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.

  • $\begingroup$ Do you have a result that implies $\mu\left(\bigcap_n E_n\right)=\lim_{n \to \infty} \mu(E_n)$? $\endgroup$ – angryavian Jun 13 '16 at 3:37
  • $\begingroup$ 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 $\bigcup_n E_n = \{0\}$. $\endgroup$ – angryavian Jun 13 '16 at 3:40
  • $\begingroup$ @angryavian, thank you very much. Yes, $E_n$ should be a union of finitely many disjoint closed intervals. $\endgroup$ – 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:

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

  • $\begingroup$ yes, for each $n$, $E_n = \cup_{k=1}^{N_n}[a_{k,n}, b_{k,n}]$. $\endgroup$ – 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.

  • $\begingroup$ $[n, +\infty)$ is not a closed interval in my definition. $\endgroup$ – LJR Jun 13 '16 at 3:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.