Fixed more typos for Scheffe.
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@@ -159,9 +159,9 @@
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\begin{lemma}[Scheffé]
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\begin{lemma}[Scheffé]
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\label{lemma:scheffe}
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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 $\fF \to f$ in $L^p(X; E)$ if and only if:
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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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\begin{enumerate}
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\item[(M)] $\fF \to g$ is locally in measure.
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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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\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{enumerate}
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\end{lemma}
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\end{lemma}
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@@ -206,7 +206,7 @@
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By (i) and (ii),
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By (i) and (ii),
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\begin{enumerate}[label=(\roman*), start=2]
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\begin{enumerate}[label=(\roman*), start=2]
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\item $\abs{\int_{A}\norm{f}_E^p d\mu - \int_{A}\norm{g}_E^p} \le 3\eps$.
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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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\end{enumerate}
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@@ -225,10 +225,10 @@
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\begin{align*}
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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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\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) \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_{\bracs{\norm{f - g}_E > \delta}} \norm{g}_E^p d\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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\end{align*}
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By (vi) and (iv), $\int_{\bracs{\norm{f - g}_E > \delta}} \norm{g}_E^p d\mu < \eps$, so
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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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\[
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\int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{(A \setminus B)}(\norm{g}_E - \delta \vee 0)^pd\mu - \eps
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\int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{(A \setminus B)}(\norm{g}_E - \delta \vee 0)^pd\mu - \eps
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\]
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\]
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