16.5 Total Variation
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 $\bracs{B_k}_{1}^{K}$, $\sum_{n \in \natp}|\mu|(A_{n}) \ge |\mu|(A)$.$\square$
Definition 16.5.2 (Total Variation of Signed Measures). Let $(X, \cm)$ be a measure space, $\mu$ be a signed measure on $(X, \cm)$, and
then:
For any $A \in \cm$,
\[|\mu|(A) = \nu(A) = \sup\bracs{\sum_{i \in I}|\mu(A_i)| \bigg | \seqi{A} \subset \cm, A = \bigsqcup_{i \in I}A_i}\]$|\mu|$ is a measure on $(X, \cm)$.
Let $\mu = \mu^{+} - \mu^{-}$ be the Jordan decomposition of $\mu$, then $|\mu| = \mu^{+} + \mu^{-}$.
and the measure $|\mu|$ is the total variation of $\mu$.
Proof. (1) and (2): Same as Definition 11.2.1.
(3): For any $A \in \cm$, $|\mu(A)| \le \mu^{+}(A) + \mu^{-}(A)$, so $(\mu^{+} + \mu^{-}) \ge |\mu|$. On the other hand, let $X = P \sqcup N$ be a Hahn decomposition of $\mu$, then
so $(\mu^{+} + \mu^{-}) \le |\mu|$.$\square$
Definition 16.5.3 (Finite Measure). Let $(X, \cm)$ be a measurable space and $\mu$ be a signed/vector measure, then $\mu$ is finite if $|\mu|$ is finite.
Definition 16.5.4 (Space of Finite Measures). Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $M(X, \cm; E)$ be the set of all finite $E$-valued vector measures on $(X, \cm)$. For each $\mu \in M(X, \cm; E)$, let $\norm{\mu}_{\text{var}}= |\mu|(X)$, then:
$M(X, \cm; E)$ equipped with $\norm{\cdot}_{\text{var}}$ is a normed vector space over $K$.
If $E$ is a Banach space, then so is $M(X, \cm; E)$.
Proof. (1): For any $\mu, \nu \in M(X, \cm; E)$ and $\seqf{A_j}\subset \cm$ such that $X = \bigsqcup_{j = 1}^{n} A_{j}$,
As this holds for all choices of $\seqf{A_j}$, $\norm{\mu + \nu}_{\text{var}}\le \norm{\mu}_{\text{var}}+ \norm{\nu}_{\text{var}}$.
(2): Let $\seq{\mu_n}\subset M(X, \cm; E)$ such that $\sum_{n \in \natp}\norm{\mu_n}_{\text{var}}< \infty$. For each $A \in \cm$, since $E$ is complete, let
then for any $\seq{A_k}\subset \cm$ and $A \in \cm$ with $A = \bigsqcup_{k \in \natp}A_{k}$,
by Fubini’s theorem.
$\square$