Uniqueness in the Riesz representation theorem for the dual of $C(X)$ in the book by Royden I'm reading the book Real Analysis 4e by Royden on the Riesz representation theorem for the dual of $C(X)$. I have two problems about the proof of this theorem. The book states the theorem this way in page 464,

Let $X$ be a compact Hausdorff space and $C(X)$ the linear space of real-valued functions on $X$, normed by maximum norm. Define the operator $T: \textrm{Radon}(X)\to[C(X)]^*$ by setting, for $v\in \textrm{Radon}(X)$, $$T_v(f)=\int_X fdv$$ for all $f\in C(X)$. Then T is a linear isometric isomorphism of $\textrm{Radon}(X)$ onto $[C(X)]^*$.

Here $\textrm{Radon}(X)$ is the collection of all signed Radon measures on $X$ equiped with the norm, $$\|\mu\|=\mu^+(X)+\mu^-(X)$$ where $\mu^+$ and $\mu^-$ are unique Randon measures according Jordan decomposition.
In its proof, 1) it begins by showing that any bounded linear functional $L$ in $[C(X)]^*$ can be represented as the difference of two positive linear functionals as $L=L_1-L_2$ with $\|L\|=L_1(1)+L_2(1)$, without proving the uniqueness of the $L_1$ and $L_2$. Then according the Riesz-Markov theorem which states that each positive linear functional $L_1$ can be uniquely expressed as $$L_1(f)=\int_X fd\mu_1$$ with a $\mu_1$ being a Radon measure, each bounded linear functional $L$ can be written as $$L(f)=L_1(f)-L_2(f)=\int_X fd\mu_1-\int_X fd\mu_2=\int_X fd(\mu_1-\mu_2)=\int_X fd\mu$$ with $\mu=\mu_1-\mu_2$ being a signed Randon measure. Thus $L=T_{\mu}$ and the operator $T$ is onto. Up to now, I can follow the argument. The following is what confuses me. It says We infer from this and Proposition 11 that the representation of $L$ as the difference of positive linear functionals is unque. The proposition 11 states that 

Let X be locally compact Hausdorff space and $\mu_1,\mu_2$ be Radon measures on $X$ for which $$\int_X fd\mu_1=\int_X fd\mu_2$$ for all compact supported continuous function $f$ in $C_c(X)$. Then $\mu_1=\mu_2$.

