Replaced mentions of normed spaces to normed vector spaces.
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@@ -3,7 +3,7 @@
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\begin{definition}[Vector Measure]
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\label{definition:vector-measure}
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Let $(X, \cm)$ be a measurable space, $E$ be a normed space over $K \in \RC$, and $\mu: \cm \to E$, then $\mu$ is a \textbf{vector measure} if:
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Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $\mu: \cm \to E$, then $\mu$ is a \textbf{vector measure} if:
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\begin{enumerate}
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\item[(M1)] $\mu(\emptyset) = 0$.
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\item[(M2)] For any $\seq{A_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}A_n} = \sum_{n \in \natp} \mu(A_n)$ where the sum converges absolutely.
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@@ -13,7 +13,7 @@
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\begin{proposition}
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\label{proposition:vector-measure-bounded}
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Let $(X, \cm)$ be a measurable space, $E$ be a normed space over $K \in \RC$, and $\mu: \cm \to E$ be a vector measure, then
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Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $\mu: \cm \to E$ be a vector measure, then
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\[
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\sup_{A \in \cm}\norm{\mu(A)}_E < \infty
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\]
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