diff --git a/src/op/vn/fc.tex b/src/op/vn/fc.tex
index f9ef440..224bfc2 100644
--- a/src/op/vn/fc.tex
+++ b/src/op/vn/fc.tex
@@ -98,39 +98,39 @@
 
     Since $\norm{E_{x, y}}_{C(\Omega(A); \complex)^*} \le \norm{x}_H\norm{y}_H$ for all $x, y \in H$, $\bracsn{E_{x, y}|x, y \in H}$ induces a bounded linear map
     \[
-    I: C(\Omega(A); \complex)^{**} \to B(H) \quad \dpn{I(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{C(\Omega(A); \complex)^*}
+    I_E: C(\Omega(A); \complex)^{**} \to B(H) \quad \dpn{I_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{C(\Omega(A); \complex)^*}
     \]
 
-    with $I(f) = \Gamma_A^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$. 
+    with $I_E(f) = \Gamma_A^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$. 
 
     For any $C \in \cb_{\Omega(A)}$, $\one_C$ is a projection in $B^\infty(\Omega(A); \complex)$. So to see that
     \[
     E: \cb_{\Omega(A)} \to B(H) \quad \dpn{E(C)x, y}{H} = E_{x, y}(C)
     \]
 
-    defines a spectral measure, it is sufficient to show that $I|_{B^\infty(\Omega(A); \complex)}$ is a *-homomorphism. 
+    defines a spectral measure, it is sufficient to show that $I_E|_{B^\infty(\Omega(A); \complex)}$ is a *-homomorphism. 
 
     Let $x, y \in H$, then as $\Gamma_A$ is a *-isomorphism, for any $f, g \in C(\Omega(A); \complex)$, 
     \begin{align*}
         \dpn{fg, E_{x, y}}{C(\Omega(A); \complex)} &= \dpn{\Gamma_A^{-1}f \cdot \Gamma_A^{-1}g \cdot x, y}{H} \\
-        &= \dpn{\Gamma_A^{-1}g \cdot x, (\Gamma_A^{-1}f)^* y}{H} = \dpn{g, E_{x, I(f)^*y}}{C(\Omega(A); \complex)}
+        &= \dpn{\Gamma_A^{-1}g \cdot x, (\Gamma_A^{-1}f)^* y}{H} = \dpn{g, E_{x, I_E(f)^*y}}{C(\Omega(A); \complex)}
     \end{align*}
 
-    As the above holds for all $g \in C(\Omega(A); \complex)$, $fE_{x, y} = E_{x, I(f)^*y}$. Now, fix $\phi \in B^\infty(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$, 
+    As the above holds for all $g \in C(\Omega(A); \complex)$, $fE_{x, y} = E_{x, I_E(f)^*y}$. Now, fix $\phi \in B^\infty(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$, 
     \begin{align*}
-        \dpn{E_{x, y}, \phi f}{C(\Omega(A); \complex)^*} &= \dpn{E_{x, I(f)^*y}, \phi}{C(\Omega(A); \complex)^*} = \dpn{I(\phi)x, I(f)^*y}{H} \\
-        &= \dpn{I(f)I(\phi)x, y}{H} = \dpn{f, E_{I(\phi)x, y}}{C(\Omega(A); \complex)}
+        \dpn{E_{x, y}, \phi f}{C(\Omega(A); \complex)^*} &= \dpn{E_{x, I_E(f)^*y}, \phi}{C(\Omega(A); \complex)^*} = \dpn{I_E(\phi)x, I_E(f)^*y}{H} \\
+        &= \dpn{I_E(f)I_E(\phi)x, y}{H} = \dpn{f, E_{I_E(\phi)x, y}}{C(\Omega(A); \complex)}
     \end{align*}
 
-    so $\phi E_{x, y} = E_{I(\phi)x, y}$ for all $\phi \in B^\infty(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^\infty(\Omega(A); \complex)$, 
+    so $\phi E_{x, y} = E_{I_E(\phi)x, y}$ for all $\phi \in B^\infty(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^\infty(\Omega(A); \complex)$, 
     \begin{align*}
-    \dpn{I(\phi \psi)x, y}{H} &= \dpn{E_{x, y}, \phi \psi}{C(\Omega(A); \complex)^*} = \dpn{E_{I(\psi) x, y}, \phi}{C(\Omega(A); \complex)^*} \\
-    &= \dpn{I(\phi)I(\psi)x, y}{H}
+    \dpn{I_E(\phi \psi)x, y}{H} &= \dpn{E_{x, y}, \phi \psi}{C(\Omega(A); \complex)^*} = \dpn{E_{I_E(\psi) x, y}, \phi}{C(\Omega(A); \complex)^*} \\
+    &= \dpn{I_E(\phi)I_E(\psi)x, y}{H}
     \end{align*}
 
