Random nonsense.

src/fa/duality/mackey.tex

@@ -78,7 +78,7 @@
 
     is a convex, circled, and closed subset. Given that $B$ is $\sigma(E, F)$-bounded, $\sup_{x \in B}|\dpn{x, \phi}{\lambda}| < \infty$ for all $x \in E$. Thus $A$ is absorbing and hence a barrel. 
 
-    Since $B$ is compact, the auxiliary space $E_B$ is a Banach space. In particular, $E_B$ is barreled by \autoref{proposition:baire-barrel}. By continuity of the inclusion map, $A \cap E_B$ is a barrel in $E_B$. Therefore there exists $\lambda > 0$ such that $B \subset \lambda A \cap E_B \subset \lambda A$.
+    Since $B$ is compact, the auxiliary space $E_B$ is a Banach space. In particular, $E_B$ is barrelled by \autoref{proposition:baire-barrel}. By continuity of the inclusion map, $A \cap E_B$ is a barrel in $E_B$. Therefore there exists $\lambda > 0$ such that $B \subset \lambda A \cap E_B \subset \lambda A$.
 \end{proof}
 
 
@@ -89,10 +89,10 @@
 
 \begin{proposition}
 \label{proposition:barreled-mackey}
-    Let $E$ be a separated barreled space over $K \in \RC$, then $E$ is a Mackey space. 
+    Let $E$ be a separated barrelled space over $K \in \RC$, then $E$ is a Mackey space. 
 \end{proposition}
 \begin{proof}
-    Let $\cf \subset E^*$ be a $\sigma(E^*, E)$-compact set and $U \in \cn_{K}(0)$ be a barrel, then $V = \bigcap_{\phi \in \cf}\phi^{-1}(U)$ is convex, circled, and closed. For each $x \in E$, $\cf(x) = \bracs{\dpn{x, \phi}{E}|\phi \in \cf}$ is bounded. Thus $V$ is absorbing and hence a barrel. Since $E$ is barreled, $V \in \cn_E(0)$. Therefore the Mackey topology is contained in the topology of $E$, and $E$ is a Mackey space. 
+    Let $\cf \subset E^*$ be a $\sigma(E^*, E)$-compact set and $U \in \cn_{K}(0)$ be a barrel, then $V = \bigcap_{\phi \in \cf}\phi^{-1}(U)$ is convex, circled, and closed. For each $x \in E$, $\cf(x) = \bracs{\dpn{x, \phi}{E}|\phi \in \cf}$ is bounded. Thus $V$ is absorbing and hence a barrel. Since $E$ is barrelled, $V \in \cn_E(0)$. Therefore the Mackey topology is contained in the topology of $E$, and $E$ is a Mackey space. 
 \end{proof}
 
 

src/fa/lc/barrel.tex

@@ -1,4 +1,4 @@
-\section{Barreled Spaces}
+\section{Barrelled Spaces}
 \label{section:barrel}
 
 \begin{definition}[Barrel]
@@ -6,7 +6,7 @@
     Let $E$ be a TVS over $K \in \RC$ and $D \subset E$, then $D$ is a \textbf{barrel} if it is convex, circled, radial, and closed. 
 \end{definition}
 
-\begin{definition}[Barreled Space]
+\begin{definition}[Barrelled Space]
 \label{definition:barreled-space}
     Let $E$ be a locally convex space over $K \in \RC$, then the following are equivalent:
     \begin{enumerate}

src/fa/lp/duality.tex

@@ -178,7 +178,7 @@ The typical argument for $L^p$ duality requires using the Radon-Nikodym theorem
     Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $H$ be a Hilbert space over $K$, $p, q \in [1, \infty]$ be Hölder conjugates such that one of the following holds:
     \begin{enumerate}[label=(\alph*)]
         \item $p \in (1, \infty)$ and $q \in (1, \infty)$.  
-        \item $p = 1$, $q = \infty$, and $\mu$ is $\sigma$-finite. 
+        \item $p = 1$, $q = \infty$, and $\mu$ is $\sigma$-finite\footnote{This should become localisable. }. 
     \end{enumerate}
 
     For each $g \in L^q(X, \cm, \mu; H)$, let

src/measure/weak-integral/definition.tex

@@ -0,0 +1,44 @@
+\section{Weak Integrals*}
+\label{section:weak-integral}
+
+\begin{definition}[Weakly Measurable]
+\label{definition:weakly-measurable}
+    Let $(X, \cm)$ be a measurable space, $E$ be a locally convex space over $K \in \RC$, and $f: X \to E$, then $f$ is \textbf{weakly measurable} if for each $\phi \in E^*$, $\phi \circ f: X \to K$ is Borel measurable. 
+\end{definition}
+
+As I know so little about weak integrals, I will Dunning-Kruger myself right now, give an opinion, and laugh about it later. My gripe with seeing the definition comes from the need to test against \textit{every} continuous linear functional. To me, this seems quite inflexible: consider integrating a distribution-valued function. \textit{Surely} it is wiser to only test this function against test functions rather than \textit{the dual of $\mathcal{D}'$ (dual with respect to $\mathcal{D}'$ with the bounded convergence topology)}. As such, it may be more productive to consider a more flexible form of testing, such as using duality. 
+
+
+\begin{definition}[Weakly Integrable*]
+\label{definition:weakly-integrable}
+    Let $(X, \cm, \mu)$ be a measure space, $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, and $f: X \to E$, then $f$ is \textbf{Dunford $\lambda$-integrable*} if:
+    \begin{enumerate}[label=(I\arabic*)]
+        \item $f$ is weakly measurable. 
+        \item For each $\phi \in F$, $\phi \circ f \in L^1(X; K)$.
+        \item For each $A \in \cm$, the mapping
+        \[
+        F \to K \quad \phi \mapsto \int_A \dpn{f(x), \phi}{\lambda} d\mu 
+        \]
+
+        is a continuous linear functional on $F$. 
+    \end{enumerate}
+
+    For each $A \in \cm$, the element $\phi \mapsto \int_A \dpn{f(x), \phi}{\lambda} d\mu$ of $F^*$ is the \textbf{Dunford $\lambda$-integral} of $f$ over $A$, denoted $\int_A^* f d\mu$. 
+
+    The function $f$ is \textbf{Pettis $\lambda$-integrable*} if it satisfies (I1), (I2), and
+    \begin{enumerate}
+        \item[(I3+)] For each $A \in \cm$, the mapping
+        \[
+        F \to K \quad \phi \mapsto \int_X \dpn{f(x), \phi}{\lambda} d\mu 
+        \]
+
+        is a $\sigma(F, E)$-continuous linear functional on $F$. 
+    \end{enumerate}
+
+    In which case, for each $A \in \cm$, $\int_A^* f d\mu$ is the \textbf{Pettis $\lambda$-integral} of $f$ over $A$. 
+\end{definition}
+
+It is at this point that I start to understand why the bidual setup is useful: existence. 
+
+
+

src/measure/weak-integral/index.tex

@@ -0,0 +1,5 @@
+\chapter{Weak Integrals*}
+\label{chap:weak-integral}
+
+\input{./definition.tex}
+