[proofplan]
Represent the [simple function](/page/Simple%20Function) on a finite disjoint Borel partition of $K$, so that the spectral integral is a finite sum of scalar multiples of spectral projections. Orthogonality of spectral projections over disjoint sets gives a Pythagorean identity for the vectors $E(\Delta_j)x$, and this yields the operator norm bound. The matrix coefficient formula follows by expanding the finite sum and comparing with the definition of the simple-function integral against $\mu_{x,y}$. Finally, real coefficients together with self-adjointness of the projections imply self-adjointness of the integral.
[/proofplan]
[step:Write the simple spectral integral on a disjoint Borel partition]
Since $s:K\to\mathbb C$ is a Borel simple function, there exist an integer $m\in\mathbb N$, pairwise disjoint Borel sets $\Delta_1,\dots,\Delta_m\in\mathcal B(K)$ whose union is $K$, and scalars $\alpha_1,\dots,\alpha_m\in\mathbb C$ such that
\begin{align*}
s=\sum_{j=1}^m \alpha_j\mathbb{1}_{\Delta_j}.
\end{align*}
By the definition of the simple spectral integral, define the operator $A\in\mathcal L(H)$ by
\begin{align*}
A:=\int_K s\,dE=\sum_{j=1}^m \alpha_jE(\Delta_j).
\end{align*}
Because $E$ is a projection-valued measure, each $E(\Delta_j)$ is an [orthogonal projection](/theorems/437), $E(K)=I$, and
\begin{align*}
E(\Delta_i)E(\Delta_j)=E(\Delta_i\cap\Delta_j)
\end{align*}
for all $i,j\in\{1,\dots,m\}$. Hence, if $i\ne j$, then $\Delta_i\cap\Delta_j=\varnothing$ and
\begin{align*}
E(\Delta_i)E(\Delta_j)=E(\varnothing)=0.
\end{align*}
[/step]
[step:Use orthogonality of spectral projections to prove the norm bound]
Fix $x\in H$. For each $j\in\{1,\dots,m\}$, define $v_j\in H$ by
\begin{align*}
v_j:=E(\Delta_j)x.
\end{align*}
If $i\ne j$, then, using self-adjointness of the projection $E(\Delta_j)$ and the disjointness relation from the previous step,
\begin{align*}
(v_i,v_j)_H=(E(\Delta_i)x,E(\Delta_j)x)_H=(E(\Delta_j)E(\Delta_i)x,x)_H=0.
\end{align*}
Thus $v_1,\dots,v_m$ are mutually orthogonal. By the Pythagorean identity for finite orthogonal families in a [Hilbert space](/page/Hilbert%20Space),
\begin{align*}
\|Ax\|_H^2=\left\|\sum_{j=1}^m \alpha_jv_j\right\|_H^2=\sum_{j=1}^m |\alpha_j|^2\|v_j\|_H^2.
\end{align*}
Since $|\alpha_j|\le \|s\|_\infty$ for every $j$ with $\Delta_j\ne\varnothing$, and empty partition pieces contribute $E(\varnothing)x=0$, it follows that
\begin{align*}
\|Ax\|_H^2\le \|s\|_\infty^2\sum_{j=1}^m \|E(\Delta_j)x\|_H^2.
\end{align*}
The same Pythagorean identity applied to
\begin{align*}
x=E(K)x=\sum_{j=1}^m E(\Delta_j)x=\sum_{j=1}^m v_j
\end{align*}
gives
\begin{align*}
\sum_{j=1}^m \|E(\Delta_j)x\|_H^2=\|x\|_H^2.
\end{align*}
Therefore
\begin{align*}
\|Ax\|_H\le \|s\|_\infty\|x\|_H.
\end{align*}
Taking the supremum over all $x\in H$ with $\|x\|_H\le 1$ gives
\begin{align*}
\|A\|_{\mathcal L(H)}\le \|s\|_\infty.
\end{align*}
[guided]
Fix a vector $x\in H$. The point of using a disjoint partition is that spectral projections over disjoint Borel sets have orthogonal ranges. For each $j\in\{1,\dots,m\}$, set
\begin{align*}
v_j:=E(\Delta_j)x.
\end{align*}
We verify orthogonality. If $i\ne j$, then $\Delta_i\cap\Delta_j=\varnothing$, so the projection-valued measure property gives
\begin{align*}
E(\Delta_j)E(\Delta_i)=E(\Delta_j\cap\Delta_i)=E(\varnothing)=0.
