Spectral Measures: Integrability I really need this as tool for other threads!
Given a Hilbert space $\mathcal{H}$.
Also a Borel space $\Omega$.
Consider a spectral measure:
$$E:\mathcal{B}(\Omega)\to\mathcal{P}(\mathcal{H}):\quad E(\sum_kA_k)\varphi=\sum_kE(A_k)\varphi$$
Regard the total measure:
$$|\langle E\varphi,\chi\rangle|(A):=\sup_\mathcal{A}\sum_{A\in\mathcal{A}}|\langle E(A)\varphi,\chi\rangle|$$
Then for every Borel function:
$$\omega\in\mathcal{B}(\Omega):\quad\left(\int|\omega|\operatorname{d|\langle E\varphi,\chi\rangle|}\right)^2\leq\int|\omega|^2\operatorname{d\|E\varphi\|}\cdot\|\chi\|^2$$
How can I proof this estimate?
 A: If $P$ is a positive operator, then $[x,y]_P = (Px,y)$ is a pseudo inner product, meaning that $(Px,x)=0$ might be true even if $x\ne 0$. For any pseudo inner product, Cauchy-Schwarz still holds:
$$
              |[x,y]_P| \le [x,x]_P^{1/2}[y,y]_P^{1/2}.
$$
Therefore, if $P$ is a positive Borel operator measure on a space $\Omega$,
\begin{align}
   \left|\sum_{n}a_n(P(E_n)x,y)\right|
      & \le \sum_{n}|a_n|(P(E_n)x,y)| \\
      & \le \sum_{n}|a_n|(P(E_n)x,x)^{1/2}(P(E_n)y,y)^{1/2} \\
      & \le \left(\sum_{n}|a_n|^2(P(E_n)x,x)\right)^{1/2}\left(\sum_{n}(P(E_n)y,y)\right)^{1/2} \\
      & \le \left(\sum_{n}|a_n|^2(P(E_n)x,x)\right)^{1/2}(P(\Omega)y,y)^{1/2}
\end{align}
In other words, if $f$ is a simple function on $\Omega$,
$$
        \left|\int f(\omega)d(P(\omega)x,y)\right| \le 
         \left(\int |f(\omega)|^2 d(P(\omega)x,x)\right)^{1/2}\|P(\Omega)\|^{1/2}\|y\|. \tag{$\dagger$}
$$
Because $P(\Omega)=I$ for a spectral measure, the above gives you what you want. But you don't need to assume a spectral measure; a positive operator measure is good enough. Therefore, if $f \in L^2(\Omega;dP((\cdot)x,x))$, there exists a unique vector--call it $\int f(\omega) dP(\omega)x$--such that
$$
        \int f(\omega)d(P(\omega)x,y)=\left(\int f(\omega)dP(\omega)x\;,\;y\right),
 \;\;\; y \in \mathcal{H}.
$$
And, if one assumes the normalization $\|P(\Omega)\|=1$,
$$
         \left\|\int f(\omega)dP(\omega)x\right\|^2 \le \int |f(\omega)|^2 d(P(\omega)x,x)
$$
Variation Integral: Because you don't want to use Radon-Nikodym, then start with a Borel set $E$ and any finite or countable Borel partition $\{E_n\}$ of $E$, and use the same arguments leading to $(\dagger)$. Start by choosing unimodular constants $\alpha_n$ such that $\alpha_n\mu_{x,y}(E_n)=|\mu_{x,y}(E_n)|$. Then
\begin{align}
      \sum_{n}|\mu_{x,y}(E_n)| & = \sum_{n}\alpha_n\mu_{x,y}(E_n) \\
 & = \sum_{n}\mu_{\alpha_n x,y}(E_n) \\
 & \le \sum_{n}\mu_{x,x}(E_n)^{1/2}\mu_{y,y}(E_n)^{1/2} \\
 & \le \left(\sum_{n}\mu_{x,x}(E_n)\right)^{1/2}\left(\sum_{n}\mu_{y,y}(E_n)\right)^{1/2} \\
 & = (\mu_{x,x}(E))^{1/2}(\mu_{y,y}(E))^{1/2}
\end{align}
Therefore the variation of the complex measure $\mu_{x,y}$ is bounded by
$$
             |\mu_{x,y}|(E) \le \mu_{x,x}(E)^{1/2}\mu_{y,y}(E)^{1/2}
$$
For any simple function $\sum_{n}\alpha_n E_n$ with disjoint Borel sets $E_n$,
\begin{align}
     \int |f|d|\mu_{x,y}| & = \sum_{n}|\alpha_n||\mu_{x,y}|(E_n) \\
   & \le \sum_{n}|\alpha_n|\mu_{x,y}(E_n)^{1/2}\mu_{y,y}(E_n)^{1/2} \\
