I didn't have closed graph theorem?

src/fa/norm/linear.tex

@@ -1,4 +1,4 @@
-\section{Linear Maps}
+\section{Linear Maps Between Normed Spaces}
 \label{section:normed-linear-maps}
 
 \begin{proposition}

src/fa/norm/multilinear.tex

@@ -1,4 +1,4 @@
-\section{Multilinear Maps}
+\section{Multilinear Maps Between Normed Spaces}
 \label{section:normed-multilinear}
 
 \begin{proposition}

src/fa/tvs/complete-metric.tex

@@ -133,3 +133,20 @@
 
     Thus by \autoref{proposition:successive-approximation-all}, $B_F(0, t) \subset T(B_E(0, r)) \in \cn_F(0)$ for all $r > r_0$. As $r_0 > 0$ is arbitrary, $T(U) \in \cn_F(0)$ for all $U \in \cn_E(0)$. Therefore $T$ is open by translation-invariance of the topology on $E$.
 \end{proof}
+
+\begin{theorem}[Closed Graph Theorem]
+\label{theorem:closed-graph}
+    Let $E, F$ be complete metric TVSs over $K \in \RC$ and $T \in \hom(E; F)$. If its graph $\Gamma(T) \subset E \times F$ is closed, then $T \in L(E; F)$. 
+\end{theorem}
+\begin{proof}
+    Given that $E$ and $F$ are both complete metric TVSs, $E \times F$ is a complete metric TVS by \autoref{proposition:product-complete}. Since $\Gamma(T) \subset E \times F$ is a closed subspace of $E \times F$, it is also a complete metric TVS over $K$ by \autoref{proposition:complete-closed}. 
+    
+    Let $\pi_1: E \times F \to E$ and $\pi_2: E \times F \to F$ be the projection maps of $E \times F$. As $\Gamma(T)$ is the graph of a function, $\pi_1|_{\Gamma(T)}: \Gamma(T) \to E$ is a continuous bijection. By the \hyperref[Open Mapping Theorem]{theorem:open-mapping}, it is an isomorphism. Therefore $T$ may be expressed as the following composition of continuous linear maps
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{\pi_1|_{\Gamma(T)}^{-1}} & \Gamma(T) \ar@{->}[r]^{\pi_2|_{\Gamma(T)}} & F
+    }
+    \]
+
+\end{proof}
+

src/measure/weak-integral/definition.tex

@@ -41,4 +41,11 @@ As I know so little about weak integrals, I will Dunning-Kruger myself right now
 It is at this point that I start to understand why the bidual setup is useful: existence. 
 
 
+\begin{proposition}
+\label{proposition:dunford-existence}
+    Let 
+\end{proposition}
+
+
+