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@@ -182,9 +182,9 @@
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\item For each $A \in \cf$, $f_n|_A = f_{A, n}$ almost everywhere.
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\end{enumerate}
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Since $Y$ is Polish, \autoref{proposition:metric-measurable-limit} implies that $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$. For each $A \in \cf$, since $\limv{n}f_{A, n}$ and $\limv{n}f_{n}|_A$ exist and are equal almost everywhere on $A$,
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Since $Y$ is Polish, \autoref{proposition:metric-measurable-limit} implies that $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$. For each $A \in \cf$, since $f_n|_A = f_{A, n}$ almost everywhere, $\limv{n}f_{A, n}$ and $\limv{n}f_{n}|_A$ exist and are equal almost everywhere on $A$. Thus
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\[
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\mu\paren{\bracs{\limv{n}f_n \text{ does not exist}} \cap A} \le \mu\bracs{\limv{n}f_{A, n} \text{ does not exist}} = 0
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\mu\paren{\bracs{\limv{n}f_n \text{ does not exist}} \cap A} \le \mu\bracs{\limv{n}f_{A, n} \text{ does not exist}} + 0 = 0
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\]
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As $\mu$ is semifinite, $\mu\bracsn{\limv{n}f_n \text{ does not exist}} = 0$, so there exists $f \in \mathcal{L}^0(X; Y)$ such that $f = \limv{n}f_n$ almost everywhere. In which case,
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