Determine the Norm of the linear functional $I_\mu(f)=\int_X f\,d\mu$ Let $X$ be a locally compact Hausdorff space and $\mu$ be a (finite) complex Radon measure on $X$. Since we have
$$\left|\int_X f\,d\mu\right|\leq \int_X |f|\,d|\mu|\leq \|f\|_\infty |\mu|(X).$$
for all $f\in C_c(X\to\mathbf{C})$, here $\|f\|_\infty:=\sup_{x\in X}|f(x)|$. The linear functional $I_\mu:C_c(X\to \mathbf{C})\to \mathbf{C}$ defined by $I_\mu(f):=\int_X f\,d\mu$ is bounded and has operator norm at most $|\mu|(X)$, here we give $C_c(X\to\mathbf{C})$ the supremum norm. How to show that $I_\mu$ is actually has operator norm $|\mu|(X)$?
 A: I'll use $\|\mu\|$ for $\lvert \mu\rvert(X)$. Given $\epsilon > 0$, choose a finite partition $\{E_1,\ldots, E_n\}$ of $X$ such that $\sum\limits_{i = 1}^n \lvert \mu(E_i)\rvert > \lvert \mu\rvert(X) - \epsilon/2$. Let $\alpha_1,\ldots, \alpha_n$ be complex numbers of unit modulus such that $\alpha_i\mu(E_i) = \lvert \mu(E_i)\rvert$, $i = 1,2,\ldots, n$. Let $s := \sum\limits_{i = 1}^n \alpha_i1_{E_i}$. Then $$\left\lvert\int_X s\, d\mu\right\rvert = \left\lvert\sum_{i = 1}^n \alpha_i\mu(E_i)\right\rvert = \sum_{i = 1}^n \lvert \mu(E_i)\rvert > \|\mu\| - \frac{\epsilon}{2}$$ By regularity of $\mu$, there exist compact sets $K_1,\ldots, K_n$ and open sets $V_1,\ldots, V_n$ such that $K_i \subset E_i \subset V_i$ and $\lvert\mu\rvert(V_i \setminus K_i) < \epsilon/(2n)$, for $i = 1,2,\ldots, n$. By Urysohn's lemma, there are functions $f_1,\ldots, f_n\in C_c(X\to \bf{R})$ such that for all $i$, $0 \le f_i \le 1$, $f_i = 1$  on $K_i$, and $\operatorname{supp}(f_i) \subset V_i$. Note that for each $i$, $1_{E_i} - f_i$ vanishes outside $V_i\setminus K_i$ and is bounded by $1$ on $V_i\setminus K_i$. Define $f := (1/n)\sum\limits_{i = 1}^n\alpha_i \,f_i$. Then $f\in C_c(X\to \bf{C})$, $\lvert f\rvert \le 1$, and 
$$\int_X\lvert s - f\rvert\, d\lvert \mu\rvert \le \left(1 - \frac{1}{n}\right)\sum_{i = 1}^n  \int_X\lvert 1_{E_i} - f_i\rvert\, d\lvert \mu\rvert \le \sum_{i = 1}^n \int_{V_i \setminus K_i} \lvert 1_{E_i} - f_i\lvert\, d\lvert \mu\rvert \le \sum_{i = 1}^n \lvert \mu\rvert(V_i\setminus K_i) \le \frac{\epsilon}{2}$$
Therefore,
\begin{align}\left\lvert \int_X f\, d\mu\right\rvert &\ge \left\lvert \int_X s\, d\mu\right\rvert - \left\lvert \int_X (s - f)\, d\mu\right\rvert
\ge \left\lvert \int_X s\, d\mu\right\rvert - \int_X \lvert s - f\rvert\, d\lvert \mu\rvert
> \left(\|\mu\| - \frac{\epsilon}{2}\right) - \frac{\epsilon}{2}
= \|\mu\| - \epsilon\end{align}
Thus $\|I_\mu\| \ge \lvert I_\mu(f)\rvert > \|\mu\| - \epsilon$. Letting $\epsilon \to 0$, we obtain $\|I_\mu\| \ge \|\mu\|$ and therefore $\|I_\mu\| = \|\mu\|$.
A: To prove the opposite inequality, note that $\mu$ is absolutely continuous with respect to $|\mu|$. By the Radon-Nikodym theorem, we have that 
$d\mu=gd|\mu|$. Moreover, it's not difficult to show that $|g(x)|=1$ for all almost every $x\in X$ (see Rudin's Real and complex analysis Theorem 6.12). Since $C_c(X\to\mathbf{C})$ is dense in $L^1(X,|\mu|)$, we can find a sequence of continuous compactly supported functions $f_n\in C_c(X\to\mathbf{C})$ such that $f_n\to g$ in $L^1(X,|\mu|$). Let $g_n=\phi(f_n)$ where $\phi:\mathbf{C}\to\mathbf{C}$ is the continous function defined by $\phi(z)=z$ if $|z|\leq 1$ and $|\phi(z)|=z/|z|$ if $|z|\geq 1$. Then $|g_n|\leq 1$ and $g_n\to g$ in $L^1(X,|\mu|)$. Thus
$$\|\mu\|=|\mu(X)|=\int_X \,d|\mu|=\int_X \overline{g}\,d\mu=\lim_{n\to\infty}\overline{g_n}\,d\mu\leq \|I_\mu\|.$$
