Show that $\lambda(E_1 \cap ... \cap E_n) > 0$ We use the Lebesgue measure $\lambda$ on $\mathbb{R}$. Let $E_t \subset (0,1)$, $t=1, ..., n$ be measurable subsets such that $\lambda(E_1)+...+\lambda(E_n) > n-1$.
How do I show that $\lambda(E_1 \cap ... \cap E_n) > 0$?

My idea is to let  $S_t=(0,1)$ \ $E_t$.  Then we also have  an inequality  for $\lambda(S_t)$. But I don't know how to express $\bigcap_t E_t$ with $S_t$.

Hint If $A, B \subseteq (0,1)$ then
$$1 \geq \lambda(A \cup B)=\lambda(A)+ \lambda(B)-\lambda(A \cap B)$$
and hence
$$
\lambda(A \cap B) \geq \lambda(A)+ \lambda(B) -1
$$
Prove now by induction that if $A_1,.., A_k \subseteq (0,1)$ then
$$
\lambda (A_1 \cap A_2 \cap ... \cap A_k)\geq \lambda (A_1) +\lambda( A_2)+... +\lambda(A_k)-k+1
$$
