# Question about measure and limsup

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!

• What have you tried so far? Why should this be true, from a intuitive point of view? – SamM Sep 13 '19 at 16:28
• Let $B_n=\bigcup_{k=n}^{\infty}A_k$. Then $$\mu(B_n) = \mu(A_n) + \mu\left(\bigcup_{k=n+1}^{\infty}A_k\setminus A_n\right) > \epsilon$$ – M. Nestor Sep 13 '19 at 16:40
• the $\limsup$ should be outside of $\mu$ shouldn't it? Like $\limsup\mu(A_n)$... $\limsup$ of a set doesn't make any sense – Pink Panther Sep 13 '19 at 17:03
• Limsup of a sequence of sets is defined in the question. There is no problem with that definition. – SamM Sep 13 '19 at 17:12

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 $$\mu\left(\limsup_{n\to\infty}A_n\right) = \mu\left(\lim_{n\to\infty}B_n\right) = \lim_{n\to\infty}\mu(B_n)\geqslant \lim_{n\to\infty}\mu(A_n)\geqslant\varepsilon.$$