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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?

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BCLC
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    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. –  Sep 21 '15 at 18:08
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    Ha, just noticed the date of this post. Someone necro'd the thread. Hopefully you have this all figured out by now. :-) –  Sep 21 '15 at 18:09
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    @Bungo I necro'd it. :P Anyway, thanks. Sort of got it with this explanation for liminf – BCLC Sep 21 '15 at 18:09

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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 $.

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martini
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  • Thanks.@martini. May you please provide some kind of reference for the limsup of a disjoint collection being $\emptyset$ ? – BCLC Jul 07 '14 at 03:34
  • Actually isn't the liminf also empty? Does this mean that the limit of a pairwise disjoint sequence of sets is empty? – BCLC Jul 07 '14 at 03:53
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    @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 $\limsup A_n$. –  Sep 21 '15 at 21:24