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!

  • $\begingroup$ 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
  • 2
    $\begingroup$ Let $B_n=\bigcup_{k=n}^{\infty}A_k$. Then $$ \mu(B_n) = \mu(A_n) + \mu\left(\bigcup_{k=n+1}^{\infty}A_k\setminus A_n\right) > \epsilon$$ $\endgroup$
    – mdnestor
    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
  • 2
    $\begingroup$ 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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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