Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X,\mathcal{A})$ be a measurable space, with measure $\mu$. Let $\{E_n\}_{n \in \mathbb{N}} \subseteq \mathcal{A}$ be a sequence of measurable sets, with $E_{n+1} \subseteq E_n, \ \forall n \in \mathbb{N}$, that is a decreasing sequence, and $\mu(E_n)=+ \infty, \ \forall n \in \mathbb{N}$. Let $E=\bigcap_{n \in \mathbb{N}} E_n$ the limit set of the sequence, $E$ is measurable for definition. Is true that $\mu(E)=+\infty$? If not in all cases, is true with $X=\mathbb{R}^{N}$, $\mathcal{A}$ the collection of Lebesgue-measurable sets, and $\mu$ the Lebesgue measure on $\mathbb{R}^N$?

share|improve this question
    
Homework? If so, please tag accordingly. –  Harald Hanche-Olsen Nov 2 '12 at 9:42
    
Is a doubt from an exercise from homework, I'll tag so, excuse me. –  Lorban Nov 2 '12 at 10:23

1 Answer 1

up vote 1 down vote accepted

Consider $E_n=(n,\infty)$ as subsets of $\Bbb R$. Similar examples can be found for general $\Bbb R^N$, in particular the set of all $\vec{x}$ such that $\lVert\vec{x}\rVert>n$.

share|improve this answer
    
Ok, with $E_n=(0,+ \infty)$ the limit set of the sequence is the empty set because natural numbers are non-limitated. Thank you for your reply. –  Lorban Nov 2 '12 at 10:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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