# Absolute value of complex Radon measure

Let $X$ be a locally compact Hausdorff space. To simplify matters we assume $X$ has a countable base. Let $\mathcal B$ be the $\sigma$-algebra generated by the set of open subsets of $X$. A Borel measure on $X$ is a mesure defined on $\mathcal B$ such that $\mu(K) \lt \infty$ for every compact subset $K$. We denote $\int_X f d\mu$ by $\mu(f)$ for an integrable function $f$. Let $\mathcal L$ be the vector space over $\mathbb C$ of continuous complex valued functions of compact support defined on $X$. We denote by $\mathcal L_+$ the set of non-negative real valued contonuous function of compact support. A linear functional $L: \mathcal L \rightarrow \mathbb C$ is called a positive linear functional if $L(f) \ge 0$ whenever $f\in \mathcal L_+$. By Riesz representation theorem, there exists a unique Borel measure $\mu$ such that $L(f) = \mu(f)$ for every $f \in \mathcal L$. We identify $L$ with $\mu$. A linear functional $L: \mathcal L \rightarrow \mathbb C$ is called a Radon measure if for every compact subset $K$ there exists a real number $M_K$ such that $|L(f)| \le M_K ||f||$ for every $f\in \mathcal L$ whose support is contained in $K$, where $||f|| = sup_{x\in X} f(x)$. It is known that a Radon measure $\mu$ can be written as $\mu = \mu_1 - \mu_2 + i(\mu_3 - \mu_4)$ where $\mu_1,\dots,\mu_4$ are positive Borel measures. For a Radon measure $\mu$ we write $|\mu|(f) = sup_{|g| \le f} |\mu(f)|$ for $f\in \mathcal L_+$. It can be proved that $|\mu|$ is a Borel measure. $|\mu|$ is called the absolute value of $\mu$. Let $\mu$ be a Radon meassure, $A$ a Borel set such that $|\mu|(A) \lt \infty$. Are the following assertions true?

1) $|\mu|(A) = sup_{K\subset A} |\mu(K)|$.

2) $|\mu|(A) = sup |\mu(A_1)| + \cdots + |\mu(A_n)|$ where $A = \cup A_i$ is a finite disjoint union of Borel sets and the sup runs through such partitions.

The motivation is that the assertions are true when $\mu$ is a positive Radon measure and 2) seems to be true when $\mu$ is bounded.

1 is certainly not true. Consider a set $X = \{a,b\}$ with two points, and a measure $\mu$ so that $a$ has measure $+1$ and $b$ has measure $-1$, i.e., $$\mu(\{a\})=1, \quad \mu(\{b\})=-1, \quad \mu(\emptyset) = \mu(X) = 0.$$ Then for any (compact) $K \subset X$ we have $|\mu(K)| \le 1$ but $|\mu|(X) = 2$.