Updated remark for dominated convergence.

src/fa/lp/ui.tex

@@ -50,7 +50,7 @@
 \label{theorem:vitali-convergence}
     Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, and $\fF \subset 2^{L^p(X; E)}$ be a filter, then $\fF$ is Cauchy in $L^p(X; E)$ if and only if:
     \begin{enumerate}
-        \item[(M)] $\fF$ is Cauchy in measure. 
+        \item[(M)] $\fF$ is locally Cauchy in measure. 
         \item[(UI)] For each $\eps > 0$, there exists $M \ge 0$ and $F \in \fF$ such that
         \[
         \sup_{f \in F}\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu < \eps
@@ -106,7 +106,7 @@
     
     Assume without loss of generality that $\mu(A) > 0$ and let $\delta = \eps\mu(A)^{-1/p}$. By (M), there exists $F_3 \in \fF$ with $F_3 \subset F_2$, such that for any $f, g \in F_3$, 
     \[
-    \mu\bracsn{\norm{f - g}_E \ge \delta} \le \paren{\frac{\eps}{2M}}^p
+    \mu(A \cap \bracsn{\norm{f - g}_E \ge \delta}) \le \paren{\frac{\eps}{2M}}^p
     \]
     
     In which case,
@@ -126,7 +126,7 @@
     Now, 
     \begin{align*}
         \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}f}_{L^p(X; E)} &\le \normn{\one_{\bracsn{\norm{f}_E \ge M}}f}_{L^p(X; E)} \\
-        &+ \normn{\one_{\bracsn{\norm{f - g}_E \ge \delta, \norm{f}_E \le M}}f}_{L^p(X; E)} \\
+        &+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta, \norm{f}_E \le M}}f}_{L^p(X; E)} \\
         &\le \eps + \eps/2 = 3\eps/2
     \end{align*}
 
@@ -140,3 +140,20 @@
     
 \end{proof}
 
+\begin{corollary}[Dominated Convergence Theorem (In Measure)]
+\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:
+    \begin{enumerate}[label=(\alph*)]
+        \item $\fF \to g$ pointwise and locally in measure.
+        \item There exists $F \in \fF$ such that $|f| \le h$ for all $f \in F$. 
+    \end{enumerate}
+    
+    then $\fF \to f$ in $L^p(X; E)$. In particular, if $p = 1$, then
+    \[
+    \lim_{f, \fF}\int f d\mu = \int g d\mu
+    \]
+\end{corollary}
+\begin{proof}
+    By \autoref{theorem:vitali-convergence}. 
+\end{proof}
+

src/measure/lebesgue-integral/complex.tex

@@ -122,9 +122,11 @@
     and $\int f d\mu = \limv{n}\int f_n d\mu$.
 \end{proof}
 
-\begin{remark}[There is no dominated convergence theorem for nets]
+\begin{remark}[Dominated Convergence Theorem for Nets?]
 \label{remark:dct-no-net}
     In analysis, one frequently encounters places where only sequential continuity is provided or required. It is my opinion that a good portion of this comes from the lack of an extension of the \hyperref[dominated convergence theorem]{theorem:dct} to nets. This limitation arises from the \hyperref[monotone convergence theorem]{theorem:mct}, where continuity from below is used.
 
     For an example, consider the Lebesgue measure on $[0, 1]$. Let $A$ be the net of all finite subsets of $[0, 1]$, directed by inclusion, then $\lim_{\alpha \in A}\one_\alpha = 1$ pointwise. However, $\int \one_\alpha = 0$ for all $\alpha \in A$.
+
+    At least, that is how I thought back in 2025. While arbitrary pointwise convergence has no reason to cooperate with the structure of a measure space, additionally supplying convergence in measure bridges the above gap. The corresponding Monotone Convergence Theorem is given at \autoref{theorem:mct-measure}, and the Dominated Convergence Theorem is given at \autoref{corollary:dct-filter} following the Vitali Convergence Theorem. 
 \end{remark}

src/measure/measurable-maps/in-measure.tex

@@ -87,7 +87,7 @@
     (3): By (2), there exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ and $f_{n_k} \to g$ almost everywhere, so $f = g$ almost everywhere. 
 \end{proof}
 
-\begin{theorem}[Monotone Convergence Theorem (In Measure)]
+\begin{theorem}[Monotone Convergence Theorem (in Measure)]
 \label{theorem:mct-measure}
     Let $(X, \cm, \mu)$ be a semifinite measure space, $\net{f} \subset \mathcal{L}^+(X, \cm)$, and $f \in \mathcal{L}^+(X, \cm)$ such that
     \begin{enumerate}[label=(\alph*)]
@@ -118,3 +118,4 @@
 
     As the above holds for all $\eps > 0$, $\lambda \in (0, 1)$, $\sup_{\alpha \in A}\int f_\alpha d\mu \ge \int \phi d\mu$. Therefore $\sup_{\alpha \in A}\int f_\alpha d\mu \ge \int f d\mu$ by \autoref{lemma:lebesgue-non-negative-strict}. 
 \end{proof}
+