Lemma 14.2.1.label Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $\dpn{E, F}{\lambda}$ be a norming duality of Banach spaces over $K$, and $f: X \to E$ be a strongly measurable function, then there exists $\seq{\phi_n}\subset \Sigma(X, \cm; F)$ such that:
- (1)
For each $n \in \natp$, $\norm{\phi_n}_{F}\le 1$.
- (2)
For every $n \in \natp$, $|\dpn{f, \phi_n}{\lambda}| \le \norm{f}_{E}$.
- (3)
$|\dpn{f, \phi_n}{\lambda}| \upto \norm{f}_{E}$ pointwise as $n \to \infty$.
Proof. Since $f(X)$ is separable, assume without loss of generality that $E$ is separable. By Proposition 12.4.1, there exists $\seq{z_n}\subset \bracsn{z \in F|\ \norm{z}_F \le 1}$ such that for each $y \in F$, $\norm{y}_{F} = \sup_{n \in \natp}\dpn{y, z_n}{\lambda}$.
For each $N \in \natp$ and $x \in X$, let $F_{N}(x) = 0 \vee \bigvee_{n = 1}^{N} |\dpn{f(x), z_n}{E}|$, then $0 \le F_{N} \le \norm{F}_{E}$ and $F_{N} \upto \norm{f}_{E}$ pointwise as $N \to \infty$.
For every $1 \le n \le N$, inductively define
Let $\phi_{N} = \sum_{n = 1}^{N} z_{n} \one_{A_{N, n}}$, then $|\dpn{f, \phi_N}{\lambda}| = F_{N}$, and
- (1)
Since $\norm{z_n}_{F} \le 1$ for all $n \in \natp$, $\norm{\phi_N}_{F} \le 1$ for each $N \in \natp$.
- (2)
For every $N \in \natp$, $|\dpn{f, \phi_N}{\lambda}| = F_{N}$, so $|\dpn{f, \phi_N}{\lambda}| \le \norm{f}_{E}$.
- (3)
As $F_{N} \upto \norm{f}_{E}$ pointwise as $N \to \infty$, $|\dpn{f, \phi_N}{\lambda}| \upto \norm{f}_{E}$ pointwise as $N \to \infty$.
$\square$
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