Fixed stack exchange citation.

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.