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108
src/fa/lp/ui.tex
108
src/fa/lp/ui.tex
@@ -144,8 +144,8 @@
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\label{corollary:dct-filter}
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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 $g, h \in L^p(X; \real)$ such that:
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\begin{enumerate}[label=(\alph*)]
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\item $\fF \to g$ locally in measure.
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\item There exists $F \in \fF$ such that $|f| \le h$ for all $f \in F$.
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\item[(M)] $\fF \to g$ locally in measure.
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\item[(D)] There exists $F \in \fF$ such that $|f| \le h$ for all $f \in F$.
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\end{enumerate}
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then $\fF \to f$ in $L^p(X; E)$. In particular, if $p = 1$, then
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@@ -154,5 +154,107 @@
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\]
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\end{corollary}
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\begin{proof}
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By \autoref{theorem:vitali-convergence}.
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Since (D) implies (UI) and (T) of the \hyperref[Vitali Convergence Theorem]{theorem:vitali-convergence}, the result follows from the Vitali Convergence Theorem.
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\end{proof}
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\begin{lemma}[Scheffé]
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\label{lemma:scheffe}
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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:
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\begin{enumerate}
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\item[(M)] $\fF \to g$ locally in measure.
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\item[(N)] $\lim_{f, \fF} \norm{f}_{L^p(X; E)} = \norm{g}_{L^p(X; E)}$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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It is sufficient to show conditions (UI) and (T) of the \hyperref[Vitali Convergence Theorem]{theorem:vitali-convergence}.
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(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 \hyperref[Dominated Convergence Theorem]{theorem:dct}, there exists $A \in \cm$ with $\mu(A) < \infty$ such that:
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\begin{enumerate}[label=(\roman*)]
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\item $\int_{A^c} \norm{g}_E^p < \eps$.
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\end{enumerate}
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Since $\norm{g}_E^p d\mu \ll \mu$, by \autoref{proposition:ac-epsilon-delta}, there exists $\delta > 0$ such that:
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\begin{enumerate}[label=(\roman*), start=1]
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\item For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_E^p d\mu < \eps$.
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\item $\int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu > \int_A \norm{g}_E^p d\mu - \eps$.
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\end{enumerate}
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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 $f \in F_2$,
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\begin{align*}
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\int_A \norm{f}_E^p d\mu &\ge \int_{A \cap \bracs{\norm{f - g}_E \le \delta}}\norm{f}_E^p d\mu \\
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&\ge \int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu - \int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_E^p d\mu
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\end{align*}
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By (iii),
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\[
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\int_A \norm{f}_E^p d\mu \ge \int_A \norm{g}_E^p d\mu - \eps - \int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_E^p d\mu
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\]
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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),
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\[
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\int_A \norm{f}_E^p d\mu \ge \int \norm{g}_E^p d\mu - 3\eps \ge \int \norm{f}_E^p d\mu - 4\eps
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\]
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and $\int_{A^c}\norm{f}_E^p d\mu \le 4\eps$.
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(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 $f \in F_1$,
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\begin{enumerate}[label=(\roman*)]
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\item $|\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| < \eps$.
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\item $\int_{A^c}\norm{f}_E^p d\mu, \int_{A^c}\norm{g}_E^p d\mu < \eps$.
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\end{enumerate}
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By (i) and (ii),
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\begin{enumerate}[label=(\roman*), start=2]
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\item $\abs{\int_{A}\norm{f}_E^p d\mu - \int_{A}\norm{g}_E^p}d\mu \le 3\eps$.
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\end{enumerate}
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Since $\norm{g}_E^p d\mu \ll \mu$, by \autoref{proposition:ac-epsilon-delta}, there exists $\delta > 0$ such that:
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\begin{enumerate}[label=(\roman*), start=3]
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\item For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_E^p d\mu < \eps$.
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\item $\int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu > \int_A \norm{g}_E^p d\mu - \eps$.
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\end{enumerate}
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By (M), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that for every $f \in F_2$,
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\begin{enumerate}[label=(\roman*), start=5]
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\item $\mu(A \cap \bracs{\norm{f - g}_E > \delta}) < \delta$.
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\end{enumerate}
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Let $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$, then for any $f \in F_2$,
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\begin{align*}
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\int_{A \setminus B}\norm{f}_E^p d\mu &\ge \int_{(A \setminus B) \cap \bracs{\norm{f - g}_E \le \delta}}\norm{f}_E^p d\mu \\
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&\ge \int_{(A \setminus B) \cap \bracs{\norm{f - g}_E \le \delta}}(\norm{g}_E - \delta \vee 0)^pd\mu \\
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&\ge \int_{(A \setminus B)}(\norm{g}_E - \delta \vee 0)^pd\mu - \int_{A \cap \bracs{\norm{f - g}_E > \delta}} \norm{g}_E^p d\mu
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\end{align*}
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By (vi) and (iv), $\int_{A \cap \bracs{\norm{f - g}_E > \delta}} \norm{g}_E^p d\mu < \eps$, so
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\[
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\int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{(A \setminus B)}(\norm{g}_E - \delta \vee 0)^pd\mu - \eps
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\]
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By (v),
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\[
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\int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{(A \setminus B)}\norm{g}_E^pd\mu - 2\eps
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\]
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Since $\mu(B) < \delta$, by (iv),
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\[
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\int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{A}\norm{g}_E^pd\mu - 3\eps
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\]
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and by (iii),
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\[
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\int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{A}\norm{f}_E^pd\mu - 6\eps
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\]
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so $\int_{B}\norm{f}_E^p d\mu \le 6\eps$ for all $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$.
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Finally, by (i) and \hyperref[Markov's Inequality]{theorem:markov-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 $f \in F_2$,
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\[
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\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu \le \int_{A \cap \bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu + \int_{A^c}\norm{f}_E^p d\mu \le 7\eps
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\]
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\end{proof}
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