I can't make sense of the following definition of supremum of sequence of functions. I can't make sense of the following definition of supremum of sequence of functions.
This definition comes from Garden of Integrals (Burk), chapter 5 (page 99):
$$\sup\left\{f_{k}(x),f_{k+1}(x),\ldots  \right \}=\bigcup_{n\geq k}\left \{ x \in  E: f_{n}(x)>c \right\}$$ 
where all $f$'s are defined on $E$. How can supremum over $f$'s can yield a union of sets taking values from $E$?
Thanks
 A: The context: Burk states the theorem: 

Theorem. If $\{f_k\}$ is a sequence of Lebesgue measurable functions defined on a measurable set $E$, with $\lim f_k=f$ pointwise on $E$, then $f$ is Lebesgue measurable on $E$.

He wants to use this to show that if $\{f_k\}$ are measurable, then "measurability is preserved under many limiting operations", for example, that $\limsup f_k$ is measurable. 
To that end, we define $g_k(x)$ to be
$$g_k(x) = \sup\{f_k(x),f_{k+1}(x),f_{k+2}(x),\ldots\}$$
because then $\limsup f_k = \lim g_k$; if each $g_k$ is measurable, then $\limsup f_k$ is measurable by the theorem. 
But I agree that there is a mess there. I think that this was a chopped-up attempt at proving that $g_k$, thus defined, is measurable. To verify that it is measurable, fix $c\in \mathbb{R}$. Then
$$\{x\in E\mid g_k(x)\gt c\} = \bigcup_{n\geq k}\{x\in E\mid f_n(x)\gt c\}.$$
Indeed, if $x$ is in the right hand side, then there exists $n\geq k$ such that $f_n(x)\gt c$, hence $g_k(x) = \sup\{ f_m(x)\mid m\geq k\}\geq f_n(x)\gt c$, so $x$ is in the left hand side.
Conversely, if $x$ is in the left hand side, then $\sup\{f_k(x),f_{k+1}(x),\ldots\}\gt c$, so there exists $n\geq k$ such that $f_n(x)\gt c$, hence $x$ is in the right hand side.
Thus, since $\{x\in E\mid g_k(x)\gt c\}$ is a countable union of measurable sets, it is measurable, so $g_k$ is measurable.
Similar comments account for the next line, where Burk writes
$$h_k(x) \equiv \inf\{f_k(x),f_{k+1}(x),\ldots\} = \bigcup_{n\geq k}\{ x\in E\mid f_k(x)\lt c\}$$
