23.3 Total Variation
Definition 23.3.1 (Total Variation of Vector Measures).label Let $(X, \cm)$ be a measure space, $\mu$ be a vector measure taking values in a normed vector space $E$, and
then:
- (1)
For any $A \in \cm$,
\[|\mu|(A) = \nu(A) = \sup\bracs{\sum_{i \in I}\norm{\mu(A_i)} \bigg | \seqi{A} \subset \cm, A = \bigsqcup_{i \in I}A_i}\] - (2)
$|\mu|$ 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 23.3.2 (Total Variation of Signed Measures).label Let $(X, \cm)$ be a measure space, $\mu$ be a signed measure on $(X, \cm)$, and
then:
- (1)
For any $A \in \cm$,
\[|\mu|(A) = \nu(A) = \sup\bracs{\sum_{i \in I}|\mu(A_i)| \bigg | \seqi{A} \subset \cm, A = \bigsqcup_{i \in I}A_i}\] - (2)
$|\mu|$ is a measure on $(X, \cm)$.
- (3)
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 14.2.1.
(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 23.3.3 (Finite Measure).label Let $(X, \cm)$ be a measurable space and $\mu$ be a signed/vector measure, then $\mu$ is finite if $|\mu|$ is finite.
Lemma 23.3.4.label Let $(X, \cm)$ be a measurable space, $K \in \RC$, $\mu: \cm \to K$ be a signed/complex measure, then:
- (1)
If $K = \real$, then $|\mu|(X) \le 2\sup_{A \in \cm}\norm{\mu(A)}_{E}$.
- (2)
If $K = \complex$, then $|\mu|(X) \le 4\sup_{A \in \cm}\norm{\mu(A)}_{E}$.
Proof. (1, scalar): Let $X = P \sqcup N$ with $P, N \in \cm$ be a Hahn decomposition of $\mu$, then
(2, scalar): By (1),
$\square$
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