Removed pretentiousness.
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@@ -37,7 +37,7 @@
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then
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then
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\begin{enumerate}
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\begin{enumerate}
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\item By definition, $\nu = \nu_a + \nu_s$.
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\item Since $X = A \sqcup B$, $\nu = \nu_a + \nu_s$.
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\item Let $E \in \cm$ with $\mu(E) = 0$, then $(1 - g)|_E = 0$ $\lambda$-almost everywhere. Thus $E \subset B$ modulo a $\lambda$-null set, and
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\item Let $E \in \cm$ with $\mu(E) = 0$, then $(1 - g)|_E = 0$ $\lambda$-almost everywhere. Thus $E \subset B$ modulo a $\lambda$-null set, and
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\[
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\[
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\nu_a(E) = \nu(E \cap A) \le \lambda(A \cap B) = 0
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\nu_a(E) = \nu(E \cap A) \le \lambda(A \cap B) = 0
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