-    and $I|_{B^\infty(\Omega(A); \complex)}$ is a homomorphism\footnote{With the same amount of writing and considerably more mental gymnastics, it can be shown that $I$ is a *-homomorphism on the full space $C(\Omega(A); \complex)^{**}$. However, it is not needed to show that $E$ is a spectral measure, and the homomorphism property falls out at the end anyways.}. 
+    and $I_E|_{B^\infty(\Omega(A); \complex)}$ is a homomorphism\footnote{With the same amount of writing and considerably more mental gymnastics, it can be shown that $I_E$ is a *-homomorphism on the full space $C(\Omega(A); \complex)^{**}$. However, it is not needed to show that $E$ is a spectral measure, and the homomorphism property falls out at the end anyways.}. 
 
-    Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_A$ is a *-isomorphism, $I(f) = \Gamma_A^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)} = \dpn{I(f)x, x}{H} \in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^\infty(\Omega(A); \real)$ and $x \in H$, $\dpn{I(\phi)x, x}{H} = \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*} \in \real$ as well. Therefore $I(\phi)$ is self-adjoint, and $I|_{B^\infty(\Omega(A); \complex)}$ is a *-homomorphism. 
+    Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_A$ is a *-isomorphism, $I_E(f) = \Gamma_A^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)} = \dpn{I_E(f)x, x}{H} \in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^\infty(\Omega(A); \real)$ and $x \in H$, $\dpn{I_E(\phi)x, x}{H} = \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*} \in \real$ as well. Therefore $I_E(\phi)$ is self-adjoint, and $I_E|_{B^\infty(\Omega(A); \complex)}$ is a *-homomorphism. 
 
     (2): By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $C(\Omega(A); \complex)^{**}$. Since the mapping $\phi \mapsto \int_{\Omega(A)}\phi dE$ is continuous from the weak* topology on $C(\Omega(A); \complex)^{**}$ to the weak operator topology on $B(H)$, 
     \[
@@ -160,20 +160,20 @@
     such that:
     \begin{enumerate}
         \item $\one(T) = I$, $\text{Id}(T) = T$, $\ol{Id}(T) = T^*$. 
-        \item The mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $C(\sigma_{B(H)}(T); \complex)^{**}$ to the weak operator topology on $C(\sigma_{B(H)}(T); \complex)$. 
+        \item The mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $C(\sigma_{B(H)}(T); \complex)^{**}$ to the weak operator topology on $A$. 
     \end{enumerate}
 
-    Moreover, there exists a unique spectral measure $E: \sigma_{B(H)}(T) \to A$
+    Moreover, there exists a unique spectral measure $E: \cb_{\sigma_{B(H)}(T)} \to A$ such that $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$ for all $\phi \in C(\sigma_{B(H)}(T); \complex)^{**}$. 
 \end{definition}
 \begin{proof}
     By the \hyperref[Spectral Theorem]{theorem:spectral-theorem-vn-1} applied to $B(H)[T]$, there exists a unique spectral measure $E$ on $\sigma_{B(H)}(T)$ such that the mapping
     \[
-    I_E: C(\sigma_{B(H)}(T); \complex)^{**} \to A \quad \phi \mapsto \int_{\sigma_A(T)} \phi dE
+    I_E: C(\sigma_{B(H)}(T); \complex)^{**} \to A \quad \phi \mapsto \int_{\sigma_{B(H)}(T)} \phi dE
     \]
 
     extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$. 
     
-    For each $\phi \in C(\sigma_{B(H)}; \complex)^{**}$, let $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$, then the mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $C(\sigma_{B(H)}(T); \complex)^{**}$ to the weak operator topology on $C(\sigma_{B(H)}(T); \complex)$ by \autoref{definition:spectral-measure-integral}.
+    For each $\phi \in C(\sigma_{B(H)}(T); \complex)^{**}$, let $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$, then the mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $C(\sigma_{B(H)}(T); \complex)^{**}$ to the weak operator topology on $A$ by \autoref{definition:spectral-measure-integral}.
 
     Finally, by uniqueness of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, and (2), the mapping $\phi \mapsto \phi(T)$ is unique. 
 \end{proof}