# intuition on lim sup and lim inf in probability spaces

So I thought I understand what the lim sup and lim inf are, especially thanks to this post, but now I am presented with a problem from basic probability theory saying:

Let $$(A_n)$$ be a sequence of events on a probability space $$(\Omega, \mathcal{F}, P)$$. Recall (from measure theory) the definitions $$\liminf_{n \to \infty} A_n := \bigcup_{n \geq 1} \bigcap_{m \geq n} A_m,$$ $$\limsup_{n \to \infty} A_n := \bigcap_{n \geq 1} \bigcup_{m \geq n} A_m.$$ Intuition:

• $$\liminf_{n \to \infty} A_n$$ consists of the elements $$\omega \in \Omega$$ that appear in almost all* sets $$A_n$$.
• $$\limsup_{n \to \infty} A_n$$ consists of the elements $$\omega \in \Omega$$ that appear in infinitely many sets $$A_n$$.

*In this context, "almost all" means that there exists an $$N \in \mathbb{N}$$ such that for all $$n \geq N$$, $$\omega \in A_n$$.

What is the difference? This looks like the exact same statement? Maybe an example would help.

Edit: One example: if $$A_n = [-2-(-1)^{n}, 2+(-1)^{n+1})$$ then $$\liminf_{n \to \infty} A_n = [-1, 1)$$ and $$\limsup_{n \to \infty} A_n=[-3, 3)$$? Maybe? Not sure.

• I would change “almost all” to “eventually all”, because that just gets the English message across better. Imagine you take a 10 hour ‘break’ from something (maybe you work maybe you don’t in that time) but then you’re stuck doing that something for the rest of eternity vs doing one hour of work every so often for the rest of eternity. Clearly these are different things. Commented Jun 8 at 8:04
• Some of the answers here might help: math.stackexchange.com/questions/107931/… Commented Jun 8 at 8:06
• @peek-a-boo thank you! this analogy is interesting, so do I understand correctly that then the $\liminf$ is the "rest of eternity" scenario and $\limsup$ is the "every so often" scenario? Commented Jun 8 at 8:10
• @arridadiyaat correct Commented Jun 8 at 8:13
• Another example/question: $A_n= p_n\mathbb{N}$ where $p_n$ is prime, i.e. $A_n$ are all multiples of the n-th prime. Then $\limsup{n\to\infty} A_n={0}$ since $m\in\limsup A_n$ implies m is divisible by infinitely many prime numbers, hence $m={0}$ but then also $\liminf A_n = {0}$? Commented Jun 9 at 11:46

$$A_n = \emptyset \qquad n \text{ even} \\ A_n = \Omega \qquad n \text{ odd}$$
then any $$\omega \in \Omega$$ is in $$A_n$$ infinitely often but none is eventually in every set, i.e. $$\lim \sup A_n = \Omega \\ \lim \inf A_n = \emptyset$$
The terminology "almost all" in this situation simply means all but finitely many (also said cofinitely many). It does not really have anything to do with the order structure of the natural numbers, but it may be phrased in the way you have written. A subset $$E\subseteq \mathbf{N}$$ is cofinite if all but finitely many naturals belong to $$E$$, and this is equivalent to the property that there exists $$N$$ so that for all $$n\ge N$$, $$n\in E$$.
• I see! like e.g. $\mathbb{Q} \in\mathbb{R}$ is infinite but not cofinite? thanks a lot! Commented Jun 9 at 11:03
• Yes, exactly. For the naturals the subset of evens $2\mathbf{N}\subset \mathbf{N}$ is another example of a non-cofinite infinite subset (its complement is the odds) Commented Jun 9 at 16:21