0
$\begingroup$

Let $\left(X,\mathcal{F},\mu\right) $ be a measure space and suppose $\left\{ A_{n}\right\} _{n=1}^{\infty} $ is a sequence of sets such that $\mu\left(A_{n}\right)\geq\varepsilon $ for some $\varepsilon>0$ and for all $n\in\mathbb{N}$ . Is this contradictory to $\mu\left(\limsup\limits _{n\to\infty}A_{n}\right)=0$ ?

I've come accustomed to thinking of Limsup as the set of $x\in X$ that belong to $A_{n}$ for an infinite number of $n$. With that in mind I don't really see any reason why this should be acontradiction. Using the more formal definition of $${\displaystyle \limsup_{n\to\infty}A_{n}=\bigcap_{n=1}^{\infty}\bigcup_{k\geq n}A_{k}}$$ also doesn't seem to provide an obvious contradiction. Also ,does it make any difference if the measure was a finite measure?

$\endgroup$
1
$\begingroup$

Using the more formal definition of limsup ... also doesn't seem to provide an obvious contradiction.

Well, it does. Note first that for every $n$, $$ \mu\left(\bigcup_{k\geqslant n}A_k\right)\geqslant\mu(A_n)\geqslant\varepsilon, $$ and deduce from this that $$ \mu\left(\limsup_{n\to\infty}A_n\right)\geqslant\varepsilon, $$ under the dominating condition that $$ \mu\left(\bigcup_{n\geqslant 1}A_n\right) $$ is finite. This condition is always satisfied when the measure $\mu$ is finite.

Recall that the measure of the union of a nondecreasing sequence of measurable sets is always the limit of the measures of the sets but that the measure of the intersection of a nonincreasing sequence of measurable sets is guaranteed to be the limit of the measures of the sets only when one of the sets has finite measure. A counterexample to keep in mind: $A_n=[n,+\infty)$ in $(\mathbb R,\mathcal B(\mathbb R))$ with the Lebesgue measure. Or, equivalently, $A_n=\{k\in\mathbb N\mid k\geqslant n\}$ in $(\mathbb N,2^\mathbb N)$ with the counting measure.

$\endgroup$
3
  • 1
    $\begingroup$ $A_n = [n,\infty) \subset \mathbb{R}$. Unless you have a condition $$\mu\left(\bigcup_{n=k}^\infty A_n\right) < \infty$$ for some $k$, there is no contradiction. $\endgroup$ – Daniel Fischer Dec 15 '13 at 16:41
  • $\begingroup$ @DanielFischer Of course, shame on me... Thanks. $\endgroup$ – Did Dec 15 '13 at 16:45
  • $\begingroup$ I had a feeling that the measure being finite is sufficient for this being true. Thanks Did and Daniel. $\endgroup$ – Serpahimz Dec 15 '13 at 16:46

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.