Lemma 15.3.2.label Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space over $K \in \RC$, $p \in (1, \infty]$, and $\phi \in L^{p}(X, \cm, \mu; E)^{*}$, then there exists a $\sigma$-finite set $A \in \cm$ such that
for all $f \in L^{p}(X, \cm, \mu; E)$.
Proof. For each $A \in \cm$, define
Let $f, g \in L^{p}(X; E)$ and $A, B \in \cm$ such that
- (1)
$X = A \sqcup B$.
- (2)
$f|_{B} = 0$, and $g|_{A} = 0$.
- (3)
$\norm{f}_{L^p(X; E)}= \norm{g}_{L^p(X; E)}= 1$.
Suppose that $p \in (1, \infty)$, then for each $t \in \real$,
Since $p > 1$, for each $t \in (0, 1)$, $(1 - t)^{p - 1}, t^{p - 1}< 1$, so $\norm{(1 - t)f + tg}_{L^p(X; E)}^{p} < 1$.
On the other hand, if $p = \infty$, then
Therefore for any $A, B \in \cm$ with $A \cap B = \emptyset$, $\norm{\phi_A}_{L^p(X; E)^*}> 0$, and $\norm{\phi_B}_{L^p(X; E)^*}> 0$,
Now, by density of simple functions,
Let $\seq{A_n}\subset \cm$ such that $\mu(A_{n}) < \infty$ for all $n \in \natp$, and $\norm{\phi_{A_n}}_{L^p(X; E)^*}\upto \norm{\phi}_{L^p(X; E)^*}$ as $n \to \infty$. Let $A = \bigcup_{n \in \natp}A_{n}$, then $A$ is $\sigma$-finite and $\norm{\phi_A}_{L^p(X; E)^*}= \norm{\phi}_{L^p(X; E)^*}$. By maximality, there exists no $B \in \cm$ with $B \cap A = \emptyset$ and $\norm{\phi_B}_{L^p(X; E)^*}> 0$, so $\phi = \phi_{A}$.$\square$
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