# Intuition behind $\limsup$ and $\liminf$ for probabilities

I've come across these limits in Fatou's lemma, this got me massively confused.

I'd be grateful if someone could explain the intuition behind limit suprema and limit infima of probabilities (or other measures). Though I understand that say $1=\sup(0,1)$ and $\inf(0,1)=0$ since the sets are open, application of this limit to probabilities I can't quite get into.

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\liminf and \limsup do exist :D – Mariano Suárez-Alvarez Feb 11 '12 at 4:26
For a sequence of real numbers $a_n$, $\limsup a_n$ is the largest limit point of the sequence and $\liminf a_n$ is the smallest limit point of the sequence. – KCd Feb 11 '12 at 4:51
It may help to note that: $\liminf x_n=\sup_{n\geq 0}\inf_{m\geq n} x_m$ and $\limsup x_n=\inf_{n\geq 0}\sup_{m\geq n} x_m$ – john w. Feb 11 '12 at 5:03
I see, but how can I related bounds on set to bounds on probabilities? – user19821 Feb 12 '12 at 0:29

Two types of limsup and liminf are at play here:

1. the liminf and limsup of sequences of real numbers,
2. the liminf and limsup of sequences of subsets of a given set.

Although these might look different, both are actually based on the same construction, available as soon as one considers an ordered set: the real line with the usual (total) order in case 1. and the set of subsets of a given set with the (partial) inclusion order in case 2. Let me now explain this in more details.

We start with the limsup of a sequence $\mathbf x=(x_n)_{n}$ of real numbers. As you know, the idea is to consider $s_n(\mathbf x)=\sup\limits_{k\geqslant n}x_k$ for every $n$, and then $s(\mathbf x)=\lim\limits_{n\to\infty}s_n(\mathbf x)=\inf\limits_{n}s_n(\mathbf x)$. One can recognize $s(\mathbf x)=\limsup x_n$ as the supremum of the set of real numbers $z$ such that $x_n\geqslant z$ for infinitely many indexes $n$.

Likewise, consider a sequence $\mathbf A=(A_n)_{n}$ of subsets of a given set $\Omega$. Introducing $s_n(\mathbf A)=\sup\limits_{k\geqslant n}A_k$ for every $n$ and $s(\mathbf A)=\lim\limits_{n\to\infty}s_n(\mathbf A)=\inf\limits_{n}s_n(\mathbf A)$ defines uniquely a set $s(\mathbf A)=\limsup A_n$.

The only difference with the case of real valued sequences is that the order relation on the set of the subsets of $\Omega$ being the inclusion, now the supremum of some sets is their union and their infimum is their intersection, instead of the supremum and the infimum in the real line when considering sequences of real numbers. Hence $s_n(\mathbf A)=\bigcup\limits_{k\geqslant n}A_k$ for every $n$ and $s(\mathbf A)=\bigcap\limits_{n}s_n(\mathbf A)$.

One can recognize $s(\mathbf A)$ as the smallest possible $B$ such that $A_n\subseteq B$ for every index $n$ except finitely many of them. If $B$ is such a set and $\omega$ is in $B$, then $\omega$ is in $A_n$ for infinitely many $n$. Conversely, one sees that $s(\mathbf A)$ is exactly the set of $\omega$ such that $\omega$ is in $A_n$ for infinitely many $n$.

Similar constructions are available for the liminf of sequences of real numbers and of subsets of $\Omega$. Now, $i(\mathbf x)=\liminf x_n$ is the infimum of the set of real numbers $z$ such that $x_n\leqslant z$ for infinitely many indexes $n$ and $i(\mathbf A)=\liminf A_n$ is the set of $\omega$ such that $\omega$ is in $A_n$ for every $n$ large enough.

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+1. This is an excellent reply! – Tim Feb 25 '12 at 1:26