I need to prove the following
The variation $|\mu|$ of a complex measure $\mu$ is finitely additive.
The following is my attempt. Basically, my idea is to prove that $|\mu|(B_1\bigcup B_2)=|\mu|(B_1)+|\mu|(B_2)$ holds for any disjoint $\mathscr{A}$-measurable sets $B_1$ and $B_2$. I would really appreciate it if someone could help me check if my work is correct or not! Thanks a lot in advance.
Note that for all finite partition $\{A_j\}_{j=1}^n$ of $B_1\bigcup B_2$ into $\mathscr{A}$-measurable sets we have\begin{align*}|\mu|(B_1) + |\mu|(B_2) &\geq \sum_{j=1}^n\left|\mu\left(A_j\bigcap B_1\right)\right| + \sum_{j=1}^n\left|\mu\left(A_j\bigcap B_2\right)\right|\\&\geq \sum_{j=1}^n|\mu(A_j)|.\end{align*}Assume to the contrary that\begin{align*}|\mu|(B_1\bigcup B_2) &= \sup\left\{\sum_{j=1}^n|\mu(A_j)|:\{A_j\}_{j=1}^n\ \text{is a finite partition of $B_1\bigcup B_2$ into $\mathscr{A}$-measurable sets}\right\} \\&> |\mu|(B_1) + |\mu|(B_2).\end{align*}Then there exists an $\epsilon>0$ such that for all finite partition $\{A_j\}_{j=1}^n$ of $B_1\bigcup B_2$ into $\mathscr{A}$-measurable sets we have $|\mu|(B_1\bigcup B_2)-\epsilon > \sum_{j=1}^n|\mu(A_j)|$; that is $|\mu|(B_1\bigcup B_2)-\epsilon$ is an upper bound of the set $\left\{\sum_{j=1}^n|\mu(A_j)|:\{A_j\}_{j=1}^n\ \text{is a finite partition of $B_1\bigcup B_2$ into $\mathscr{A}$-measurable sets}\right\}$, which contradicts the fact that $|\mu|(B_1\bigcup B_2)$ is the least upper bound of this set. Thus,$$|\mu|\left(B_1\bigcup B_2\right) \leq |\mu|(B_1) + |\mu|(B_2).$$Now we prove the reverse direction. Note that for all finite partition $\{E_i\}_{i=1}^n$ of $B_1$ into $\mathscr{A}$-measurable sets and for all finite partition $\{F_j\}_{j=1}^m$ of $B_2$ into $\mathscr{A}$-measurable sets we have that $\{E_i\}_{i=1}^n$ and $\{F_j\}_{j=1}^m$ together form a finite partition of $B_1\bigcup B_2$ into $\mathscr{A}$-measurable sets, and so$$|\mu|\left(B_1\bigcup B_2\right) \geq \sum_{i=1}^n|\mu(E_i)| + \sum_{j=1}^m|\mu(F_j)|.$$Let $\{F_j\}_{j=1}^m$ be any finite partition of $B_2$ into $\mathscr{A}$-measurable sets. We first prove that $|\mu|(B_1\bigcup B_2)\geq|\mu|(B_1)+\sum_{j=1}^m|\mu(F_j)|$. Assume to the contrary that\begin{align*}|\mu|(B_1) &= \sup\left\{\sum_{i=1}^n|\mu(E_i)|:\{E_i\}_{i=1}^n\ \text{is a finite partition of $B_1$ into $\mathscr{A}$-measurable sets}\right\} \\&> |\mu|(B_1\bigcup B_2) - \sum_{j=1}^m|\mu(F_j)|.\end{align*}Then for all finite partition $\{E_i\}_{i=1}^n$ of $B_1$ into $\mathscr{A}$-measurable sets there exists an $\epsilon>0$ such that $|\mu|(B_1)-\epsilon>|\mu|(B_1\bigcup B_2)-\sum_{j=1}^m|\mu(F_j)| \geq\sum_{i=1}^n|\mu(E_i)|$; that is, $|\mu|(B_1)-\epsilon$ is an upper bound of the set $\left\{\sum_{i=1}^n|\mu(E_i)|:\{E_i\}_{i=1}^n\ \text{is a finite partition of $B_1$ into $\mathscr{A}$-measurable sets}\right\}$, which contradicts the fact that $|\mu|(B_1)$ is the least upper bound of this set. So for all finite partition $\{F_j\}_{j=1}^m$ of $B_2$ into $\mathscr{A}$-measurable sets,$$|\mu|\left(B_1\bigcup B_2\right) \geq |\mu|(B_1) + \sum_{j=1}^m|\mu(F_j)|.$$Next we show that $|\mu|(B_1\bigcup B_2)\geq|\mu|(B_1)+|\mu|(B_2)$. Assume to the contrary that\begin{align*}|\mu|(B_2) &= \sup\left\{\sum_{j=1}^m|\mu(F_j)|:\{F_j\}_{j=1}^m\ \text{is a finite partition of $B_2$ into $\mathscr{A}$-measurable sets}\right\} \\&> |\mu|(B_1\bigcup B_2) - |\mu|(B_1).\end{align*}Then for all finite partition $\{F_j\}_{j=1}^m$ of $B_2$ into $\mathscr{A}$-measurable sets there exists an $\epsilon>0$ such that $|\mu|(B_2)-\epsilon>|\mu|(B_1\bigcup B_2)-|\mu|(B_1) \geq \sum_{j=1}^m|\mu(F_j)|$; that is, $|\mu|(B_2)-\epsilon$ is an upper bound of the set $\left\{\sum_{j=1}^m|\mu(F_j)|:\{F_j\}_{j=1}^m\ \text{is a finite partition of $B_2$ into $\mathscr{A}$-measurable sets}\right\}$, which contradicts the fact that $|\mu|(B_2)$ is the least upper bound of this set. So$$|\mu|\left(B_1\bigcup B_2\right) \geq |\mu|(B_1) + |\mu|(B_2).$$Therefore, we have proved that$$|\mu|\left(B_1\bigcup B_2\right) = |\mu|(B_1) + |\mu|(B_2).$$
Some related definition:
Definition$\quad$ Let $(X,\mathscr{A})$ be a measurable space. A complex measure on $(X,\mathscr{A})$ is a function $\mu$ from $\mathcal{A}$ to $\mathbb{C}$ that satisfies $\mu(\emptyset)=0$ and is $\textit{countably additive}$, in the sense that\begin{align*} \mu\left(\bigcup_{n=1}^{\infty}A_n\right) = \sum_{n=1}^{\infty}\mu(A_n)\end{align*}holds for each infinite sequence $\{A_n\}$ of disjoint sets in $\mathscr{A}$. Note that by definition a complex measure has only complex values and so has no infinite values.