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Added missing justification.
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Bokuan Li
2026-06-14 23:09:35 -04:00
parent 33c6a6775d
commit 5b1e0f86e6
1 changed files with 9 additions and 2 deletions
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src/measure/vector/ac.tex
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@@ -77,11 +77,18 @@
\norm{\nu(A) - \sum_{n = 1}^N \nu(A_n)}_E &= \norm{\nu(A) - \sum_{n = 1}^N \nu(A_n)}_E &=
\norm{\nu\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E \\ \norm{\nu\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E \\
&= &=
\norm{\limv{k}\nu_k\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E \limv{k}\norm{\nu_k\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E
\end{align*} \end{align*}
Since $\seq{\nu_k}$ is uniformly absolutely continuous,
\[
\lim_{N \to \infty}\sup_{k \in \natp}\norm{\nu_k\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E = 0
\]
so $\nu$ is a vector measure. Absolute continuity is equivalent to uniform continuity as a mapping $\cm_n \to E$, and uniform absolute continuity is equivalent to uniform equicontinuity as mappings $\cm \to E$. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\nu \in UC(\cm_0; E)$, so $\nu \ll \mu$. so $\nu$ is a vector measure.
Absolute continuity is equivalent to uniform continuity as a mapping $\cm_n \to E$, and uniform absolute continuity is equivalent to uniform equicontinuity as mappings $\cm \to E$. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\nu \in UC(\cm_0; E)$, so $\nu \ll \mu$.
\end{proof} \end{proof}