Complex Measures: Norm Given a measure space $\Omega$.
Consider a complex measure:
$$\mu:\Sigma(\Omega)\to\mathbb{C}:\quad\mu\left(\biguplus_kA_k\right)=\sum_k\mu(A_k)$$
Regard the total variation:
$$|\mu|(A):=\sup_{\uplus_kA_K=A}|\mu(A_k)|$$
Then the norm is finite:
$$\|\mu\|:=|\mu|(\Omega)<\infty$$
Is there an elegant proof?
 A: Don't know about "elegant", but there's a straightforward just-follow-the-inequalities proof.
First, if all you want is to show that norm is finite, you can assume $\mu$ is actually real valued. Now say the norm is infinite. So you can find disjoint sets $E_n$ with $\sum |\mu(E_n)|$ arbitrarily large. Noting that $|x|=x^+-x^-$ for $x$ real, either you can find disjoint $E_n$ with $\mu(E_n)>0$ and $\sum \mu(E_n)$ arbitrarily large or you can find disjoint $E_n$ with $\mu(E_n)<0$ and $\sum-\mu(E_n)$ arbitrarily large. Assume the first case.
Taking unions of our disjoint sequences, we can find sets $E$ with $\mu(E)>0$ and $\mu(E)$ arbitrarily large.
So we can find sets $E_n$ with $\mu(E_1)>1$ and $$\mu(E_{n+1})>3\mu(E_n).$$
The meaning of the notation "$E_n$" has been changing from line to line. Fix that; from now on the $E_n$ satisfy that last inequality. Disjunctificate the $E_n$: Let $F_1=E_1$ and $$F_{n+1}=E_{n+1}\setminus\bigcup_{j\le n}E_j.$$Now the $F_n$ are disjoint and the previous inequality implies that $\mu(F_n)>0$ and $$\mu(F_n)\to+\infty.$$So if $F=\bigcup F_n$ then $$\mu(F_n)=+\infty.$$In particular, $\mu(F)$ is unreal, contradiction.
A: For all $C \subset \Omega$ you can define $||\mu(C)|| = \sup_{A\subset C} |\mu(A)|$, and have the immediate inequality $||\mu||(C_1 \sqcup C_2) \le ||\mu||(C_1) + ||\mu||(C_2)$. 
Assume that $||\mu||(\Omega) = \infty$. We'll construct inductively a decreasing sequence of subsets from $\Sigma$  $  B_1 \supset\ldots  B_n \supset \ldots$ such that $|\mu(B_{n+1})| \ge |\mu(B_{n}| + 1$ and $||\mu||(B_n) = \infty$ for all $n$  Let's get $B_1$ first. Since $||\mu||(\Omega) = \infty$ there exists $A \subset \Omega$ with $|\mu(A)|\ge 2|\mu(\Omega)| + 1$. Now $\mu(\Omega) = \mu(A) + \mu(\Omega \backslash A)$ so both $|\mu(A)|$, $|\mu(\Omega \backslash A)|$ are $\ge |\mu(\Omega)| + 1$. Moreover, at least one of the $||\mu||(A)$, $||\mu||(\Omega \backslash A)$ is infinity. We choose $B_1$ to be that one. Getting $B_n$ inside $B_{n-1}$ once $B_{n-1}$ is found is the same argument. 
Once we have this sequence $B_n$ we consider $A_n \colon= B_n \backslash B_{n+1}$. We have $|\mu(A_n)| \ge 1$ and $A_n$ are disjoint. However, for every countable family of disjoint subsets $(A_n)$ the series $\sum_n \mu(A_n)$ is convergent so $\mu(A_n) \to 0$. We got a contradiction. 
Note that we have used very little: $\mu$ is defined on a ring of subset of $\Omega$ with values in an abelian group with an absolute value. Now if $\mu$ takes values in a finite dimensional normed space we can consider the total variation $|||\mu|||(A)$ as
$\sup \sum_n |\mu(A_n)|$ over all countable partitions of A . That turns out to be finite as well, the proof is just a bit more involved, I'll add some detail later.
