0
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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 $n.$ $\endgroup$ – DanielWainfleet Mar 3 at 9:40
  • $\begingroup$ In my previous comment the phrase "i.e. $x\in A_n$ " should be "i.e. $x\in \lim \inf A_n$". $\endgroup$ – DanielWainfleet Mar 3 at 9:48
1
$\begingroup$

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 $\liminf A_n = \left\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1 \right\} =\limsup A_n.$

$\endgroup$
  • $\begingroup$ Good! So I think I got it right. Your explanation helped me too, it makes more sense now. Thanks :) $\endgroup$ – M.Gonzalez Mar 3 at 4:34
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – M.Gonzalez Mar 4 at 4:29

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.