Spectral Theorem - $AB = BA \implies B\Phi(f) = \Phi(f)B$ Consider the construction of the Borel Functional Calculus for a self adjoint operator $A$ as descibred here:
Continuity of the functional calculus form of the Spectral Theorem
or better yet, here: 
http://www.math.mcgill.ca/jakobson/courses/ma667/mendelsontomberg-spectral.pdf on page 4.
Assume the continuous functional calculus was established.
I want to show: $AB = BA \implies B\Phi(f) = \Phi(f)B$ where $f \in \mathbb{B}(\mathbb{R})$ the bounded Borel functions on $\mathbb{R}$.
See my suggestion for a simpler proof below please.

I think my last answer has a problem in it (see comments), I'll leave it there in case someone comes upon this. Here's my new suggested solution:
Step 1 is to prove $\forall x \in H (x, B\Phi(f) x) = (x, \Phi(f) B x)$.
Step 2 is to notice that the above implies $\forall x,y \in H$ $(x, B\Phi(f) y) = (x, \Phi(f) B y)$ by the polarization identity.
Step 3 the above of course implies that $\Phi(f) B = B\Phi(f)$.
So we only need a proof for 1:
Set $x \in H$. 
Define $\phi_k = B^*x +i^kx$ for $k \in \{0,1,2,3\}$, $\psi_k = x +i^kB^*x$ for $k \in \{0,1,2,3\}$.
Define $\mu = \sum_{k = 0}^{3} |\mu_{\phi_k}| + |\mu_{\psi_k}|$. Where $\mu_{\phi_k}$, $\mu_{\psi_k}$ are the spectral measures associated to the corresponding vectors. As each is a regular Borel measure it follows that $\mu$ is a regular Borel measure.
Choose $\{f_n\} \subset C(\sigma(A))$ s.t $\int f_n d\mu \to \int f d\mu$. It follows that $\int f_n \mu_{\psi_k} \to \int f \mu_{\psi_k}$, and $\int f_n \mu_{\phi_k} \to \int f \mu_{\phi_k}$ for all $k \in \{0,1,2,3\}$. See $\int_X f_n d\mu_1 \to \int_X fd\mu_1 $ and $\int_X f_n d\mu_2\to \int_X f d\mu_2$ for a bounded Borel function $f$
So we have that $(x, B\Phi(f)x) = (B^*x, \Phi(f)x) \overset{definition}{=} \frac{1}{4} \sum_{k = 0}^{3} i^k \int f d\mu_{\phi_k} = lim_n \frac{1}{4} \sum_{k = 0}^{3} i^k \int f_n d\mu_{\phi_k} = lim_n(B^*x, \Phi(f_n)x) \overset{f_n \in C(\sigma(A))}{=} lim_n(x, \Phi(f_n)Bx) = lim_n \frac{1}{4} \sum_{k = 0}^{3} i^k \int f_n d\mu_{\psi_k} = \frac{1}{4} \sum_{k = 0}^{3} i^k \int f d\mu_{\psi_k} = (x, \Phi(f)Bx)$.
Comments are really appreciated!