\end{align*}
Since $E(\Delta_j)$ is an orthogonal projection, it is self-adjoint. Hence
\begin{align*}
(v_i,v_j)_H=(E(\Delta_i)x,E(\Delta_j)x)_H=(E(\Delta_j)E(\Delta_i)x,x)_H=0.
\end{align*}
Thus the vectors $v_1,\dots,v_m$ are mutually orthogonal.
Now expand $Ax$. By the definition of $A$,
\begin{align*}
Ax=\sum_{j=1}^m \alpha_jE(\Delta_j)x=\sum_{j=1}^m \alpha_jv_j.
\end{align*}
Because the vectors $v_j$ are mutually orthogonal, the finite Pythagorean identity applies to the family $(\alpha_jv_j)_{j=1}^m$ and gives
\begin{align*}
\|Ax\|_H^2=\left\|\sum_{j=1}^m \alpha_jv_j\right\|_H^2=\sum_{j=1}^m |\alpha_j|^2\|v_j\|_H^2.
\end{align*}
The scalar $\|s\|_\infty$ bounds every coefficient that occurs on a nonempty partition piece. If $\Delta_j=\varnothing$, then $v_j=E(\varnothing)x=0$, so the corresponding term is zero regardless of the displayed coefficient. Therefore
\begin{align*}
\|Ax\|_H^2\le \|s\|_\infty^2\sum_{j=1}^m \|v_j\|_H^2.
\end{align*}
It remains to identify the last sum. Since the sets $\Delta_1,\dots,\Delta_m$ form a partition of $K$, finite additivity of the projection-valued measure gives
\begin{align*}
E(K)x=\sum_{j=1}^m E(\Delta_j)x=\sum_{j=1}^m v_j.
\end{align*}
Also $E(K)=I$, so $E(K)x=x$. Applying Pythagoras once more to the orthogonal family $(v_j)_{j=1}^m$ yields
\begin{align*}
\sum_{j=1}^m \|v_j\|_H^2=\left\|\sum_{j=1}^m v_j\right\|_H^2=\|x\|_H^2.
\end{align*}
Combining the two estimates gives
\begin{align*}
\|Ax\|_H^2\le \|s\|_\infty^2\|x\|_H^2.
\end{align*}
Taking square roots gives
\begin{align*}
\|Ax\|_H\le \|s\|_\infty\|x\|_H.
\end{align*}
Since this holds for every $x\in H$, the definition of the operator norm gives
\begin{align*}
\|A\|_{\mathcal L(H)}\le \|s\|_\infty.
\end{align*}
[/guided]
[/step]
[step:Expand matrix coefficients and identify the scalar simple integral]
Fix $x,y\in H$. By linearity of the finite sum and the definition of $\mu_{x,y}$,
\begin{align*}
(Ax,y)_H=\left(\sum_{j=1}^m \alpha_jE(\Delta_j)x,y\right)_H=\sum_{j=1}^m \alpha_j(E(\Delta_j)x,y)_H=\sum_{j=1}^m \alpha_j\mu_{x,y}(\Delta_j).
\end{align*}
By the definition of the integral of a simple function with respect to the finite complex measure $\mu_{x,y}$, whose existence is given by [citetheorem:8404],
\begin{align*}
\int_K s\,d\mu_{x,y}=\sum_{j=1}^m \alpha_j\mu_{x,y}(\Delta_j).
\end{align*}
Therefore
\begin{align*}
\left(\left(\int_K s\,dE\right)x,y\right)_H=\int_K s\,d\mu_{x,y}.
\end{align*}
[/step]
[step:Use real coefficients to prove self-adjointness]
Assume that $s$ is real-valued. We may choose the simple-function representation above with $\alpha_j\in\mathbb R$ for every $j\in\{1,\dots,m\}$. Each $E(\Delta_j)$ is an orthogonal projection, hence $E(\Delta_j)^*=E(\Delta_j)$. Using finite linearity of the adjoint operation and the fact that $\overline{\alpha_j}=\alpha_j$, we obtain
\begin{align*}
A^*=\left(\sum_{j=1}^m \alpha_jE(\Delta_j)\right)^*=\sum_{j=1}^m \overline{\alpha_j}E(\Delta_j)^*=\sum_{j=1}^m \alpha_jE(\Delta_j)=A.
\end{align*}
Thus $A=\int_K s\,dE$ is self-adjoint. This proves all asserted properties of the simple spectral integral.
[/step]
