I do not understand some passages of this proof. Can someone help me?
Let $X\ne\emptyset$ and $\mathcal{A}$ a $\sigma$-algebra on $X$. On $\mathbb{R}$ we consider the Borel $\sigma$-algebra $\mathcal{B}$.
Definition. An application \begin{equation} f\colon X\to\mathbb{R} \end{equation} is $\textit{measurable}$ if $f^{-1}(B)\in\mathcal{A}$ for all $B\in\mathcal{B}$.
Theorem. If $\{f_n\}$ is a sequence of measurable functions on $X$ in $\mathbb{R}$, then are measureble (on the set where they assume finite value) \begin{equation} \begin{split} 1)&\quad \sup f_n(x)\\ 2)&\quad \inf f_n(x)\\ 3)&\quad \limsup f_n(x) \end{split} \end{equation}
proof. $1)$ Let $f(x)\colon=\sup f_n(x)$ . Observe that the set on which $f(x)$ assumes finite value is measurable, in fact \begin{equation} A=\{x:f(x)<\infty\}=\bigcup_{m=1}^{\infty}\{x:f_n(x)\le m,\;\forall n\in\mathbb{N}\}. \end{equation} Now, we show that $f(x)$ is measurable: \begin{equation} \{x\in A:f(x)\le a\}=\bigcap_{n\in\mathbb{N}}\{x\in A:f_n(x)\le a\}\in\mathcal{A} \end{equation} for all $a\in\mathbb{R}$.
$2)$ $\inf f_n=-\sup(-f_n(x))$.
$3)$ We observe that the set on which $\sup_{k\ge n}f_k(x)<\infty$ does not depend on $n$ (why?). Let $A=\{x:\sup_{k\ge n}f_k<\infty\}$ and we place $g_n=\sup_{k\ge n}f_k(x)$ which is measurable on $A$ for $1)$. The set $A'=\{x:\inf_n(\sup_{k\ge n}f_k(x))>-\infty\}\subset A$ (why?). On $A'$ $\limsup f_n(x)$ is measurable for $1)$ and $2)$.
Thanks!
$A_n=\{\sup_{k\ge n} f_k=g_n<\infty\}$ is independent of $n$ because $\mathbb R =\{f_k<\infty\}$ for all $k$. $A'$ is obtained as a subset of $A$ by definition (you are working with $g_n:A\to\mathbb R$ ). $\lim_{n\to\infty}\sup_{k\ge n} = \inf_{n}\sup_{k\ge n} $ and part $2)$ proves that the left-hand-side is measurable. It might be helpful to try to prove a similar statement for sequences $f_n:X\to \mathbb R\cup \{-\infty,\infty\}$.