# Set sequences $\liminf$ and $\limsup$ - correct?

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!

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

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

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

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.

• 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! – M.Gonzalez Mar 4 at 4:29