The notion of limsup for sets I was working through The Borel-Cantelli lemma from Real Analysis problem book and ran into the following comment:
Let $\{E_k\}_{k\geq1}$ is a countable family of measurable subsets of $\mathbb{R}^d$ and that $\sum \limits_{k=1}^{\infty}m(E_k)<\infty$.
Let $$E=\{x\in \mathbb{R}^d: x\in E_k, \text{for  infinitely many} \ k\}=\limsup\limits_{k\to \infty}(E_k).$$
I know the notion of $\limsup$ and $\liminf$ for sequences from $[-\infty,+\infty]$. But here $E_k$ are sets (NOT numbers!). How did they get that $$\{x\in \mathbb{R}^d: x\in E_k, \text{for  infinitely many} \ k\}=\limsup\limits_{k\to \infty}(E_k)?$$
This equality seems to me quite weird.
Would be very grateful for explanation!
EDIT: Is there some essential difference between $\limsup$ of sequence of sets and reals?
 A: We've seen in another answer that
$$ \limsup_{n \to \infty} E_n = \bigcap_{n \ge 1} \bigcup_{m \ge n} E_m.$$
One can define $\limsup$ more generally (on complete lattices) using the operations join $\vee$ and meet $\wedge$, in the symbolically obvious way:
$$ \limsup_{n \to \infty} x_n = \bigwedge_{n \ge 1} \bigvee_{m \ge n} x_m.$$
Here, $\{ x_i \}$ is a sequence in some complete lattice. 
Conveniently, given some set $X$, its power set $\mathcal P(X)$ is a complete lattice when $\vee$ is defined as taking unions, and $\wedge$ is defined as taking intersections. $\mathbb R$ is also a complete lattice when $\vee$ is taken to be your usual $\sup$, and $\wedge$ is taken to be your usual $\inf$.
Hence, we can view these two seemingly different definitions of $\limsup$ as instances of the same definition, only applied in slightly different settings (i.e. lattices).
A: You should probably just consider
$$\limsup\limits_{k\to \infty}(E_k) = \{x\in \mathbb{R}^d: x\in E_k, \text{for  infinitely many } k\}$$
to be a definition of the left-hand side for the moment.
Theoretical Economist provides a (correct) theoretical motivation for this definition, but it also has a nice interpretation in terms of pointwise limsups of functions:
If we define the sequence of functions
$$ f_k(x) = \begin{cases} 1 & \text{if } x\in E_k  \\ 0 & \text{otherwise } \end{cases} $$
-- that is, $f_k$ is the indicator function for $E_k$ -- then
$$ f(x) = \limsup_{k\to\infty} f_k(x) $$
is exactly the indicator function for the set $\limsup_{k\to\infty} E_k$.
Note that for a sequence of numbers $a_k$ that are all either $0$ or $1$, $\limsup_{k\to\infty} a_k$ is $1$ if and only $a_k=1$ for infinitely many $k$, and is $0$ otherwise.
A: For a sequence of sets $E_{1},E_{2},\ldots$ define their limit superior by
$$
\limsup_{n}E_{n}=\bigcap_{n\geq1}\bigcup_{m\geq n}E_{m}.
$$
Compare this to the definition of the limit superior of a sequence of numbers $x_{1},x_{2},\ldots$
$$
\limsup_{n}x_{n}=\inf_{n\geq1}\sup_{m\geq n}x_{m}.
$$
The definitions above, while similar, are certainly not the same.
In particular, $\limsup_{n}E_{n}$ is a set while $\limsup_{n}x_{n}$ is a number.
Let's prove that $\limsup_{n}E_{n}=E$ where
$$
E=\left\{ x\colon x\in E_{n}\text{ for infinitely many }n\right\} .
$$
Suppose $x$ is in $E$.
By definition, there exist positive integers $n_{1}<n_{2}<\cdots$ such that $x$ is in $E_{n_{k}}$ for all $k$.
Now, let $n$ be arbitrary.
Since $n_n \geq n$ and $x$ is in $E_{n_n}$, it follows that $x$ is in $$\bigcup_{m\geq n} E_{m} = E_n \cup E_{n+1} \cup \cdots \cup E_{n_n} \cup E_{n_n+1} \cup \cdots$$
Since $n$ was arbitrary, $x$ is in $\limsup_{n}E_{n}$.
Can you complete the argument for the converse?
A: First, I think you are not sure what the limit of { a sequence of sets } is.  It is a similar concept to a sequence of real numbers. However, It is not the same! 
What I think visually is a set as a circle (Venn diagram),  a set E is contained by a set $E_1$, then $E_1$ is contained by $E_2$, and so on.  and all the sets are gradually increasing and approaching a set $E_k$ which is the limit .  
What is in $E_k$?,ie {x: x $\in E_{k}$,for infinitely many k}. We can also write it as unions of all sets. 
Secondly, lim sup $E_k$ can be just simplified as a limit for the connivence of understandings. if $E_k$ is the limit for a sequence of sets, it is also the limit for a subsequence of sets. this subsequence happens to be all upper bounds of $E_k$ in this case.
This is not rigorous, but hope this will help you to understand. 
