Polished A-A and added new lines for broken enumerates.
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@@ -71,7 +71,8 @@
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\]
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As this holds for all $\eps > 0$, $\mu^*(E) \le \sum_{n \in \natp}\mu^*(E_n)$.
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\end{enumerate}
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\end\{enumerate\}
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Therefore $\mu^*: 2^E \to [0, \infty]$ is an outer measure.
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To see that all Borel sets are $\mu^*$-measurable, let $E \subset X$ and $U \in \topo$. First suppose that $E$ is open. Let $f \prec E \cap U$, then for any $g \prec E \setminus \supp{f}$, $\supp{f} \cap \supp{g} = \emptyset$ and $f + g \prec E$, so
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