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 $\mu$.$\square$
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