Lemma 15.4.5 (Scheffé).label Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g \in L^{p}(X; E)$, then $\fF \to g$ in $L^{p}(X; E)$ if and only if:
- (M)
$\fF \to g$ locally in measure.
- (N)
$\lim_{f, \fF}\norm{f}_{L^p(X; E)}= \norm{g}_{L^p(X; E)}$.
Proof. It is sufficient to show conditions (UI) and (T) of the Vitali Convergence Theorem.
(T): Let $\eps > 0$. By (N), there exists $F_{1} \in \fF$ such that $|\norm{f}_{L^p(X; E)}^{p} - \norm{g}_{L^p(X; E)}^{p}| < \eps$ for all $f \in F_{1}$. By the Dominated Convergence Theorem, there exists $A \in \cm$ with $\mu(A) < \infty$ such that:
- (i)
$\int_{A^c}\norm{g}_{E}^{p} < \eps$.
Since $\norm{g}_{E}^{p} d\mu \ll \mu$, by Proposition 22.4.3, there exists $\delta > 0$ such that:
- (ii)
For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_{E}^{p} d\mu < \eps$.
- (iii)
$\int_{A}[(\norm{g}_{E} - \delta) \vee 0]^{p} d\mu > \int_{A} \norm{g}_{E}^{p} d\mu - \eps$.
By (M), there exists $F_{2} \in \fF$ with $F_{2} \subset F_{1}$ such that for each $f \in F_{2}$, $\mu(A \cap \bracs{\norm{f - g}_E > \delta}) < \delta$. In which case, for any $f \in F_{2}$,
By (iii),
and by (ii), $\int_{A} \norm{f}_{E}^{p} d\mu \ge \int_{A}\norm{g}_{E}^{p} d\mu - 2\eps$. Finally, by (i),
and $\int_{A^c}\norm{f}_{E}^{p} d\mu \le 4\eps$.
(UI): Let $\eps > 0$. By (N) and (T), there exists $F_{1} \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that for every $f \in F_{1}$,
- (i)
$|\norm{f}_{L^p(X; E)}^{p} - \norm{g}_{L^p(X; E)}^{p}| < \eps$.
- (ii)
$\int_{A^c}\norm{f}_{E}^{p} d\mu, \int_{A^c}\norm{g}_{E}^{p} d\mu < \eps$.
By (i) and (ii),
- (iii)
$\abs{\int_{A}\norm{f}_E^p d\mu - \int_{A}\norm{g}_E^p}d\mu \le 3\eps$.
Since $\norm{g}_{E}^{p} d\mu \ll \mu$, by Proposition 22.4.3, there exists $\delta > 0$ such that:
- (iv)
For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_{E}^{p} d\mu < \eps$.
- (v)
$\int_{A}[(\norm{g}_{E} - \delta) \vee 0]^{p} d\mu > \int_{A} \norm{g}_{E}^{p} d\mu - \eps$.
By (M), there exists $F_{2} \in \fF$ with $F_{2} \subset F_{1}$ such that for every $f \in F_{2}$,
- (vi)
$\mu(A \cap \bracs{\norm{f - g}_E > \delta}) < \delta$.
Let $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$, then for any $f \in F_{2}$,
By (vi) and (iv), $\int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_{E}^{p} d\mu < \eps$, so
By (v),
Since $\mu(B) < \delta$, by (iv),
and by (iii),
so $\int_{B}\norm{f}_{E}^{p} d\mu \le 6\eps$ for all $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$.
Finally, by (i) and Markov’s Inequality, there exists $M \ge 0$ such that $\mu\bracs{\norm{f}_E \ge M}< \delta$ for all $f \in F_{2}$. Therefore for any $f \in F_{2}$,
$\square$
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