Adjusted wording for the sequential statement.

refs.bib

@@ -218,7 +218,7 @@
 
 @misc {StackRadonDual,
     title = {How to understand C(X)'' = bounded Borel measurable functions?},
-    author = {GEdgar (https://math.stackexchange.com/users/442/gedgar)},
+    author = {Edgar, G.A.},
     howpublished = {Mathematics Stack Exchange},
     note = {URL:https://math.stackexchange.com/q/392719 (version: 2013-05-15)},
     eprint = {https://math.stackexchange.com/q/392719},

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 $x$. 

src/measure/index.tex

@@ -8,5 +8,4 @@
 \input{./measurable-maps/index.tex}
 \input{./lebesgue-integral/index.tex}
 \input{./bochner-integral/index.tex}
-\input{./differentiation/index.tex}
 \input{./notation.tex}

src/measure/vector/fin.tex

@@ -33,7 +33,7 @@
 \end{proof}
 
 
-While the space of bounded Borel functions on $(X, \cm)$ forms a subspace of the dual of $M(X, \cm; \complex)$, it may not be immediately clear that they are insufficient. Before moving on to an explicit description of this dual, it is beneficial to consider a simple "example". 
+While the space of bounded Borel functions on $(X, \cm)$ forms a subspace of the dual of $M(X, \cm; \complex)$, it may not be immediately clear that they are insufficient. Before moving on to an explicit description of this dual, it is beneficial to consider the following "example". 
 
 Let $X = [0, 1]$, equipped with its Borel $\sigma$-algebra, and $\mu$ be the Lebesgue measure on $X$, then for any $x \in [0, 1]$, $\mu$ is mutually singular with the delta mass at $x$. Therefore the closure of $\text{span}\bracs{\delta_x|x \in X}$ in $M_R(X, \cm; \complex)$ is a proper closed subspace of $M(X, \cm; \complex)$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $\phi \in M(X, \cm; \complex)^*$ such that $\dpn{\delta_x, \phi}{M(X, \cm; \complex)} = 0$ for all $x \in X$, but $\dpn{\mu, \phi}{M(X, \cm; \complex)} = 1$. If there exists a bounded Borel function $f: X \to \complex$ such that $\dpn{\nu, \phi}{M(X, \cm; \complex)} = \int f d\nu$ for all $\nu \in M(X, \cm; \complex)$, then $f(x) = \dpn{\delta_x, \phi}{M(X, \cm; \complex)} = 0$ for all $x \in [0, 1]$, which is impossible. Therefore $\phi$ cannot be represented as a bounded Borel function. 
 
@@ -82,7 +82,7 @@ In any case, the above example shows that a linear functional on $M(X, \cm; \com
     Let $M = \sup\bracs{\norm{\nu_J}_{\text{var}}| J \subset I \text{ countable}}$, then there exists $J \subset I$ countable such that $\norm{\nu_J}_{\text{var}} = M$. For each $i \in I \setminus J$, 
     \[
     \norm{\nu_J}_{\text{var}} + \normn{\nu_a^{(i)}}_{\text{var}}
-    = \norm{\nu_{J} + \nu_a^{(i)}}_{\text{var}} = \normn{\nu_{J \cup \bracs{i}}}_{\text{var}} \le M
+    = \normn{\nu_{J} + \nu_a^{(i)}}_{\text{var}} = \normn{\nu_{J \cup \bracs{i}}}_{\text{var}} \le M
     \]
 
     By maximality of $M$, $\normn{\nu_a^{(i)}}_{\text{var}} = 0$, and by maximality of $\seqi{\mu}$, $\nu - \nu_J = 0$. Let $g \in [l^1(I); L^1(\mu_i; H)]$ be defined by $g_i = f_i$ for each $i \in I$, then $\nu = \nu_J = \mu_g$, and the mapping is surjective. 
@@ -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},
     \[
     \sup_{n \in \natp}\norm{f_n}_u \le \sup_{n \in \natp}\norm{f_n}_{\mathscr{M}^*} < \infty
     \]