How to show a Borel Operator Measure dilates to a Spectral Measure? Does anyone know a simple proof of the following theorem stating that a positive Borel operator measure $P$ on $\mathbb{R}$ can be written as $V^{\star}EV$ for a Borel spectral measure $E$?

Theorem (Operator Measure Dilation): Let $P$ be a function from the Borel subsets of $\mathbb{R}$ into the bounded linear operators $\mathcal{L}(\mathcal{H})$ on a complex Hilbert space $\mathcal{H}$ such that $\mu_{x}(S)=(P(S)x,x)$ is a positive Borel measure for each $x\in\mathcal{H}$, with $P(\mathbb{R})=I$. Then there is complex Hilbert space $\mathcal{K}$, a spectral measure $E$ from the Borel subsets of $\mathbb{R}$ into $\mathcal{L}(\mathcal{K})$, and an isometry $V : \mathcal{H}\rightarrow\mathcal{K}$ such that $P(S)=V^{\star}E(S)V$.

 A: It is convinient to use the following notation $\mu_{x, y}(S) := \left<P(S)x, y\right>$ and shortly $\mu_{x}$ for $\mu_{x,x}$.
Let $B(\mathbb{R})$ denote the abelian C*-algebra of bounded Borel functions on $\mathbb{R}$.
Define a map $\Phi_P \colon B(\mathbb{R}) \rightarrow \mathcal{L}(\mathcal{H})$ by
$$\left<\Phi_P(f)x, y\right> = \int_{\mathbb{R}} f \ \mathrm{d}\mu_{x,y}.$$
Note that $\Phi_P$ is well-defined, since 
$$\left<\Phi_P(f)x, x\right> = \int_{\mathbb{R}} f \mathrm{d} \mu_x \leq \|f\|_{\infty}\|x\|,$$
(the assumption that $P(\mathbb{R})=I$ was used) and moreover $\Phi_P$ is a bounded linear map itself. 
Positivity of $\mu_x$ implies that $\Phi_P$ is a positive map, furthermore it is an easy exercise to show that $\Phi_P$ is a completely positive map since $B(\mathbb{R})$ is an abelian C* algebra. 
Now let us apply the Stinespring factorisation theorem:
Therefore there exists a Hilbert space $\mathcal{K}$ and a unital *-homomorphism
$$\pi \colon B(\mathbb{R}) \rightarrow \mathcal{B}(\mathcal{K})$$
such that
$$ \Phi_P(f) = V^*\pi(f)V,$$
where $V \in \mathcal{B}(\mathcal{H}, \mathcal{K})$. Moreover, $V$ is an isometry and
$$ I = \Phi_P(1) = V^* \pi(1)V = V^*V.$$
Using the fact that $\pi$ is a unital *-homomorphism, we obtain that there is a unique spectral measure $E \colon B(\mathbb{R}) \rightarrow \mathcal{B}(\mathcal{K})$ such that
$$ \pi(f) = \int_{\mathbb{R}} f \ \mathrm{d}E.$$
Finally, let $S$ be an arbitrary Borel subset of $\mathbb{R}$. Then
$$ \left<P(S)x, y \right>  = \int_{\mathbb{R}} \mathbf{1}_S \ \mathrm{d}\mu_{x,y} = \left< \Phi_P(\mathbf{1}_S)x, y\right> $$ $$= \left< V^*\pi(\mathbf{1}_S)Vx, y\right>=  \left< V^*\int_{\mathbb{R}} \mathbf{1}_S \ \mathrm{d}EVx, y\right> =\left< V^*E(S)Vx, y\right>$$
for all $x, y \in \mathcal{H}$.
