how does limsup and liminf for a sequence of sets, apply to probability theory. any real world examples would be much appreciated



 limsup and liminf for a sequence of sets
For a sequence of subsets $A\_n$ of a set $X$, the $\limsup A\_n$ $= \cap\_{N=1}^\infty ( \cup\_{n\ge N} A\_n )$ and $\liminf A\_n$ $= \cup\_{N=1}^\infty (\cap\_{n \ge N} A\_n)$.


If $ x \in \limsup A\_n$ then $x$ is in all of the $\cup\_{n\ge N} A\_n$, which means no matter how large you pick $N$ you will find an $A\_n$ with $n>N$ of which $x$ is a member. Thus members of $\limsup A\_n$ are those elements of $X$ that are members of infinitely many of the $A\_n$'s. If $A\_n$ are thought of as events (in the sense of probability) $\limsup A\_n$ will be another event. It corresponds exactly to the occurance of infinitely many of the $A\_n$'s. This is why $\limsup A\_n$ is sometimes written $x \in A\_n$ infinitely often.


Similarly, if $x\in \liminf A\_n$ then $x$ is in one of $\cap\_{n\ge N} A\_n$, which means $x \in A\_n$ for all $n > N$. Thus, for $x$ to be in the $\liminf$, it must be in all of the $A\_n$, with finitely many exceptions. This is how the phrase "ultimately all of them" comes up.


Both of these operations, similar to their counterparts in metric spaces, concern the tail of the sequence $\{A\_n\}$. I.e., neither changes if an initial portion of the sequence is truncated. As a previous response pointed out, often the sets $A\_n$ are defined to track the deviation of a sequence of random variables from a candidate limit by setting $A\_n = \{x: |Y\_n(x) -Y(x)| \ge \epsilon\}$. The members of $\limsup A\_n$ then represents those sequences that every now and then deviate $\epsilon$ away from $Y(x)$, which is solely determined by the tail of the sequence $Y\_n$.


Here is a conceptual game that can be partially understood using these concepts: We have a deck of cards, on the face of each card an integer is printed; thus the cards are $\{1,2,3...\}.$ At the nth round of this game, the first $n^2$ cards are taken, they are shuffled. You pick one of them. If your pick is 1, you win that round. Let $A\_n$ denote the event that you win the nth round. The complement $A\_n^c$ of $A\_n$ will represent that you lose the $n^{th}$ round. The event $\limsup A\_n$ represents those scenarios in which you win infinitely many rounds. The complement of this event is $\liminf A\_n^c$, and this represents those scenarios in which you ultimately lose all of the rounds. By the Borel Cantelli Lemma $P(\limsup A\_n)$ $=0$ or equivalently $P(\liminf A\_n^c)=1$. Thus, a player of this game will deterministically experience that there comes a time, after which he never wins.


