Construction of $\mu$ and $\mathfrak M$ in Riesz Representation Theorem (Rudin) Rudin defines

For every open set $V$, $$\mu(V) = \sup\{\Lambda f: f\prec V\}\ \  \ldots \text{(1)}$$ where $f\prec V$ means that $f\in C_c(X)$, $0\le f\le 1$, $\text{supp}(f) \subset V$.

$\color{red}{\text{Question 1. }}$Then they go on to say, it is clear that if $V_1\subset V_2$, then $\mu(V_1) \le \mu(V_2)$ according to the above definition. Why is that? $$\mu(V_1) = \sup\{\Lambda f: f\prec V_1\}, \mu(V_2) = \sup\{\Lambda f: f\prec V_2\}$$
Thus far, we don't know that $\Lambda f = \int_x f d\mu$, since $\mu$ was constructed just now.
The author now concludes using the above definition of $\mu(V)$ for open sets, that

$$\begin{equation}\mu(E) = \inf\{\mu(V): E\subset V, V\text{ is open}\}\end{equation} \ \  \ldots \text{(2)}$$ if $E$ is an open set, and it is consistent with (1) to define $\mu(E)$ by (2) for every $E\subset X$.

$\color{red}{\text{Question 2. }}$Where does this come from? I see it is consistent with the previous definition because of this answer.

Let $\mathfrak M_F$ be the class of all $E\subset X$ which satisfy $\mu(E) < \infty$ and $\mu(E) = \sup\{\mu(K): K\subset E, K\text{ is compact}\}$. Finally let $\mathfrak M$ be the class of all $E\subset X$ such that $E\cap K\in\mathfrak M_F$ for every compact $K$. It is evident that $\mu$ is monotone, i.e. $\mu(A) \le \mu(B)$ if $A\subset B$ - and that $\mu(E) = 0$ implies $E\in \mathfrak M_F$ and $E\in\mathfrak M$. Thus (e) holds, and so does (c), by definition.

$\color{red}{\text{Question 3. }}$Why is $\mu$ monotone? This is not obvious to me yet. Also, suppose $\mu(E) = 0$. Then, $\mu(E) < \infty$ - but why is $\mu(E) = \sup\{\mu(K): K\subset E, K\text{ is compact}\}$? Basically, how do we know that $E\in \mathfrak M_F$ and $E\in\mathfrak M$? If everything above holds, I do agree that (e) and (c) hold.
Thanks a lot! I have provided pictures the text below for context:





 A: 
Question 1:
If $V_1$ and $V_2$ are open and $V_1\subset V_2$, then
$$
\{\Lambda f:f\in C_c(X),\, f\prec V_1\}\subset \{\Lambda f:f\in C_c(X),\, f\prec V_2\}
$$
for if $\operatorname{supp}(f)\subset V_1$, then $\operatorname{supp}\subset V_2$. The rest of this question follows from properties of the supremum of sets. So
\begin{align}
\widetilde{\mu}(V_1)\leq \widetilde{\mu}(V_2)
\end{align}

Question 2 is not difficult to check. To clarify, lets define
$$
\begin{align}
\widetilde{\mu}(V):&=\sup\{\Lambda f: f\in C_c(X),\, f\prec V\} \tag{1}\label{one}
\end{align}
$$
for $V$ open, and
$$
\begin{align}
\mu(E):&=\inf\{\widetilde{\mu}(E): V\,open, \, E\subset V\} \tag{2}\label{two}
\end{align}
$$
What Rudin claims is that $\mu(E)=\widetilde{\mu}(V)$ when $E$ is open. This is immediate consequence of definition \eqref{one} for open sets, for
$E\in\{V\,open, \, E\subset V\}$ when $E$ is open; moreover, when $E$ is open, the $\inf$ is in fact a $\min$.

Question 3:
Once that it has been established that $\mu$ extends the definition of $\widetilde{\mu}$ to any subset in $X$, i.e., $\mu=\widetilde{\mu}$ when restricted to open sets, let's use $\mu$ henceforth.
Also, notice that $\mu$ is monotone, for if $E_1\subset E_2$, the
$$
\{\mu(V):\, V\,open,\, E_2\subset V\}\subset \{\mu(V):\, V\,open,\, E_1\subset V\}
$$
and so, by the properties of $\inf$, $\mu(E_1)\leq\mu(E_2)$.
The rest is to restrict $\mu$  to  special classes of sets: $\mathfrak M_F$ and later on $\mathfrak M$. This is to produce a measure that is inner regular and outer regular (Rudin uses the Lebesgue-Carathéodory notion of measurability and first proves that $\mu$ is an outer measure, and that the sets in $\mathfrak M$ are measurable, and $\mathfrak M$ contains the Borel sets.)

