How to show that the measure of $\cap_n E_n $ is not $0$?

Question

Let $E_n$ ($n \in \mathbb{Z}_{\geq 1}$) be the union of a finite set of closed intervals and the sum of the lengths of the intervals is large than or equal to a fixed positive number $a > 0$. Suppose that $E_1 \supset E_2 \supset \cdots$. How to show that the measure of $\cap_n E_n $ is not $0$? Thank you very much.

Edit: There is $a>0$ such that for every $n$, $E_n=I_1+\cdots+I_{M_n}$ for some positive integer $M_n$, $|I_1|+\cdots+|I_{M_n}|\geq a$, and $I_1,\ldots,I_{M_n}$ are closed intervals. Here $I_1, \ldots, I_{M_n}$ are disjoint.

Answer 1

I'm not sure if I'm understanding the question. But for $n \in \mathbb{N}$ the set $E_n$ has the form $ E_n = \bigcup_{k=1}^{N_n}[a_{k}^{n}, b_{k}^{n}]$ ? In that case, if you have a finite measure you may use the following:

$$\begin{eqnarray} \mu\left (\bigcap_{n=1}^{\infty}E_n \right ) &=& \lim_{n \to \infty} \mu\left (E_n \right )\end{eqnarray}$$

Answer 2

What about $E_n=[n,+\infty)$? Is $[n,+\infty)$ a closed interval in your definition? If you have the condition that $E_1$ has a finite measure, then the conclusion is correct.