Adjusted wording for the sequential statement.
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Bokuan Li
2026-06-16 21:09:48 -04:00
parent ff2218c79b
commit 597a92b006
2 changed files with 5 additions and 7 deletions

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@@ -93,8 +93,7 @@ In any case, the above example shows that a linear functional on $M(X, \cm; \com
\]
\end{proof}
Despite the fact that it does not cover the full dual space, the bounded Borel functions still forms a subspace where weak-* convergence has a convenient description.
Despite not covering the full dual space, the bounded Borel functions still form a sequentially weak-* closed subspace with a convenient description for sequential convergence.
\begin{proposition}
\label{proposition:measure-l-infinity-dominated-convergence}
@@ -103,14 +102,14 @@ Despite the fact that it does not cover the full dual space, the bounded Borel f
\item[(P)] For each $x \in X$, $\bracs{x} \in \cm$, and the delta mass $\delta_x$ is in $\mathscr{M}$.
\end{enumerate}
Then, for any bounded measurable functions $\bracsn{f_n: X \to E^*|n \in \natp}$ and $f: X \to E^*$, the following are equivalent:
then for any sequence $\seq{f_n: X \to E^*}$ of bounded strongly measurable functions, the following are equivalent:
\begin{enumerate}
\item For each $\mu \in \mathscr{M}$, $\limv{n}\int f_n d\mu = \int f d\mu$.
\item For each $x \in X$, $\limv{n}f_n(x) = f(x)$, and $\sup_{n \in \natp}\norm{f_n}_u < \infty$.
\item For each $\mu \in \mathscr{M}$, $\limv{n}\int f_n d\mu$ exists.
\item There exists a bounded strongly measurable function $f: X \to E^*$ such that $f_n \to f$ pointwise and $\sup_{n \in \natp}\norm{f_n}_u < \infty$.
\end{enumerate}
\end{proposition}
\begin{proof}
(1) $\Rightarrow$ (2): By (P), for each $x \in X$, $\limv{n}f_n(x) = f(x)$. By the \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness},
(1) $\Rightarrow$ (2): By (P), for each $x \in X$, $\limv{n}f_n(x)$ exists. By the \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness},
\[
\sup_{n \in \natp}\norm{f_n}_u \le \sup_{n \in \natp}\norm{f_n}_{\mathscr{M}^*} < \infty
\]