21.4 Absolute Continuity
Definition 21.4.1 (Absolutely Continuous).label Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, and $\nu$ be a signed measure or vector veasure on $(X, \cm)$, then $\nu$ is absolutely continuous with respect to $\mu$, denoted $\nu \ll \mu$, if for every $E \in \cm$ with $\mu(E) = 0$, $\nu(E) = 0$ as well.
Lemma 21.4.2.label Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, and $\nu$ be a signed measure or vector measure on $(X, \cm)$, then $\nu \ll \mu$ if and only if $|\nu| \ll \mu$.
Proposition 21.4.3.label Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, $E$ be a normed space over $K \in \RC$, and $\nu: \cm \to E$ be a vector measure, then the following are equivalent:
- (1)
$\nu$ is absolutely continuous with respect to $\mu$.
- (2)
For each $\eps > 0$, there exists $\delta > 0$ such that for any $A \in \cm$ with $\mu(A) < \delta$, $\norm{\nu(A)}_{E} < \eps$.
Proof, [Theorem 3.15, Fol99]. $(\neg 2) \Rightarrow (\neg 1)$: Using Lemma 21.4.2, assume without loss of generality that $\nu$ is a positive finite measure.
By assumption, there exists $\eps > 0$ and $\seq{A_n}\subset \cm$ such that for each $n \in \natp$, $\mu(A_{n}) < 2^{-n}$ and $\norm{\nu(A_n)}_{E} \ge \eps$. For each $N \in \natp$, let $B_{N} = \bigcup_{n \ge N}A_{n}$, then $\mu(B_{N}) < 2^{-N+1}$. By Proposition 20.1.5,
but by Proposition 21.2.3,
so (1) does not hold.$\square$
Definition 21.4.4 (Uniformly Absolutely Continuous).label Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, $E$ be a normed space over $K \in \RC$, and $\mathcal{U}\subset M(X, \cm; E)$ be a family of $E$-valued vector measures on $(X, \cm)$, then $\mathcal{U}$ is uniformly absolutely continuous with respect to $\mu$ if for every $\eps > 0$, there exists $\delta > 0$ such that for all $A \in \cm$ with $\mu(A) < \delta$, $\norm{\nu(A)}_{E} < \eps$ for all $\nu \in \mathcal{U}$.
Theorem 21.4.5 (Vitali-Hahn-Saks).label Let $(X, \cm, \mu)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, $E$ be a normed vector space over $K \in \RC$, and $\seq{\nu_n}\subset M(X, \cm; E)$ be $E$-valued vector measures on $(X, \cm)$ such that:
- (a)
For each $n \in \natp$, $\nu$ is absolutely continuous with respect to $\mu$.
- (b)
For each $A \in M(X, \cm; E)$, $\nu(A) = \limv{n}\nu_{n}(A)$ exists.
then
- (1)
$\seq{\nu_n}$ is uniformly absolutely continuous with respect to $\mu$.
- (2)
The mapping $\nu$ is a $E$-valued vector measure absolutely continuous with respect to $\mu$.
Proof, [Theorem II.2.1, Yos12]. Let $\cm_{0}$ be the euivalence classes of essentially equal sets in $\cm$, equipped with the Fréchet-Nikodym metric with respect to $\mu$. For each $n \in \natp$, $\nu_{n} \ll \mu$ by assumption (a), and $\nu_{n} \in UC(\cm_{0}; E)$.
(1): Let $\eps > 0$. For each $N \in \natp$, let
then since $\seq{\nu_n}\subset UC(\cm_{0}; E)$, $A_{N} \subset \cm_{0}$ is closed. By assumption (b), $\cm_{0} = \bigcup_{N \in \natp}A_{N}$. By the Baire Category Theorem, there exists $N \in \natp$, $A \in \cm_{0}$, and $\delta > 0$ such that for every $m, n \ge N$ and $B \in \cm_{0}$ with $\mu(A \Delta B) \le \delta$, $\norm{\nu_m(B) - \nu_n(B)}_{E} \le \eps$.
Let $B \in \cm$ with $\mu(B) = \mu(\emptyset \Delta B) \le \delta$ and write $B = (A \cup B) \setminus (A \setminus B)$, then $\mu((A \cup B) \Delta B) \le \delta$ and $\mu((A \setminus B) \Delta B) \le \delta$. Thus
and $\seq{\mu_n}$ is uniformly absolutely continuous with respect to $\mu$.
(2): Let $\seq{A_n}\subset \cm$ and $A \in \cm$ with $A = \bigsqcup_{n \in \natp}A_{n}$, then for each $N \in \natp$,
Since $\seq{\nu_k}$ is uniformly absolutely continuous,
so $\nu$ is a vector measure.
Absolute continuity is equivalent to uniform continuity as a mapping $\cm_{n} \to E$, and uniform absolute continuity is equivalent to uniform equicontinuity as mappings $\cm \to E$. By the Arzelà-Ascoli Theorem, $\nu \in UC(\cm_{0}; E)$, so $\nu \ll \mu$.$\square$
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