I don't understand how the uniqueness of the decomposition of $L$ can follow from this proposition and the fact that $T$ is onto:-(.
2) The proof continues by showing that $$\|L\|=\mu_1(X)+\mu_2(X)=\|\mu\|$$ which I understand. It then immediately jumps to the conclusion that Therefore T is an isometric isomorphism, which I have a little problem about.
I tried to fill the missing details about the final jump myself. If I accept the argument before the final jump, then I think the proof, in summary, shows that any bounded linear functional $L$ can uniquely determine a signed Radon measure $\mu$ and thus the operator $T$ is one-to-one and onto. Also, obviously, $T$ is linear. Therefore $T$ has a linear inverse $T^{-1}$. The last formula  $\|L\|=\|\mu\|$ leads to the fact that $T$ and $T^{-1}$ both have a finite/bounded norm according the definition of the norm of operators ($\|T\|=sup_{\mu}\frac{\|T_{\mu}\|}{\|\mu\|}$ in this theorem). Thus, both  $T$ and $T^{-1}$ are bounded linear operators, i.e., continuous linear operators. Thus  $T$ is an isomorphism. $\|T\|$ happens to be $1$. This means it is also isometric isomorhphism which preserves the norms of $L$ and $\mu$. Is my understanding about the final jump right ?
 A: I've lost track a little of what is in question here. I think that if we show the minimality of the constructed decomposition of $L$ as the difference of two positive linear functionals, the rest of the proof will be clear. If not, please request further elaboration.
Since I don't have access to Royden's book at the moment, I don't know how he constructs the decomposition. I expect it will be the standard method, though, which I first will sketch:
Let $P = \{ f \in C(X) : f \geqslant 0\}$. Then we start by constructing a functional $\Lambda_0 \colon P \to [0,+\infty)$ that dominates $L$. We define
$$\Lambda_0(f) = \sup \{ L(g) : g \in C(X), \, \lvert g\rvert \leqslant f\}.$$
Since $\lvert \pm f\rvert \leqslant f$ for $f\in P$, we have $\Lambda_0(f) \geqslant \max \{ L(f), L(-f)\} = \lvert L(f)\rvert \geqslant 0$, and since $\lvert L(g)\rvert \leqslant \lVert L\rVert\cdot\lVert g\rVert_\infty \leqslant \lVert L\rVert\cdot \lVert f\rVert_\infty$ for $\lvert g\rvert \leqslant f$, we have
$$0 \leqslant \Lambda_0(f) \leqslant \lVert L\rVert\cdot \lVert f\rVert_\infty$$
for $f\in P$.
Next one shows that $\Lambda_0(t\cdot f) = t\cdot \Lambda_0(f)$ for $t \in [0,+\infty)$ and $\Lambda_0(f + g) = \Lambda_0(f) + \Lambda_0(g)$ for $f,g \in P$. Thus one can extend $\Lambda_0$ to a positive linear functional $\Lambda_1$ on $C(X)$ by defining
$$\Lambda_1(f) = \Lambda_0(f^+) - \Lambda_0(f^-)$$
for $f \in C(X)$. One verifies that $\lVert\Lambda_1\rVert = \Lambda_1(1) = \lVert L\rVert$, and that $L_1 := \frac{1}{2}(\Lambda_1 + L)$ and $L_2 := \frac{1}{2}(\Lambda_1 - L)$ are positive linear functionals. Then one has the decomposition $L = L_1 - L_2$, and $\lVert L_1\rVert + \lVert L_2\rVert = L_1(1) + L_2(1) = \Lambda_1(1) = \lVert L\rVert$.
It remains to show the minimality of this decomposition. Suppose one has positive linear functionals $M_1, M_2$ with $L = M_1 - M_2$. We want to show that $M_1 - L_1 = M_2 - L_2$ are positive. So let $f \in P$, and $g \in C(X)$ with $\lvert g\rvert \leqslant f$. Then
\begin{align}
L(g) &= M_1(g) - M_2(g) \\
&= M_1(g^+ - g^-) - M_2(g^+ - g^-) \\
&= M_1(g^+) - M_1(g^-) - M_2(g^+) + M_2(g^-) \\
&\leqslant M_1(g^+) + M_1(g^-) + M_2(g^+) + M_2(g^-) \\
&= M_1(\lvert g\rvert) + M_2(\lvert g\rvert) \\
&\leqslant M_1(f) + M_2(f),
\end{align}
and thus $\Lambda_1(f) = \Lambda_0(f) \leqslant M_1(f) + M_2(f)$, from which we obtain
$$2L_1(f) = \Lambda_1(f) + L(f) \leqslant M_1(f) + M_2(f) + L(f) = M_1(f) + M_2(f) + M_1(f) - M_2(f) = 2M_1(f),$$
which is equivalent to the desired $M_1 - L_1 \geqslant 0$.
And by the minimality, we have
\begin{align}
\lVert M_1\rVert + \lVert M_2\rVert &= M_1(1) + M_2(1)\\
&= L_1(1) + L_2(1) + \bigl((M_1-L_1)(1) + (M_2 - L_2)(1)\bigr)\\
&\geqslant L_1(1) + L_2(1) = \lVert L\rVert,
\end{align}
with equality if and only if $0 = (M_1-L_1)(1) + (M_2-L_2)(1) = \lVert M_1 - L_1\rVert + \lVert M_2 - L_2\rVert$, i.e. $M_1 = L_1$ and therefore also $M_2 = L_2$.
While we do not have uniqueness of the decomposition of $L$ as the difference of two positive functionals (except in the trivial case $C(X) = \{0\}$, which happens for $X = \varnothing$) without side conditions, we have a unique decomposition with the additional condition that $\lVert L_1\rVert + \lVert L_2\rVert = \lVert L\rVert$.
The existence of this decomposition together with the Riesz-Markov theorem gives the existence of a Radon measure $\mu = \mu_1 - \mu_2$ with $T_\mu = T_{\mu_1} - T_{\mu_2} = L_1 - L_2 = L$ such that $\lVert\mu\rVert \leqslant \lVert\mu_1\rVert + \lVert\mu_2\rVert = \lVert L_1\rVert + \lVert L_2\rVert = \lVert L\rVert$. Since on the other hand one clearly has $\lVert T_\nu\rVert \leqslant \lVert\nu\rVert$ for every Radon measure $\nu$, it follows that $\lVert\mu\rVert = \lVert L\rVert$.
