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I'm trying to understand more about the limits of sequences of sets in Measure Theory.

Given a sequence of sets $\{A_n\}_{n\in \mathbb{N}} = \{ A_1,A_2, \ldots \}$, what does $\sup_n \{ A_n \}$ represent?

The reason why I'm asking is because I'd like to derive what $\limsup_{n\rightarrow\infty}A_n$ means… and this should formally be $\lim_{n\rightarrow\infty} \sup\{A_k | k \geq n \}$ - right?

2 Answers 2

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The notation $\sup_n\{A_n\}$ is ambiguous, and I would avoid using it without more context. In the context of the $\limsup$ or $\liminf$ of sets, we are taking the partial order on sets by inclusion: $A \leq B$ if $A \subseteq B$. Then the supremum of a sequence $\{A_n\}_n$ is the smallest possible upper bound for every element of the sequence - the smallest set containing each $A_n$. This must be $\sup_n A_n = \bigcup_n A_n.$

Note that not every partially ordered set has well-defined suprema and infima - this kind of poset is called a lattice.

For a sequence of real numbers, the definition of limit superior is ${\limsup} \,\{a_n\}_n = \lim_{n \rightarrow \infty} \sup_{m \geq n} a_m.$ The notation $\lim_{n \rightarrow \infty} A_n$ doesn't make sense for a sequence $\{A_n\}_n$ of sets. But observe that if $B_n = \bigcup_m A_{m \geq n}$ then $B_n$ is a nested sequence: $B_{n+1} \subseteq B_n$, since $B_{n+1}$ is the union over a smaller set of indices. Since $B_n$ is getting smaller and smaller, we might define the "limit" of $B_n$ to be small: the set of $x$ so contained in every $B_n$, or $\cap_n B_n$. Putting this together, a reasonable analog for the $\lim \sup$ of a sequence of real numbers applied to sets is

$\limsup A_n := \bigcap_{n=1}^\infty \bigcup_{m =n}^\infty A_m.$ Taking apart this definition, we see that $x \in \lim \sup A_n$ if and only if for all $n \in \mathbb{N}$ there exists $m \geq n$ so that $x \in A_m$: $\forall n \in \mathbb{N}\, \exists m \in \mathbb{N} \,m \geq n \text{ and } x \in A_m.$

This says that no matter how large of an $n$ you choose, I can find a larger $m$ so that $x \in A_m$. Another way of saying this is that there is no upper bound $n$ to the set of $m$ so that $x \in A_m$; and this is equivalent to saying that $x \in A_m$ for infinitely many $m$.

See the Wikipedia article for more info.

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    @JairTaylor Thanks!2012-10-27
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To complement the answer of Jair Taylor: We can relate the notion of $\sup$ and $\limsup$ to the usual notions for real numbers by working with indicator functions: $x\in\bigcup_n A_n\iff x\in A_n\textrm{ for some }n\iff 1_{A_n}(x)=1\textrm{ for some }n\iff \sup_n 1_{A_n}(x)=1$

and

$\lim_{n\to\infty}\sup 1_{A_n}(x)=1\textrm{ iff } \lim_n 1_{\bigcup_{m=n}^\infty {A_m}}(x)=1\iff \inf_n 1_{\bigcup_{m=n}^\infty {A_m}}(x)=1\iff x\in\bigcap_n \bigcup_{m=n}^\infty A_n\iff x\in A_n\textrm{ for infinitely many }n.$

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    I mean the limit of the suprema. Now \limsup_{n\rightarrow\infty}1_{A_n}(x) = \lim_{n\rightarrow\infty} \sup_{k} \{1_{A_k}(x) | k > n\} only means that there are *infinitely many* $n$ such that $A_n$ contains $x$, while $x\in\bigcap_n A_n$ means $x\in A_n$ for *all* $n$.2012-10-28