21.7 Spaces of Finite Measures
Definition 21.7.1 (Space of Finite Measures).label 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:
- (1)
$M(X, \cm; E)$ equipped with $\norm{\cdot}_{\text{var}}$ is a normed vector space over $K$.
- (2)
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.$\square$
While the space of bounded Borel functions on $(X, \cm)$ forms a subspace of the dual of $M(X, \cm; \complex)$, it may not be immediately clear that they are insufficient. Before moving on to an explicit description of this dual, it is beneficial to consider a simple ”example”.
Let $X = [0, 1]$, equipped with its Borel $\sigma$-algebra, and $\mu$ be the Lebesgue measure on $X$, then for any $x \in [0, 1]$, $\mu$ is mutually singular with the delta mass at $x$. Therefore the closure of $\text{span}\bracs{\delta_x|x \in X}$ in $M_{R}(X, \cm; \complex)$ is a proper closed subspace of $M(X, \cm; \complex)$. By the Hahn-Banach Theorem, there exists $\phi \in M(X, \cm; \complex)^{*}$ such that $\dpn{\delta_x, \phi}{M(X, \cm; \complex)}= 0$ for all $x \in X$, but $\dpn{\mu, \phi}{M(X, \cm; \complex)}= 1$. If there exists a bounded Borel function $f: X \to \complex$ such that $\dpn{\nu, \phi}{M(X, \cm; \complex)}= \int f d\nu$ for all $\nu \in M(X, \cm; \complex)$, then $f(x) = \dpn{\delta_x, \phi}{M(X, \cm; \complex)}= 0$ for all $x \in [0, 1]$, which is impossible. Therefore $\phi$ cannot be represented as a bounded Borel function.
Aside from abusing the Hahn-Banach theorem, a more explicit construction of such a functional is via the atomic decomposition. However, it is not included here due to time constraints.
In any case, the above example shows that a linear functional on $M(X, \cm; \complex)$ may act as different Borel functions on different families of measures that are mutually singular to each other. The following theorem will make this relation precise.
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,
$\square$
Despite the fact that it does not cover the full dual space, the bounded Borel functions still forms a subspace where weak-* convergence has a convenient description.
Proposition 21.7.3.label Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $\mathscr{M}\subset M(X, \cm; E)$ be a closed subspace such that:
- (P)
For each $x \in X$, $\bracs{x}\in \cm$, and the delta mass $\delta_{x}$ is in $\mathscr{M}$.
Then, for any bounded measurable functions $\bracsn{f_n: X \to E^*|n \in \natp}$ and $f: X \to E^{*}$, the following are equivalent:
- (1)
For each $\mu \in \mathscr{M}$, $\limv{n}\int f_{n} d\mu = \int f d\mu$.
- (2)
For each $x \in X$, $\limv{n}f_{n}(x) = f(x)$, and $\sup_{n \in \natp}\norm{f_n}_{u} < \infty$.
Proof. (1) $\Rightarrow$ (2): By (P), for each $x \in X$, $\limv{n}f_{n}(x) = f(x)$. By the Uniform Boundedness Principle,
(2) $\Rightarrow$ (1): By the Dominated Convergence Theorem.$\square$
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