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 $\ce$.$\square$
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