I don't understand $\limsup_n\:A_n$ in Probability Theory, why should it be equal to "infinitely many $A_n$ occur"?

Question

The definition for limsup is:

$$\bigcap _{n=1}^{\infty }\bigcup _{k=n}^{\infty }\:A_k$$

So this means that $$\left(A_1\cup \:A_2\cup \:A_3\cup \:...\right)\cap \left(A_2\cup \:\:A_3\cup \:\:...\right)\cap \:\left(A_3\cup \:\:\:...\right)\cap \:...$$ which is equal to the last element of this infinite sequence (let's call it $A_{\infty }$)

For example let $A_n\:=\:\left[0,\frac{1}{n}\right]$, then $$\left(A_1\cup \:A_2\cup \:A_3\cup \:...\right)\cap \left(A_2\cup \:\:A_3\cup \:\:...\right)\cap \:\left(A_3\cup \:\:\:...\right)\cap \:... = \left\{0\right\}$$ as $$\left[0,\frac{1}{\infty }\right]\:=\:\left[0,\:0\right]\:=\:\left\{0\right\}$$

The problem here with Probability Theory and its understanding of limsup is that we can create such kind of scenarios that "$A_{\infty }$" won't be equal to "ifinitely many $A_n$ occur".

For example let $A_{n}=\left\{\text{person number n wins the lottery}\right\}$, "$A_{\infty }$" (and thus $\limsup_n\:A_n$) would mean that the "last person of this infinite sequence will win the lottery", and not that "infinitely many $A_n$ occur" which is equal to "everyone wins" (as this is clearly not the case here, because only the $\infty$th person wins)

I really need help to understand this concept

Answer 1

There is no such "last person". What we have is $A_i$ where $i=1,2,3,\dots$ (i.e. $i\in \mathbf Z_+$)

So in your example $\lim \sup A_n=\emptyset$.

Understanding of $\lim\sup$:

Let's say $a\in \lim \sup A_n=\bigcap_m\bigcup_{n\ge m} A_n$. That means $a$ occurs in $\bigcup_{n\ge m} A_n$ for every $m$. If $a$ doesn't occur in $A_1,A_2,\dots$ for infinite times, say, $a\in A_{i_1},A_{i_2},A_{i_3},\dots,A_{i_k}$. Then a large integer $m_0$ (large enough so that $m_0>i_1,i_2,\dots,i_k$) will lead to contradiction since $a\not\in \bigcup_{n\ge m_0} A_n$.

Answer 2

I think it's better to consider a harder example here - since your example of $A_n=[0,1/n]$ has lots and lots of special properties that make it a bit hard to generalize from (e.g. it's a nested sequence of compact sets).

Consider, for instance, the sequence (not worrying about non-unique decimal representations): $$A_n=\{x : \text{the }n^{th}\text{ decimal digit of }x\text{ after the decimal point is }7\}$$ Each of these sets is just some union of intervals, but the way they fit together is complicated - essentially, from a probabilistic point of view, these are independent events. Note that there is surely no "last" term in this sequence because there is no last place value in a decimal expansion.

Let's consider the $\limsup$ of this in parts. First, for any $n$, consider $$(A_n\cup A_{n+1}\cup A_{n+2}\cup\ldots)$$ Literally, this is the set of $x$ such that there is a $7$ in the $n^{th}$ place after the decimal point or at the $(n+1)^{th}$ place or the $(n+2)^{th}$ or so on - otherwise said, it's the set of $x$ such that there is a $7$ somewhere at or after the $n^{th}$ place.

If you intersect these unions over all $n$, you get the $\limsup$, which then says, "The set of $x$ such that, for every $n$, there is a $7$ in the decimal expansion of $x$ either at or after the $n^{th}$ place."

This condition is the same as "infinitely many $7$'s". Note that if there were infinitely many $7$'s, there would have to be a $7$ after any $n$ places, because there could only have been finitely many $7$'s before that number of places. Conversely, if $x$ is in our $\limsup$, then there must be at least one place with a $7$, but there also must be another $7$ somewhere further down the line (by taking $n$ to be the place value after the first $7$) and another $7$ after that and so on - giving an infinite sequence of $7$'s. This reasoning generalizes to the idea that the $\limsup$ of a sequence of sets is the set of $x$ which are contained in infinitely many sets of that sequence.

Just to dispel any idea of "the infinite last term", you might want to consider the sequence $A_n=[1/{2n},1/n]$. You should note first that no term is in infinitely many of these sets - these are all sets of positive numbers, but $1/n$ will eventually be less than any positive number. You can see this from the $\limsup$ too: the union $(A_n\cup A_{n+1}\cup A_{n+2}\cup\ldots)$ is just equal to $(0,1/n]$ and then intersecting these over all $n$ gives the empty set. Note that, if you tried to say that the sequence had an imaginary "last term" or "limit" $A_{\infty}=[1/\infty,1/\infty]=[0,0]$, you would get the wrong answer - such notions have nothing to do with computing $\limsup$.

Answer 3

Use elements to understand it. Set $B_n:= \bigcup_{k=n}^{\infty }A_k$ and note three things:

1. $\limsup_{n\to \infty }A_n= \bigcap_{n=1}^{\infty }B_n$

2. $B_1 \supset B_2\supset \ldots \supset B_k\supset \ldots $, that is, the sequence of sets $(B_n)_n$ is decreasing.

3. Every $B_n$ is the union of infinitely many $A_k$.

Then if $x\in \bigcap_{n=1}^{\infty }B_n$ it means that $x\in B_n$ for all $n\in {\mathbb N}$. Now suppose that $x$ belongs only to a finite number of $A_k$, then as the $B_n$ are shrinking and they contains less and less $A_k$ each time then there is some $N\in {\mathbb N}$ such that $x\notin B_n$ for all $n\geqslant N$. This has shown that if $x$ belong to a finite number of $A_n$ then it cannot belong to $\limsup_{n\to \infty }A_n$, therefore if $x\in \limsup_{n\to \infty }A_n$ it necessarily belongs to an infinite number of $A_n$.

By the other hand if $x$ belongs to an infinite number of $A_n$ then $x\in B_n$ for all $n\in {\mathbb N}$, and so it belongs to the limit superior of the $A_n$. Then we had shown that $x\in \limsup_{n\to \infty }A_n$ if and only if $x$ belongs to infinitely many $A_n$.