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I'm studying set theory for probability and statistics, and it's important, in order to work with a $\sigma$-algebra, to discuss the concept of $\liminf$ and $\limsup$ for sequences of sets.

But in doing so I'm not sure I got it well. For example:

When $A_n = \{(-1)^n\}$ we know: $$\liminf A_n = \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_n = \emptyset $$ because no elements appear in all sets $A_n$ except for a finite number of them. However: $$\limsup A_n = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty}A_n = \{-1, 1\}$$because elements $-1,1$ will end up appearing in all sets (and in the union of them) so the sequence does not converge.

Also, for a sequence such as:

$$ A_n = \left\{(x,y) \in \mathbb{R}^2 : \left(x-\frac{(-1)^n}{n}\right)^2 + y^2 \leq 1 \right\} $$

one can evaluate:

$$ \limsup A_n = \left\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1 \right\}= \liminf A_n$$ because, for the $\limsup$, the intersection of all unions starting with index $k=n$ will end up resulting in the circle centered in $(0,0)$. In a similar way, for the $\liminf$, the union of all intersections will end up coming as close as the circle centered in $(0,0)$, because $A_{n+1}\cap A_n \subset A_{n+2}\cap A_{n+1}$.

Is my track of thought logical? I am a bit afraid I didn't get the concept well...

Thank you!

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    $\begingroup$ You can say that $x\in \lim \inf A_n$ iff $\{n: x\not \in A_n\}$ is finite. i.e. $x\in A_n$ iff $x$ belongs to $A_n$ for all but finitely many $n$..... And $x\in \lim\sup A_n$ iff $x$ belongs to $A_n$ for infinitely many $n.$ $\endgroup$ Mar 3, 2019 at 9:40
  • $\begingroup$ In my previous comment the phrase "i.e. $x\in A_n$ " should be "i.e. $x\in \lim \inf A_n$". $\endgroup$ Mar 3, 2019 at 9:48

2 Answers 2

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To help your understanding, you can also consider the characterization of $\liminf A_n$ and $\limsup A_n$ using sequences in $A_n$

$\liminf A_n$ contains all the points $a$ such that exists a sequence $(a_n)$ converging towards $a$ with $a_n \in A_n.$

$\limsup A_n$ contains all the points $a$ such that exists a subsequence $(a_n')$ converging towards $a$ with $a_n' \in A_n'$

With your fist example, i.e., $A_n = \{(-1)^n\}$, clearly $\liminf A_n = \emptyset$ and $\limsup A_n = \{-1,1\}$

With your second example, $A_n = \left\{(x,y) \in \mathbb{R}^2 : \left(x-\frac{(-1)^n}{n}\right)^2 + y^2 \leq 1 \right\}$. Again, clearly $\liminf A_n = \left\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1 \right\} =\limsup A_n.$

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  • $\begingroup$ Good! So I think I got it right. Your explanation helped me too, it makes more sense now. Thanks :) $\endgroup$ Mar 3, 2019 at 4:34
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The following thoughts helped me to understand the limsup and liminf. And I hope it would also help you. Define $E_{j} = \bigcup \limits_{n=j}^{\infty} A_{n}$, then $~\limsup A_{n} = \bigcap \limits_{j=1}^{\infty} E_{j}$.

Similarly, Define $F_{j} = \bigcap \limits_{n=j}^{\infty}A_{n}$ , then $~\liminf A_{n} = \bigcup \limits_{j=1}^{\infty} F_{j}$.

Notice that $F_{j}$ is a sequence of increasing sets, i.e. $F_{1}\subset F_{2}\subset F_{3} \subset~ ...$ and $E_j$ is a sequence of decreasing sets, i.e. $E_{1} \supset E_{2} \supset E_{3} \supset~ ...$

This becomes very handy, when you want to calculate the measure of these liminf and limsup sets, as you can use continuity of measure.

Now if you have a positive measure $\mu$, consider the measure of the following set

$\mu(\bigcap \limits_{j=1}^{m}E_{j}) = \mu(E_{m})$. (*)

Since $E_{j}$ is decreasing, this measure is also decreasing. The measure of the limsup is the limit the measure in (*). If you think about it, this measure behaves like a limsup of a function, i.e. it is non-increasing as m increases.

I hope this would help.

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  • $\begingroup$ The proof I had seen for the fact that a sequence of sets has a limit if and only if $\liminf = \limsup$ uses the first paragraphs of your answer. But it was nice to see how it can be useful when calculating the measure of of $\liminf$ and $\limsup$ sets. Thanks! $\endgroup$ Mar 4, 2019 at 4:29

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