diff --git a/src/fa/order/lattice.tex b/src/fa/order/lattice.tex
index 000d8e3..9996c96 100644
--- a/src/fa/order/lattice.tex
+++ b/src/fa/order/lattice.tex
@@ -65,7 +65,6 @@
     Let $(E, \le)$ be an ordered vector space, then $E$ is a \textbf{vector lattice} if for any $x, y \in E$, $x \vee y = \sup\bracs{x, y}$ and $x \wedge y = \inf\bracs{x, y}$ exist. 
 \end{definition}
 
-
 \begin{definition}[Absolute Value]
 \label{definition:order-absolute-value}
     Let $(E, \le)$ be a vector lattice and $x \in E$, then $|x| = x \vee -x$ is the \textbf{absolute value} of $x$. 
diff --git a/src/measure/vector/fin.tex b/src/measure/vector/fin.tex
index e02d459..799abcb 100644
--- a/src/measure/vector/fin.tex
+++ b/src/measure/vector/fin.tex
@@ -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},
     \[
     \sup_{n \in \natp}\norm{f_n}_u \le \sup_{n \in \natp}\norm{f_n}_{\mathscr{M}^*} < \infty
     \]