Justify the inequalities Notation: $X,Y$ are locally compact Hausdorff spaces with first countability. $C_0(X)$ denotes complex-valued continuous functions on $X$ which vanish at infinity. The space is endowed with the supremum norm $\|f \|_{\infty} = \sup_{x \in X}|f(x)|.$
Let $\phi:C_0(X) \rightarrow C_0(Y)$ be a linear isomorphism and norm-increasing with $\|\phi\| < 2.$
Its adjoint is denoted as  $\phi^*:C_0(Y)^* \rightarrow C_0(X)^*.$
For any $x \in X,$ $\mu_x$ is a unit positive mass measure concentrated at $\{x\}$ only, that is, $\mu\{x \} =1$ and $\mu(X \setminus \{ x \}) = 0.$
In Cambern's paper entitled 'A Generalized Banach Stone Theorem,', he quoted the following in the the proof: 

Let $x \in X$ be any point of $X$ and let $\{ U_n:n\in \mathbb{N} \}$ be a neigbourhood of basis $x,$ with $U_{n+1} \subseteq U_n$ for all $n.$
  For each $n$, choose a function $f_{x,n} \in C_0(X)$ with $f_{x,n}(x) = \| f_{x,n}\|_{\infty} = 1$ and $f_{x,n}(x^{\prime}) = 0$ for all $x^{\prime} \in X \setminus U_n.$ Then $\lim_{n}\int f_{x,n} d\mu$ exists for all $\mu \in C_0(X)^*,$ and hence $\lim_n \int \phi(f_{x,n}) d\mu$ exists for all $\mu \in C_0(Y)^*.$
We claim there is at least one $y \in Y$ such that $\lim_n |\phi(f_{x,n})(y)| > M.$ For if not, since the moduli $|\phi(f_{x,n})|$ are uniformly bounded by $2$,  we would obtain 
  $$1 = \lim_n \int f_{x,n} d\mu_x = \lim_n \int \phi(f_{x,n}) d({\phi^*}^{-1} \mu_x) = \int \left( \lim_n \phi(f_{x,n}) \right)d({\phi^*}^{-1} \mu_x) \leq \int \left| \lim_n \phi(f_{x,n}) \right| d\left|({\phi^*}^{-1} \mu_x)\right| \leq M.$$

Question: I have trouble obtaining the two inequalities 
$$\int \left( \lim_n \phi(f_{x,n}) \right)d({\phi^*}^{-1} \mu_x) \leq \int \left| \lim_n \phi(f_{x,n}) \right| d\left|({\phi^*}^{-1} \mu_x)\right| \leq M.$$
Any hint would be appreciated.
 A: The first inequality is of the form
$$\int f \, d\mu \leq \int |f| \, |d\mu|$$
Let us insert a couple of inequalities:
$$
\int f \, d\mu 
\leq \left| \int f \, d\mu \right| 
\leq \int \left| f \, d\mu \right| 
= \int |f| \, |d\mu|
$$
The first inequality is just the real number inequality, $x \leq |x|$. The second inequality is an integral analog of the triangle inequality $|x+y|\leq|x|+|y|$, and the last equality is just a measure analog of $|xy| = |x| |y|$.
The second inequality depends on
$
| \lim_n \phi(f_{x,n}) | 
= \lim_n | \phi(f_{x,n}) |
\leq M
$
and
$
d\left|({\phi^*}^{-1} \mu_x)\right|
$
being a unit positive mass measure.
A: The first inequality is fairly straightforward. Replacing the function by its absolute value and the measure by its total variation is going to increase (or at least not decrease) the integral.
For the second inequality, we need two facts:
i)  This statement is proved by contradiction, so we are assuming 
$|\lim_n(\phi(f_{x,n}))(y)| = \lim_n|(\phi(f_{x,n}))(y)|\leq M$ for all values of $y$.
ii) $\lVert (\phi^*)^{-1} \rVert = 1$ as pointed out at the top of page 398
Then we have
\begin{align}
\int \left| \left( \lim_n \phi(f_{x,n})  \right) \right| d |\phi^{*-1}\mu_x |
&\leq
\sup_y \left| \left( \lim_n \phi(f_{x,n}) (y) \right) \right| \lVert \phi^{*-1} \rVert \lVert \mu_x \rVert \\
&\leq
M
\end{align}
Edit: More details
\begin{align}
\int \left| \left( \lim_n \phi(f_{x,n})  \right) \right| d |\phi^{*-1}\mu_x |
&\leq
\sup_y \left| \left( \lim_n \phi(f_{x,n}) (y) \right) \right|
\int  d |\phi^{*-1}\mu_x |\\
&\leq
\sup_y \left| \left( \lim_n \phi(f_{x,n}) (y) \right) \right|
|\phi^{*-1}\mu_x |(Y)\\
&\leq
\sup_y \left| \left( \lim_n \phi(f_{x,n}) (y) \right) \right| \lVert \phi^{*-1} \rVert \lVert \mu_x \rVert \\
&\leq
M
\end{align}
