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Author SHA1 Message Date
Bokuan Li
84fb052c78 Fixed more typos for Scheffe.
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2026-06-21 00:29:02 -04:00
Bokuan Li
19e2d2df51 Typo fixes in Scheffe. 2026-06-21 00:24:59 -04:00
Bokuan Li
8c5400bf88 Adjusted the in measure version of DCT. 2026-06-21 00:22:48 -04:00
Bokuan Li
5ce01bd9f3 Added draft of Scheffe's lemma. 2026-06-21 00:21:11 -04:00

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@@ -144,8 +144,8 @@
\label{corollary:dct-filter} \label{corollary:dct-filter}
Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g, h \in L^p(X; \real)$ such that: Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g, h \in L^p(X; \real)$ such that:
\begin{enumerate}[label=(\alph*)] \begin{enumerate}[label=(\alph*)]
\item $\fF \to g$ locally in measure. \item[(M)] $\fF \to g$ locally in measure.
\item There exists $F \in \fF$ such that $|f| \le h$ for all $f \in F$. \item[(D)] There exists $F \in \fF$ such that $|f| \le h$ for all $f \in F$.
\end{enumerate} \end{enumerate}
then $\fF \to f$ in $L^p(X; E)$. In particular, if $p = 1$, then then $\fF \to f$ in $L^p(X; E)$. In particular, if $p = 1$, then
@@ -154,5 +154,107 @@
\] \]
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
By \autoref{theorem:vitali-convergence}. Since (D) implies (UI) and (T) of the \hyperref[Vitali Convergence Theorem]{theorem:vitali-convergence}, the result follows from the Vitali Convergence Theorem.
\end{proof} \end{proof}
\begin{lemma}[Scheffé]
\label{lemma:scheffe}
Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g \in L^p(X; E)$, then $\fF \to g$ in $L^p(X; E)$ if and only if:
\begin{enumerate}
\item[(M)] $\fF \to g$ locally in measure.
\item[(N)] $\lim_{f, \fF} \norm{f}_{L^p(X; E)} = \norm{g}_{L^p(X; E)}$.
\end{enumerate}
\end{lemma}
\begin{proof}
It is sufficient to show conditions (UI) and (T) of the \hyperref[Vitali Convergence Theorem]{theorem:vitali-convergence}.
(T): Let $\eps > 0$. By (N), there exists $F_1 \in \fF$ such that $|\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| < \eps$ for all $f \in F_1$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct}, there exists $A \in \cm$ with $\mu(A) < \infty$ such that:
\begin{enumerate}[label=(\roman*)]
\item $\int_{A^c} \norm{g}_E^p < \eps$.
\end{enumerate}
Since $\norm{g}_E^p d\mu \ll \mu$, by \autoref{proposition:ac-epsilon-delta}, there exists $\delta > 0$ such that:
\begin{enumerate}[label=(\roman*), start=1]
\item For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_E^p d\mu < \eps$.
\item $\int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu > \int_A \norm{g}_E^p d\mu - \eps$.
\end{enumerate}
By (M), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that for each $f \in F_2$, $\mu(A \cap \bracs{\norm{f - g}_E > \delta}) < \delta$. In which case, for any $f \in F_2$,
\begin{align*}
\int_A \norm{f}_E^p d\mu &\ge \int_{A \cap \bracs{\norm{f - g}_E \le \delta}}\norm{f}_E^p d\mu \\
&\ge \int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu - \int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_E^p d\mu
\end{align*}
By (iii),
\[
\int_A \norm{f}_E^p d\mu \ge \int_A \norm{g}_E^p d\mu - \eps - \int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_E^p d\mu
\]
and by (ii), $\int_A \norm{f}_E^p d\mu \ge \int_A\norm{g}_E^p d\mu - 2\eps$. Finally, by (i),
\[
\int_A \norm{f}_E^p d\mu \ge \int \norm{g}_E^p d\mu - 3\eps \ge \int \norm{f}_E^p d\mu - 4\eps
\]
and $\int_{A^c}\norm{f}_E^p d\mu \le 4\eps$.
(UI): Let $\eps > 0$. By (N) and (T), there exists $F_1 \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that for every $f \in F_1$,
\begin{enumerate}[label=(\roman*)]
\item $|\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| < \eps$.
\item $\int_{A^c}\norm{f}_E^p d\mu, \int_{A^c}\norm{g}_E^p d\mu < \eps$.
\end{enumerate}
By (i) and (ii),
\begin{enumerate}[label=(\roman*), start=2]
\item $\abs{\int_{A}\norm{f}_E^p d\mu - \int_{A}\norm{g}_E^p}d\mu \le 3\eps$.
\end{enumerate}
Since $\norm{g}_E^p d\mu \ll \mu$, by \autoref{proposition:ac-epsilon-delta}, there exists $\delta > 0$ such that:
\begin{enumerate}[label=(\roman*), start=3]
\item For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_E^p d\mu < \eps$.
\item $\int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu > \int_A \norm{g}_E^p d\mu - \eps$.
\end{enumerate}
By (M), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that for every $f \in F_2$,
\begin{enumerate}[label=(\roman*), start=5]
\item $\mu(A \cap \bracs{\norm{f - g}_E > \delta}) < \delta$.
\end{enumerate}
Let $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$, then for any $f \in F_2$,
\begin{align*}
\int_{A \setminus B}\norm{f}_E^p d\mu &\ge \int_{(A \setminus B) \cap \bracs{\norm{f - g}_E \le \delta}}\norm{f}_E^p d\mu \\
&\ge \int_{(A \setminus B) \cap \bracs{\norm{f - g}_E \le \delta}}(\norm{g}_E - \delta \vee 0)^pd\mu \\
&\ge \int_{(A \setminus B)}(\norm{g}_E - \delta \vee 0)^pd\mu - \int_{A \cap \bracs{\norm{f - g}_E > \delta}} \norm{g}_E^p d\mu
\end{align*}
By (vi) and (iv), $\int_{A \cap \bracs{\norm{f - g}_E > \delta}} \norm{g}_E^p d\mu < \eps$, so
\[
\int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{(A \setminus B)}(\norm{g}_E - \delta \vee 0)^pd\mu - \eps
\]
By (v),
\[
\int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{(A \setminus B)}\norm{g}_E^pd\mu - 2\eps
\]
Since $\mu(B) < \delta$, by (iv),
\[
\int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{A}\norm{g}_E^pd\mu - 3\eps
\]
and by (iii),
\[
\int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{A}\norm{f}_E^pd\mu - 6\eps
\]
so $\int_{B}\norm{f}_E^p d\mu \le 6\eps$ for all $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$.
Finally, by (i) and \hyperref[Markov's Inequality]{theorem:markov-inequality}, there exists $M \ge 0$ such that $\mu\bracs{\norm{f}_E \ge M} < \delta$ for all $f \in F_2$. Therefore for any $f \in F_2$,
\[
\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu \le \int_{A \cap \bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu + \int_{A^c}\norm{f}_E^p d\mu \le 7\eps
\]
\end{proof}