Theorem 21.7.2.label Let $(X, \cm)$ be a measurable space, $H$ be a Hilbert space over $K \in \RC$, and $\mathscr{M}\subset M(X, \cm; H)$ be a closed subspace such that:
- (A)
For each $\mu \in \mathscr{M}$ and $\nu \in M(X, \cm; H)$ with $\nu \ll \mu$, $\nu \in \mathscr{M}$.
Let $|\mathscr{M}| = \bracsn{|\mu|: \mu \in \mathscr{M}}$, then for any maximal family $\seqi{\mu}\subset |\mathscr{M}|$ of pairwise mutually singular measures,
- (1)
For each $f \in [l^{1}(I); L^{1}(\mu_{i}; H)]$, let
\[\mu_{f}: \cm \to H \quad A \mapsto \sum_{i \in I}\int_{A} f_{i} d\mu_{i}\]then the mapping
\[[l^{1}(I); L^{1}(\mu_{i}; H)] \to \mathscr{M}\quad f \mapsto \mu_{f}\]is an isometric isomorphism.
- (2)
$\mathscr{M}^{*}$ is isometrically isomorphic to $[l^{\infty}(I); L^{\infty}(\mu_{i}; H)]$.
In other words,
- (3)
There exists a semifinite measure space $(\Omega, \cn, \mu)$ such that $\mathscr{M}\iso L^{1}(\Omega; H)$ and $\mathscr{M}^{*} \iso L^{\infty}(\Omega; H)$.
Proof, [(ht]. (1): By (A), $\mu_{f} \in \mathscr{M}$ for all $f \in [l^{1}(I); L^{1}(\mu_{i}; H)]$.
On the other hand, let $\nu \in \mathscr{M}$. For each $i \in I$, let $\nu = \nu_{a}^{(i)}+ \nu_{s}^{(i)}$ be the Lebesgue decomposition of $\nu$ with respect to $\mu_{i}$, then by the Radon-Nikodym theorem, there exists $f_{i} \in L^{1}(\mu_{i}; H)$ such that $\nu_{a}^{(i)}(dx) = f_{i} \mu_{i}(dx)$.
For each countable $J \subset I$, define
Let $M = \sup\bracs{\norm{\nu_J}_{\text{var}}| J \subset I \text{ countable}}$, then there exists $J \subset I$ countable such that $\norm{\nu_J}_{\text{var}}= M$. For each $i \in I \setminus J$,
By maximality of $M$, $\normn{\nu_a^{(i)}}_{\text{var}}= 0$, and by maximality of $\seqi{\mu}$, $\nu - \nu_{J} = 0$. Let $g \in [l^{1}(I); L^{1}(\mu_{i}; H)]$ be defined by $g_{i} = f_{i}$ for each $i \in I$, then $\nu = \nu_{J} = \mu_{g}$, and the mapping is surjective.
(2): By Theorem 14.3.4 and Theorem 14.2.4,
$\square$
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