# Limit of a decreasing sequence of measurable sets.

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 $\mathbb{R}^N$?

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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

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 $\lVert\vec{x}\rVert>n$.
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