Polished A-A and added new lines for broken enumerates.
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@@ -52,7 +52,8 @@
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\begin{enumerate}
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\item[(a)] $f_n \to f$ strongly pointwise.
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\item[(b)] There exists $g \in L^1(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$.
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\end{enumerate}
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\end\{enumerate\}
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then $\int f d\mu = \limv{n}\int f_n d\mu$.
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\end{theorem}
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@@ -76,7 +77,8 @@
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\begin{enumerate}
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\item For any $x \in E$ and $A \in \cm$, $I_\lambda(x \cdot \one_A) = x \mu(A)$.
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\item For any $f \in L^1(X, |\mu|; E)$, $\normn{I_\lambda f}_{G} \le \norm{\lambda}_{L^2(E, F; G)} \cdot \norm{f}_{L^1(X, |\mu|; E)}$.
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\end{enumerate}
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\end\{enumerate\}
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For any $f \in L^1(X; E)$, $I_\lambda f = \int \lambda(f, d\mu)$ is the \textbf{Bochner integral} of $f$ with respect to $\mu$ and $\lambda$.
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\end{definition}
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@@ -96,7 +98,8 @@
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\begin{enumerate}
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\item[(a)] $f_n \to f$ strongly pointwise.
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\item[(b)] There exists $g \in L^1(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$.
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\end{enumerate}
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\end\{enumerate\}
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then $\int \lambda(f , d\mu) = \limv{n}\int \lambda(f_n, d\mu)$.
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\end{theorem}
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@@ -11,7 +11,8 @@
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\begin{enumerate}
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\item[(a)] For each $n \in \natp$, $\norm{f_n}_E \le \norm{f}_E$.
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\item[(b)] $\norm{f_n(x) - f(x)}_E \to 0$ pointwise as $n \to \infty$.
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\end{enumerate}
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\end\{enumerate\}
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\end{enumerate}
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If the above holds, then $f$ is a \textbf{strongly measurable} function.
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@@ -39,7 +40,8 @@
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\begin{enumerate}
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\item For any strongly measurable functions $f, g: X \to E$ and $\lambda \in K$, $\lambda f + g$ is strongly measurable.
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\item For any strongly measurable functions $\bracs{f_n: X \to E|n \in \natp}$ and $f: X \to E$, if $f_n \to f$ strongly pointwise, then $f$ is strongly measurable.
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\end{enumerate}
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\end\{enumerate\}
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\end{proposition}
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\begin{proof}
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