why $2^{-n}h_n(x)=t_n(x)$ except in $V_n- K_n?$ Rudin RCA- Lusin's theorem I have some  confusion regarding Rudin RCA book, page number $55$

Lusin's  theorem  : Suppose  $f$ is a complex   measurable  function on $X$ .$\mu(A) < \infty$ , $f(x)= 0$if  $x \notin A $ and $\epsilon >0$.The  there  exists a $g \in C_c(X)$ such that $ \mu(\{x : f(x) \neq g(x)\} <\epsilon$

In the theorem of the proof it  is written that

By Urysohn's lemma ,there are function $h_n$ such that  $K_n\prec h_n \prec V_n.$ Define  $$g(x) =\sum_{n=1}^{\infty}2^{-n}h_n(x) $$ for all $ x \in X$
This series converge uniformly on X , so $g$  is contnious.Also ,the support of  $g$ lie in $\bar V$. since $2^{-n}h_n(x)=t_n(x)$  except in $V_n- K_n$

My confusion : why $2^{-n}h_n(x)=t_n(x)$  except in $V_n- K_n?$ why not  $2^{-n}h_n(x)=t_n(x)$  except in $X- V_n?$
My thinking : $h_n(x)= 2^nt_n(x)=\begin{cases} 1 \ \text{if  x} \in V_n \\ 0 \ \text{if  x} \notin V_n  \end{cases} = \begin{cases} 1 \ \text{if  x} \in V_n \\ 0 \ \text{if  x} \in  X- V_n  \end{cases}$
So  I think it should be $2^{-n}h_n(x)=t_n(x)$  except in $X- V_n$
 A: This is a long comment trying to explain the the arguments used in the reference given by the OP to the proof of Lusin's theorem:

*

*It is assumed that there is a set $A$ such that $f=0$ on $X\setminus A$.

*Only the case where $\mu$ is a real finite measure, $0\leq f<1$ and $A$ compact is analyzed for the general conclusion follows from this.

One uses the standard  nondecreasing dyadic sequence $s_n$ ($s_0=0$) that converges to $f$. Set $t_n=s_n-s_{n-1}$. Since $f=0$ on $X\setminus A$,  $2^nt_n$ is the indicator function of a set $A_n\subset A$; hence $$f=\sum_nt_n=\sum_n2^{-n}(2^nt_n)=\sum_n2^n\mathbb{1}_{A_n}$$
Let $V$ be an open set with compact closure $\overline{V}$ such that $A\subset V$ (this can be done because $X$ is locally compact and Hausdorff). By regularity of Borel measures, there are compact and open sets $K_n$ and $V_n$ such that $K_n\subset A_n\subset V_n\subset V$ such that $\mu(V_n\setminus K_n)<2^{-n}\varepsilon$. Applying Urysohn's lemma gives a sequence of functions $h_n\in\mathcal{C}_{00}(X)$ such that $K_n\prec h_n\prec V_n$. Then
$$g:=\sum_n2^{-n}h_n,$$
being a convergent sequence of continuous functions, is a continuous function on $X$, and $\operatorname{supp}(g)\subset \overline{V}$. Furthermore,   $g(x)=0$ for all $x\in X\setminus\bigcup_nV_n$, and $\{g=0\}\subset\{f=0\}$.
Notice that $h_n=1$ on $K_n$; hence, for $x\in K_n$, $2^{-n}h_n(x)=2^{-n}\mathbb{1}_{A_n}(x)=2^{-n}t_n(x)$. Thus, on $\bigcup_nK_n$ we have  $f=g$. On $\Big(\bigcup_nV_n\setminus\bigcup_nK_n\Big)\subset\bigcup_n(V_n\setminus K_n)$ it may be that $g\neq f$; however, $\mu\Big(\bigcup_n(V_n\setminus K_n)\Big)\leq\sum_n\mu(V_n\setminus K_n)<\sum_n2^{-n}\varepsilon=\varepsilon$. Finally notice that  $\{f\neq g\}\subset\bigcup_n V_n\setminus K_n$.
