What does the supremum of a sequence of sets represent? 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?
 A: 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 
$$\lim_{n\to\infty}\sup 1_{A_n}(x)=1\textrm{ iff } \lim_n 1_{\bigcup_{m=n}^\infty {A_m}}(x)=1\iff \inf_n 1_{\bigcup_{m=n}^\infty {A_m}}(x)=1\iff x\in\bigcap_n \bigcup_{m=n}^\infty A_n\iff x\in A_n\textrm{ for infinitely many }n.$$
A: 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.
A: I am adding an other way of understanding it, i was too a little bit confuse, i hope it will help futur reader.
Let $X_n(\omega)$ a sequence of real r.v. and we suppose that $sup_{n \in \mathbb{N}}(X_n(\omega))=S(\omega)=S\in \mathbb{R}$ exists .
That means by definition:
1-$\forall n,\omega \Rightarrow X_n(\omega) \leq S$
2-It exists at least one $n' \in \mathbb{N}$ and at least one $\omega \in \Omega$ s.t. : $a<X_{n'}(\omega) \leq S$
Now we note: $\bigcup_{n \in \mathbb{N}}^{}\left \{ \omega \in \Omega: X_n(\omega)>a \right \}=\bigcup_{n \in \mathbb{N}}^{}\left \{ \omega \in \Omega:S \geq X_n(\omega)>a \right \}=\bigcup_{n \in \mathbb{N}}^{}\left \{ \omega \in \Omega : S=sup_{n\in \mathbb{N}}(X_{n }(\omega)) \geq a \right \}=\left \{ \omega \in \Omega : S=sup_{n\in \mathbb{N}}(X_{n }(\omega)) \geq a \right \}$
