^\mathsf{c}$, and obviously measure of $\bigcup_{1 \le k \le n}{{E_k}^\mathsf{c}}$ being a subset of $(0,1)$ is bounded between 0 and 1. I am having a hard time trying to exploit the lower bound of the finite sum of measures. I thought of using inclusion-exclusion but that would probably create other terms as well.
How can I prove this ? Thanks.
Hint: let $F_k = (0,1) \backslash E_k$. What do you know about $\sum_k \mu(F_k)$?