0
$\begingroup$

Let $(A_n)_{n=1}^{\infty}$ be a pairwise disjoint collection.

$\limsup A_n = \emptyset$?!

This is in relation to @StephenMontgomery-Smith's hint here.

Trying prove $\sigma$-additivity from continuity of measure, I ended up with

$\mu(A) - \sum_{n=1}^{\infty} \mu(A_n) = \mu(\limsup A_n)$ (where $A = \bigcup_{n=1}^{\infty} A_n$). I guess $\limsup A_n = \emptyset$. Otherwise, how do I show that its measure is zero?

Here is my attempt to prove that $\limsup A_n = \emptyset$.

$\limsup A_n = \bigcap_{k=1}^{\infty} \bigcup_{n=k}^{\infty} A_n$

$=\bigcap_{k=1}^{\infty} \bigcup_{n=k+1}^{\infty} A_n$

$=\bigcap_{k=1}^{\infty} A \setminus B_k$ (where $B_n = \bigcup_{k=1}^{n} A_k$)

$=\bigcap_{k=1}^{\infty} A \cap B_k^{C}$

$=(A \cap B_1^{C}) \cap (A \cap B_2^{C}) \cap ...$

$=A \cap (B_1^{C} \cap B_2^{C} \cap ...)$

$=A \cap (B_1 \cup B_2 \cup ...)^{C}$

$=A \cap A^{C}$

$=\emptyset$

If this is correct, which part makes use of the fact that $(A_n)_{n=1}^{\infty}$ be a pairwise disjoint collection?

$\endgroup$
  • 1
    $\begingroup$ Nonrigorous but intuitive reasoning: $\limsup A_n$ consists of all points $x$ which are contained in infinitely many of the $A_n$. But if the $A_n$ are pairwise disjoint, then no point can be in more than one of them. $\endgroup$ – Bungo Sep 21 '15 at 18:08
  • 1
    $\begingroup$ Ha, just noticed the date of this post. Someone necro'd the thread. Hopefully you have this all figured out by now. :-) $\endgroup$ – Bungo Sep 21 '15 at 18:09
  • 1
    $\begingroup$ @Bungo I necro'd it. :P Anyway, thanks. Sort of got it with this explanation for liminf $\endgroup$ – BCLC Sep 21 '15 at 18:09
3
$\begingroup$

It is correct and you use the the disjointness when you say $$ \bigcup_{k \ge n+1} A_k = A\setminus (A_1 \cup\cdots \cup A_n) $$ If the $ A_i $s are not disjoint, we only have $\supseteq $.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks.@martini. May you please provide some kind of reference for the limsup of a disjoint collection being $\emptyset$ ? $\endgroup$ – BCLC Jul 7 '14 at 3:34
  • $\begingroup$ Actually isn't the liminf also empty? Does this mean that the limit of a pairwise disjoint sequence of sets is empty? $\endgroup$ – BCLC Jul 7 '14 at 3:53
  • 1
    $\begingroup$ @BCLC: $\liminf A_n \subseteq \limsup A_n$, so if $\limsup A_n$ is empty, then automatically $\liminf A_n$ is also empty. Note that $\liminf A_n$ consists of all points $x$ which are in all but finitely many $A_n$, whereas $\limsup A_n$ consists of all points of $x$ which are in infinitely many $A_n$. The $\liminf$ condition is stronger than the $\limsup$ condition, which is why $\liminf A_n$ is a subset of $\limsup A_n$. $\endgroup$ – Bungo Sep 21 '15 at 21:24

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.