Definition 16.5.1 (Total Variation of Vector Measures). Let $(X, \cm)$ be a measure space, $\mu$ be a vector measure taking values in a normed vector space $E$, and
then:
For any $A \in \cm$,
\[|\mu|(A) = \nu(A) = \sup\bracs{\sum_{i \in I}\norm{\mu(A_i)} \bigg | \seqi{A} \subset \cm, A = \bigsqcup_{i \in I}A_i}\]$|\mu|$ is a measure on $(X, \cm)$.
and the measure $|\mu|$ is the total variation of $\mu$.
Proof. (1): Since all finite partitions are partitions, $|\mu| \le \nu$. On the other hand, let $\seqi{A}\subset \cm$ and $A = \bigsqcup_{i \in I}A_{i}$, then for any $J \subset I$ finite,
so
As this holds for all $\seqi{A}$ such that $A = \bigsqcup_{i \in I}A_{i}$, $|\mu|(A) \ge \nu(A)$.
(2): Let $A \in \cm$ and $\seq{A_n}$ such that $A = \bigsqcup_{n \in \natp}A_{n}$. For each $n \in \natp$, let $\seq{B_{n, k}}\subset \cm$ such that $A_{n} = \bigsqcup_{k \in \natp}B_{n, k}$, then for any $N \in \natp$,
As the above inequality is independent of the choice of $\bracsn{B_{n, k}|n, k \in \natp}$, $|\mu|(A) \ge \sum_{n = 1}^{N} |\mu|(A_{n})$ for all $N \in \natp$. Thus $|\mu|(A) \ge \sum_{n \in \natp}|\mu|(A_{n})$.
On the other hand, let $\bracs{B_k}_{1}^{K} \subset \cm$ such that $A = \bigsqcup_{k = 1}^{K} B_{k}$, then for each $N \in \natp$,
so
As the above holds for all choices of $\bracs{B_k}_{1}^{K}$, $\sum_{n \in \natp}|\mu|(A_{n}) \ge |\mu|(A)$.$\square$