22.5 Localisable Measures
Definition 22.5.1 (Essential Supremum).label Let $(X, \cm, \mu)$ be a measure space, $\ce \subset \cm$, and $S \in \cm$, then $S$ is an essential upper bound of $\ce$ if for any $E \in \ce$, $\mu(E \setminus S) = 0$. If in addition, for any essential upper bound $T$ of $\ce$, $\mu(S \setminus T) = 0$, then $S$ is an essential supremum of $\ce$.
Definition 22.5.2 (Localisable).label Let $(X, \cm, \mu)$ be a measure space, then $X$ is localisable if:
- (1)
$\mu$ is semifinite.
- (2)
For every $\ce \subset \cm$, there exists an essential supremum $A$ of $\ce$.
Definition 22.5.3 (Decomposable).label Let $(X, \cm, \mu)$ be a measure space and $\seqi{A}\subset \cm$, then $\seqi{A}$ is a decomposition of $(X, \cm, \mu)$ if:
- (1)
For each $i \in I$, $\mu(A_{i}) < \infty$.
- (2)
$X = \bigsqcup_{i \in I}X_{i}$.
- (3)
$\cm = \bracs{E \subset X|E \cap A_i \in \cm \forall i \in I}$.
- (4)
For each $E \in \cm$, $\mu(E) = \sum_{i \in I}\mu(E \cap A_{i})$.
If $(X, \cm, \mu)$ admits a decomposition, then it is decomposable.
Lemma 22.5.4.label Let $(X, \cm, \mu)$ be a finite measure space, $\ce \subset \cm$, and $\mathcal{S}$ be the set of all essential upper bounds of $\ce$, then:
- (1)
$\mathcal{S}\ne \emptyset$.
- (2)
For any $S \in \cm$, $S \in \mathcal{S}$ if and only if $\mu(S \cap E) = \mu(E)$ for all $E \in \ce$.
- (3)
For any $\seq{S_n}\subset \mathcal{S}$, $\bigcap_{n \in \natp}S_{n} \in \mathcal{S}$.
- (4)
There exists $S \in \mathcal{S}$ such that $\mu(S) = \inf\bracs{\mu(T)|T \in \mathcal{S}}$.
- (5)
For any $S \in \mathcal{S}$ with $\mu(S) = \inf\bracs{\mu(T)|T \in \mathcal{S}}$, $S$ is an essential supremum of $\ce$.
Proof. (1): $X \in \mathcal{S}$.
(2): Since $\mu$ is finite, for any $E \in \ce$, $\mu(S \cap E) = \mu(E)$ if and only if $\mu(E \setminus S) = 0$.
(3): Firstly, for any $S, T \in \mathcal{S}$ and $E \in \ce$,
so $S \cap T \in \mathcal{S}$. Now let $\seq{S_n}\subset X$ and $E \in \ce$, then since $\mu$ is finite,
by continuity from above. By (2), $\bigcap_{n \in \natp}S_{n} \in \mathcal{S}$.
(4): Let $M = \inf\bracs{\mu(T)|T \in \mathcal{S}}$ and $\seq{S_n}\subset \mathcal{S}$ such that $\limv{n}\mu(S_{n}) = M$, then by continuity from above,
By (3), $\bigcap_{n \in \natp}S_{n} \in \mathcal{S}$, therefore the minimum is achieved.
(5): Let $R \in \mathcal{S}$. By (3), $S \cap R \in \mathcal{S}$ with $\mu(S \cap R) = \inf\bracs{\mu(T)|T \in \mathcal{S}}= \mu(S)$, so
and $S$ is the essential supremum of $\ce$.$\square$
Proposition 22.5.5.label Let $(X, \cm, \mu)$ be a decomposable measure space, then $X$ is localisable.
Proof. Let $\ce \subset \cm$ and $\seqi{A}\subset \cm$ be a decomposition of $X$.
For each $i \in I$, let $\ce_{i} = \bracs{E \cap A_i|E \in \ce}$. By Lemma 22.5.4, there exists an essential upper bound $S_{i} \in \cm$ of $\ce_{i}$ contained in $A_{i}$ with respect to the restricted measure $\mu|_{A_i}$. In other words,
- (i)
$S_{i}$ is an essential upper bound of $\ce_{i}$.
- (ii)
For any essential upper bound $T \in \cm$ of $\ce_{i}$ with $T \subset A_{i}$, $\mu(S_{i} \setminus T) = 0$.
Now, let $S = \bigsqcup_{i \in I}S_{i}$, then since $X = \bigsqcup_{i \in I}A_{i}$ and $S_{i} \subset A_{i}$ for all $i \in I$, $S \cap A_{i} = S_{i} \in \cm$ for all $i \in I$, so $S \in \cm$. For any $E \in \ce$, $E \cap A_{i} \in \ce_{i}$. Passing through the decomposition, (i) implies that,
and $S$ is an essential upper bound of $\ce$.
Finally, let $T \in \cm$ be an essential upper bound of $\ce$, then $T \cap A_{i}$ is an essential upper bound of $\ce_{i}$ for all $i \in I$. The decomposition and (ii) then shows that
therefore $S$ is an essential supremum of $\ce$.$\square$
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