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

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.

• Do you have a result that implies $\mu\left(\bigcap_n E_n\right)=\lim_{n \to \infty} \mu(E_n)$? – angryavian Jun 13 '16 at 3:37
• Also, is each $E_n$ a union of finitely many disjoint closed intervals? One could "cheat" by having for example $E_n = [0,1/n] = [0,1/n] \cup [0,1/n] \cup \cdots \cup [0,1/n]$, and then the "sum of the lengths of the intervals" can be made arbitrarily large, but $\bigcup_n E_n = \{0\}$. – angryavian Jun 13 '16 at 3:40
• @angryavian, thank you very much. Yes, $E_n$ should be a union of finitely many disjoint closed intervals. – LJR Jun 13 '16 at 3:59

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:
$$\begin{eqnarray} \mu\left (\bigcap_{n=1}^{\infty}E_n \right ) &=& \lim_{n \to \infty} \mu\left (E_n \right )\end{eqnarray}$$
• yes, for each $n$, $E_n = \cup_{k=1}^{N_n}[a_{k,n}, b_{k,n}]$. – LJR Jun 13 '16 at 3:50
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.
• $[n, +\infty)$ is not a closed interval in my definition. – LJR Jun 13 '16 at 3:51