Working on the real line $(\mathbb{R})$, let $\mu : \mathscr{M} \rightarrow [0, +\infty]$ represent the Lebesgue measure ($\mathscr{M}$ is the set of measurable subsets of $\mathbb{R}$).

I want to find a sequence of sets such that

$$\mu \left(\bigcup_{n = 1}^{\infty} E_n \right) = 2 \ \text{and} \ \lim_{n \to \infty} \mu(E_n) = 1.$$

I'm not sure I understand correctly what a sequence of sets is. Could the sequence of sets be finite, the infinity symbol is confusing me here. How should I think/solve this problem?

  • $\begingroup$ Take $E_1=[0,1]$ and $E_n=[1+\frac{1}{n},2]$ for $n\ge 2$ $\endgroup$ – Svetoslav Mar 2 '16 at 13:47
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    $\begingroup$ Even simpler, $E_1=[0,2]$ and $E_n=[0,1]$ for $n\geq 2$. $\endgroup$ – Gabriel Romon Mar 2 '16 at 13:50

A sequence of sets is just a countable collection of sets indexed by natural numbers. That is, for each natural number $n$ the sequence specifies a set $E_n$. Since for each particular $n$, $\mu(E_n)$ is a real number, we can form a sequence of measures $\mu(E_n)$, and ask for its limit (as a sequence of reals).

The union of the sequence of sets takes nothing from the indexing; it is just the set $\bigcup\limits_{n=1}^{\infty}E_n$ that contains some $x$ iff $x \in E_n$ for some $n$.

For your problem, take $E_n = [1-\frac{1}{n}, 2]$.

$$\lim\limits_{n\rightarrow\infty} \mu(E_n)=\lim\limits_{n\rightarrow\infty} (1+\frac{1}{n})=1$$

Now, since for any $n$, $E_n \subset E_1$, we conclude


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    $\begingroup$ Thank you for your detailed and clear explanation. It was extremely helpful and understandable. $\endgroup$ – Von Kar Mar 2 '16 at 14:07
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    $\begingroup$ @VonKar Glad to be of service. PS Thank you for my 3k reputation achievement :) $\endgroup$ – lisyarus Mar 2 '16 at 14:11

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