22.4 Scaffolds
Definition 22.4.1 (Scaffold*).label Let $(X, \cm, \mu)$ be a measure space and $\cf \subset \bracs{A \in \cm|\mu(A) < \infty}$ be an ideal, then $\cf$ is a scaffold for $\mu$ if for all $E \in \cm$,
and the quadruple $(X, \cm, \cf, \mu)$ is a scaffolded measure space.
For any semifinite measure space $(X, \cm, \mu)$, $\cf = \bracs{A \in \cm|\mu(A) < \infty}$ is the canonical scaffold for $\mu$, and $(X, \cm, \mu)$ will be equipped with this scaffold unless specified otherwise.
Example 22.4.2.label Let $X$ be a LCH space, $\mu$ be a Radon measure, and $\mathcal{K}$ be the collection of compact subsets of $X$, then $\mathcal{K}$ is a scaffold for $\mu$.
Lemma 22.4.3 (Gluing Lemma for Measures).label Let $(X, \cm)$ be a measurable space, $\cf \subset \cm$ be an ideal, and $\bracsn{\mu_A}_{A \in \cf}$ such that:
- (a)
For each $A \in \cf$, $\mu_{A}$ is a finite measure on $A$.
- (b)
For each $A, B \in \cf$ and $E \in \cm$, $\mu_{A}(E \cap A \cap B) = \mu_{B}(E \cap A \cap B)$.
Let
then:
- (1)
$\mu$ is a measure.
- (2)
$\cf$ is a scaffold for $(X, \cm, \mu)$.
- (3)
For each $E \in \cm$, $\mu(A \cap E) = \mu_{A}(A \cap E)$.
Proof. (3): Let $E \in \cm$, then $\mu_{A}(A \cap E) \le \mu(A \cap E)$ by definition. On the other hand, for any $B \in \cf$,
Since the above holds for all $B \in \cf$, $\mu(A \cap E) \le \mu_{A}(A \cap E)$.
(1): Let $\seq{E_n}\subset \cm$ be pairwise disjoint, then for each $A \in \cm$,
so $\mu\paren{\bigsqcup_{n \in \natp}E_n}\le \sum_{n \in \natp}\mu(E_{n})$.
On the other hand, let $n \in \natp$, $\seqf{A_k}\subset \cf$, and $A = \bigcup_{k = 1}^{n} A_{k} \in \cf$, then
As this holds for all choice of $\seq{A_k}\subset \cf$,
and as the above holds for all $n \in \natp$, $\sum_{n \in \natp}\mu(E_{n}) \le \mu\paren{\bigsqcup_{n \in \natp}E_n}$.
(2): By definition of $\mu$.$\square$
Corollary 22.4.4.label Let $(X, \cm, \mu)$ be a measure space, $\cf \subset \bracs{A \in \cm|\mu(A) < \infty}$ be an ideal, and
then
- (1)
$\mu_{\cf}$ is a measure on $(X, \cm)$.
- (2)
$\cf$ is a scaffold for $\mu_{\cf}$/
and $\mu_{\cf}$ is the $\cf$-scaffolded part of $\mu$.
Proof. For each $A \in \cf$ and $E \in \cm$, let $\mu_{A}(E) = \mu(E \cap A)$, then $\bracsn{\mu_A}_{A \in \cf}$ is a family of measures satisfying Lemma 22.4.3. Therefore $\mu_{\cf}$ as defined is a measure, and $\cf$ is a scaffold for $\mu_{\cf}$.$\square$
Lemma 22.4.5.label Let $(X, \cm, \cf, \mu)$ be a scaffolded measure space and $f \in L^{+}(X)$, then for any $E \in \cm$,
Proof. For any $F \in \cm$,
so by linearity, the above holds for all simple functions in $L^{+}(X)$.
Now, for each simple function $\phi \in \Sigma^{+}(X)$ with $\phi \le f$,
As the above holds for all $\phi \in \Sigma^{+}(X)$ with $\phi \le f$,
$\square$
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