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b68c050225
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@@ -135,7 +135,7 @@
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\item $I_g(f) + I_g(f') \le I_g(f + f') + \eps$.
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\item $I_g(f) + I_g(f') \le I_g(f + f') + \eps$.
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\end{enumerate}
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\end{enumerate}
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In which case, $I(f) + I(f') \le I(f + f') + \eps$. Since this holds for all $\eps > 0$,
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In which case, $I(f) + I(f') \le I(f + f') + 3\eps$. Since this holds for all $\eps > 0$,
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\begin{enumerate}[start=4, label=(\roman*)]
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\begin{enumerate}[start=4, label=(\roman*)]
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\item For each $f, f' \in C_c^+(G)$, $I(f + g) \ge I(f) + I(g)$.
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\item For each $f, f' \in C_c^+(G)$, $I(f + g) \ge I(f) + I(g)$.
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\end{enumerate}
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\end{enumerate}
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@@ -178,9 +178,11 @@
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By \autoref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.
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By \autoref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.
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Let $A \in \cb_X$ and $\eps > 0$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $U \in \cn^o(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C_c(X; [0, 1])$ such that $f|_K = 1$ and $\supp{f} \subset U$. In which case, for any $y \in E$,
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Let $A \in \cb_X$ and $\eps > 0$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $U \in \cn^o(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C_c(X; [0, 1])$ such that $f|_K = 1$ and $\supp{f} \subset U$. In which case, for any $y \in E$,
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\[
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\begin{align*}
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\norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} \le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E
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\norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} &\le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \\
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\]
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&\le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E
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\end{align*}
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\end{proof}
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\end{proof}
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\begin{theorem}[Lusin]
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\begin{theorem}[Lusin]
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