   & \le \left(\sum_{n}|\alpha_n|^2\mu_{x,x}(E_n)\right)^{1/2}\left(\sum_{n}\mu_{y,y}(E_n)\right)^{1/2} \\
   & \le \left(\int |f|^2d\mu_{x,x}\right)^{1/2}\|y\|
\end{align}
A: Precalculus
For simple functions:
$$\sigma\in\mathcal{B}(\Omega):\quad\int\sigma\operatorname{dE}:=\sum_k\sigma_kE(A_k)\quad(\sigma=\sum_k\sigma_k1_k)$$
It holds the relation:
$$\|\{\int\sigma\operatorname{dE}\}\varphi\|^2=\sum_{kl}\sigma_k\overline{\sigma_l}\langle E(A_k)\varphi,E(A_l)\varphi\rangle\\
=\sum_k|\sigma_k|^2\|E(A_k)\varphi\|^2=\int|\sigma|^2\operatorname{d\|E\varphi\|^2}$$
Especially one obtains:
$$\|\{\int\sigma\operatorname{dE}\}\varphi\|^2\leq\|\sigma\|_\infty\|\varphi\|\implies\|\{\int\sigma\operatorname{dE}\}\|\leq\|\sigma\|_\infty$$
For bounded functions:*
$$\omega\in\mathcal{B}(\Omega):\quad\int\omega\operatorname{dE}:=\lim_n\int\sigma_n\operatorname{dE}\quad(\|\omega\|_\infty<\infty)$$
So the relation remains:
$$\|\{\int\omega\operatorname{dE}\}\varphi\|^2=\lim_n\|\{\int\sigma_n\operatorname{dE}\}\varphi\|^2\\=\lim_n\int|\sigma_n|^2\operatorname{d\|E\varphi\|^2}=\int|\omega|^2\operatorname{d\|E\varphi\|^2}$$
*Convergence: Operatornorm!
Integration
By Radon-Nikodym:
$$\langle E(A)\varphi,\chi\rangle=\int_A\dfrac{\operatorname{d\langle E\varphi,\chi\rangle}}{\operatorname{d|\langle E\varphi,\chi\rangle|}}\operatorname{d|\langle E\varphi,\chi\rangle|}=:\int_A(\ldots)\operatorname{d|\langle E\varphi,\chi\rangle|}$$
For simple functions:
$$\langle\{\int\sigma\operatorname{dE}\}\varphi,\chi\rangle=\langle\{\sum_k\sigma_kE(A_k)\}\varphi,\chi\rangle\\
=\sum_k\sigma_k\langle E(A_k)\varphi,\chi\rangle=\int\sigma(\ldots)\operatorname{d|\langle E\varphi,\chi\rangle|}$$
For bounded functions:*
$$\langle\{\int\omega\operatorname{dE}\}\varphi,\chi\rangle=\lim_n\langle\{\int\sigma_n\operatorname{dE}\}\varphi,\chi\rangle\\
=\lim_n\int\sigma_n(\ldots)\operatorname{d|\langle E\varphi,\chi\rangle|}=\int\omega(\ldots)\operatorname{d|\langle E\varphi,\chi\rangle|}$$
*Convergence: Uniform!
Estimate
For the derivative:*
$$|\dfrac{\operatorname{d\langle E\varphi,\chi\rangle}}{\operatorname{d|\langle E\varphi,\chi\rangle|}}|\equiv1\implies|(\dfrac{\operatorname{d\langle E\varphi,\chi\rangle}}{\operatorname{d|\langle E\varphi,\chi\rangle|}})^{-1}|\equiv1$$
For bounded functions:
$$0\leq\int|\omega|\operatorname{d|\langle E\varphi,\chi\rangle|}=\int|\omega|(\ldots)^{-1}(\ldots)\operatorname{d|\langle E\varphi,\chi\rangle|}=\langle\{\int|\omega|(\ldots)^{-1}\operatorname{dE}\}\varphi,\chi\rangle$$
By Cauchy-Schwarz:
$$|\langle\{\int|\omega|(\ldots)^{-1}\operatorname{dE}\}\varphi,\chi\rangle|\leq\|\{\int|\omega|(\ldots)^{-1}\operatorname{dE}\}\varphi\|\cdot\|\chi\|$$
And one arrives at:
$$\|\{\int|\omega|(\ldots)^{-1}\operatorname{dE}\}\varphi\|^2=\int|\omega|^2\cdot1\operatorname{d\|E\varphi\|^2}=\int|\omega|^2\operatorname{d\|E\varphi\|^2}$$
For unbounded functions:
$$(\int|\omega|\operatorname{d|\langle E\varphi,\chi\rangle|})^2=\lim_n(\int|\omega_n|\operatorname{d|\langle E\varphi,\chi\rangle|})^2\\
\leq\lim_n\int|\omega_n|^2\operatorname{d\|E\varphi\|^2}\cdot\|\chi\|^2=\int|\omega|^2\operatorname{d\|E\varphi\|^2}\cdot\|\chi\|^2$$
*Choice: Everywhere!
