Given $E_i\subset (0,1)$ is Lebesgue measurable for each $n=1,2,...,n$ and $\sum\limits_{i=1}^{n}m(E_i)>n-1$. We want to show that $m(\bigcap\limits^{n}_{i=1} E_i)>0$.
What I understand is that $m(E_i)\le1$ for each $i=1,2,...,n$. But I could not proceed further. Any hint will be appreciated.
We will assume that $(X,\mathcal{F},m)$ is a probability space, where $\mathcal{F}$ is a $\sigma$-algebra on $X$. This is the same as saying $E_k \subset (0,1)$ and $m(X) = 1$.
Suppose for a contradiction that $m(\cap_{k=1}^n E_k) = 0$. Then, since $(X,\mathcal{F},m)$ is a probability space, it follows that $m(\cup_{k=1}^n E_k^c) = 1$. By subadditivity of $m$, we have that $1 = m(\cup_{k=1}^n E_k^c) \leq \sum_{k=1}^n m(E_k^c)$ and since $m(X) = 1$, we get that $\sum_{k=1}^n m(E_k^c) = 1$. Now, for $1 \leq k \leq n$, we have that $m(E_k \sqcup E_k^c) = m(X) = 1$ and so summing these, we get $$n = \sum_{k=1}^n m(E_k \sqcup E_k^c) = \sum_{k=1}^n m(E_k) + \sum_{k=1}^n m(E_k^c) = \sum_{k=1}^n m(E_k) + 1.$$ What can you deduce from here?