Lusin's theorem in Rudin's RCA (https://i.stack.imgur.com/AxsLz.png)
I can't understand the proof of Lusin's theorem in RCA( theorem 2.24).
I have understand that "Then $2^n t_n$ is the characteristic function..." Next, Rudin says fix an open set $V$ s.t. $A\subset V$, there are compact sets $K_n$ and open sets $V_n$ s.t. $K_n\subset T_n\subset V_n\subset V$ and $\mu (V_n-K_n)<2^{-n}\epsilon$.
I have questions; 
1) Why take $K_n\subset T_n\subset V_n\subset V$?
Because of theorem 2.7( Suppose $U$ is open in a locally compact Hausdorff space $X$, $K\subset U$, and $K$ is compact. Then there is an open set V with compact closure s. t. $K\subset V\subset \bar{V}\subset U$), we can check the existence of $V_n$ s.t. $K_n\subset V_n\subset V$. Note that we can take $K_n$ in such a way since X is a locally compact. By a regularity of measure (now suppose some measure described in Riesz-Markov-Kakutani theorem), that is $\mu (E)=\rm{sup}{ \mu (K) | K\subset E}$ for compact $K$, for every open set $E$, and for every measurable set $E$, note $T_n$ is a measurable set, we take $K_n, V_n$ s.t.  $K_n\subset T_n\subset V_n\subset V$. Am I right?
2) Why $\mu (V_n-K_n)<2^{-n}\epsilon$?
If X is a locally compact, σ-compact Hausdorff space and if measurable sets $M$ and a measure $\mu$ have properties described in the Riesz-Markov-Kakutani theorem (theorem 2.14), there is a closed set $F$ and open set $V$ s.t. $F\subset E \subset V$ and $\mu (V-F) <\epsilon$(theorem 2.17 (a)). Note that this claim holds even if a closed set is replaced a compact set since X is Hausdorff. Thus, $\mu (V_n-K_n)<2^{-n}\epsilon$ holds. But, in Lusin's theorem, X is not supposed σ-compact. How justified this inequality?
 A: First, Rudin says (in p.55) that $\mu$ satisfies the properties in Theorem 2.14. In particular, $\mu$ satisfies the following (see p.41):
(b) $\mu(K)<\infty$ for every compact set $K$ in $X$.
(c) For every $E\in \mathfrak{M}$, we have $\mu(E)=\inf \{\mu (V):E \subset V, V ~\text{open} \}$.
(d) For every open set $E$, and for every open set $E \in \mathfrak{M}$ with $\mu (E) <\infty$, we have $E\in \mathfrak{M}$, we have $\mu(E)=\sup \{\mu (K):K \subset E, K ~\text{compact} \}$.
Now let us see how your questions (1) and (2) hold. Note that the existence of an open set $V$ containing $A$ with $\bar {V}$ compact follows easily from the assumptions that $X$ is locally compact and $A$ is compact (use the compactness argument).
Since $s_n$ is measurable for each $n$ (by Theorem 1.17), so is $t_n$, and hence $T_n \subset A$ is a measurable set, i.e., we have $T_n \in \mathfrak{M}$. Since we are first assuming that $A$ is compact (see the first sentence in the proof of 2.24), we have $\mu(A) <\infty$ by (b). Thus, $\mu (T_n)$ is also finite, so we can apply both (c) and (d) to $T_n$. Therefore we have an open set $V_n$ and a compact set $K_n$ such that $K_n \subset T_n \subset V_n$, $\mu(V_n)<\mu(T_n)+2^{-(n+1)}\epsilon$, and $\mu(K_n)>\mu(T_n)-2^{-(n+1)}\epsilon$ (Note that in particular, $V_n,T_n,K_n$ all have finite measure). Since $V$ is an open set containing $T_n \subset A$, we may obviously assume that $V_n \subset V$, by replacing $V_n$ by $V_n \cap V$ if necessary. Then it follows that $K_n \subset T_n \subset V_n \subset V$, and
$$\mu(V_n-K_n)=\mu(V_n-T_n)+\mu(T_n-K_n)<2^{-(n+1)}\epsilon+2^{-(n+1)}\epsilon=2^{-n}\epsilon,$$
as desired.
