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?


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.

  • $\begingroup$ \limsup and \liminf are built-in LaTeX commands, actually. $\endgroup$ – Asaf Karagila Oct 27 '12 at 18:39
  • $\begingroup$ @Jair Taylor Thanks for this! It helps to know that limit properties of sequences do not naturally carry over to sets... That said, according to the definition you provided, is it true that $\liminf_n A_n \subset \inf_{n} A_n = \cap_n A_n \subset \cup_n A_n = \sup_n A_n \subset \limsup_n A_n$? $\endgroup$ – Elements Oct 27 '12 at 18:41
  • $\begingroup$ @AsafKaragila: Thanks! I tried this at first but I thought it didn't work for some reason. $\endgroup$ – Jair Taylor Oct 27 '12 at 18:43
  • $\begingroup$ \bigcup and \bigcap yield better results in displayed equations. $\endgroup$ – Did Oct 27 '12 at 18:45
  • $\begingroup$ @Elements: You have it backwards. $\cap_n A_n \subseteq \liminf_n A_n \subseteq \limsup_n A_n \subseteq \cup_n A_n$. $\endgroup$ – Jair Taylor Oct 27 '12 at 18:46

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: $$x\in\bigcup_n A_n\iff x\in A_n\textrm{ for some }n\iff 1_{A_n}(x)=1\textrm{ for some }n\iff \sup_n 1_{A_n}(x)=1$$


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

  • $\begingroup$ In the second equation, did you mean to write $\limsup_{n\rightarrow\infty}1_{A_n}(x) = 1$ iff \lim_n $1_{\cap_{m=n}^\infty A_m}(x) = 1$... If we use the normal definition of limit superior, then $\limsup_{n\rightarrow\infty}1_{A_n}(x) = \lim_{n\rightarrow\infty} \sup_{k} \{1_{A_k}(x) | k > n\}$... so this should require that $x \in A_k$ for all $k > n$ as $n\rightarrow\infty$. In other words, $x\in\cap_{n=1}^\infty A_n$ - right? Or am I confused because you meant $\lim_{n\rightarrow\infty}(\sup)$ and not $\limsup{n\rightarrow\infty}$ $\endgroup$ – Elements Oct 28 '12 at 4:27
  • $\begingroup$ 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 $n$. $\endgroup$ – Michael Greinecker Oct 28 '12 at 10:20

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