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Let $(X, M, \mu)$ be a measure space and $\{A_n\}$ be a sequence of measurable sets.

I want to show that if $\mu$ is a finite measure and $\mu(A_n) > \epsilon>0$ for each n, then $\mu ($lim sup $A_N)≥\epsilon $.

Since lim sup $A_n$ = $\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k$, we have $$ \mu(\text{lim sup } A_n) ≤ \mu(\bigcup_{k=n}^\infty A_k)$$ for all $n$, but I'm not sure where to go from here to show the desired result. I would appreciate any help on how to proceed. Thanks in advance!

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  • $\begingroup$ What have you tried so far? Why should this be true, from a intuitive point of view? $\endgroup$
    – SamM
    Sep 13, 2019 at 16:28
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    $\begingroup$ Let $B_n=\bigcup_{k=n}^{\infty}A_k$. Then $$ \mu(B_n) = \mu(A_n) + \mu\left(\bigcup_{k=n+1}^{\infty}A_k\setminus A_n\right) > \epsilon$$ $\endgroup$
    – M. Nestor
    Sep 13, 2019 at 16:40
  • $\begingroup$ the $\limsup$ should be outside of $\mu$ shouldn't it? Like $\limsup\mu(A_n)$... $\limsup$ of a set doesn't make any sense $\endgroup$ Sep 13, 2019 at 17:03
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    $\begingroup$ Limsup of a sequence of sets is defined in the question. There is no problem with that definition. $\endgroup$
    – SamM
    Sep 13, 2019 at 17:12

1 Answer 1

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Let $B_n=\bigcup_{k=n}^\infty A_k$, then $\limsup_{n\to\infty} A_n = \bigcap_{n=1}^\infty B_n$. Since $B_n\supset A_n$, $\mu(B_n)\geqslant \mu(A_n)$. Moreover, $B_n\supset B_{n+1}$, so $\{B_n\}$ is a decreasing sequence of sets. By continuity from above (here we use the fact that $\mu$ is a finite measure), we have $$ \mu\left(\limsup_{n\to\infty}A_n\right) = \mu\left(\lim_{n\to\infty}B_n\right) = \lim_{n\to\infty}\mu(B_n)\geqslant \lim_{n\to\infty}\mu(A_n)\geqslant\varepsilon. $$

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