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On page 14.

Suppose $\{f_n\}$ is a sequence of extended-real functions on a set $X$. Then $\sup\limits_nf_n$ and $\limsup\limits_n f_n$ are the functions defined on $X$ by $$ \left(\sup_nf_n\right)(x)=\sup_n(f_n(x)), $$ .....

How to understand these two $n$s of $\sup\limits_n(f_n(x))$? May it be $$ \sup \{f_n(x),f_{n+1}(x), \cdots\}? $$ But in Theorem 1.14 on the same page, there is another notation: $\sup\limits_{n\geq 1}f_n$, this theorem says

If $f_n:X\to [-\infty, +\infty]$ is measurable, for $n=1,2,3,\ldots,$ and $$ g=\sup_{n\geq 1} f_n,~~~~h=\limsup_{n\to \infty} f_n, $$ then $g$ and $h$ are measurable.

In the proof, there is a $g^{-1}((\alpha, \infty]=\cup_{n=1}^\infty f_n^{-1}((\alpha, \infty])$. From here I guess that $\left(\sup\limits_{n\geq 1}f_n\right)(x)=\sup\{f_1(x), f_2(x), \cdots\}$, is it right?

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The subscript of $\sup$ is meant to denote the variable for which the supremum is taken, and when it isn't clear from context, where that variable ranges. That is: \begin{align} \sup_n(f_n(x))&:=\sup\{f_n(x): n\in\Bbb N\}\quad\text{we 'guess' the range from context}\\ \sup_{n\geq k}(f_n(x))&:=\sup\{f_n(x): n\geq k\}\quad\text{we are explicitly given the range} \end{align}

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