[proofplan]
We first define the operator integral for bounded Borel simple functions and prove its algebraic rules by refining finite Borel partitions. The key estimate is the simple-function bound
\begin{align*}
\left\|\int_K s\,dE\right\|_{\mathcal L(H)}\le \|s\|_\infty,
\end{align*}
which follows from orthogonality of the projections assigned to disjoint Borel sets. We then approximate arbitrary bounded Borel functions uniformly by simple functions, use the estimate to obtain an operator-norm limit independent of the approximating sequence, and pass the simple-function identities to the limit by norm-continuity of addition, multiplication, and the adjoint operation.
[/proofplan]
[step:Define the integral for simple Borel functions and check it is well defined]
Let $\mathcal S_b(K)$ denote the set of bounded Borel simple functions $s:K\to\mathbb C$. If $s\in\mathcal S_b(K)$ has a representation
\begin{align*}
s=\sum_{j=1}^{m} a_j\mathbb 1_{\Delta_j},
\end{align*}
where $m\in\mathbb N$, $a_1,\dots,a_m\in\mathbb C$, and $\Delta_1,\dots,\Delta_m\in\mathcal B(K)$ form a finite Borel partition of $K$, define
\begin{align*}
\Phi_E(s):=\int_K s\,dE:=\sum_{j=1}^{m} a_jE(\Delta_j)\in\mathcal L(H).
\end{align*}
This definition is independent of the chosen disjoint representation. Indeed, suppose also
\begin{align*}
s=\sum_{\ell=1}^{n} b_\ell\mathbb 1_{\Omega_\ell},
\end{align*}
where $n\in\mathbb N$, $b_1,\dots,b_n\in\mathbb C$, and $\Omega_1,\dots,\Omega_n\in\mathcal B(K)$ form a finite Borel partition of $K$. Refining both representations by the finite Borel partition
\begin{align*}
\Gamma_{j\ell}:=\Delta_j\cap\Omega_\ell
\end{align*}
and using finite additivity of the projection-valued measure on disjoint Borel sets gives
\begin{align*}
E(\Delta_j)=\sum_{\ell=1}^{n}E(\Gamma_{j\ell})
\end{align*}
and
\begin{align*}
E(\Omega_\ell)=\sum_{j=1}^{m}E(\Gamma_{j\ell}).
\end{align*}
On every nonempty $\Gamma_{j\ell}$, the equality of the two scalar simple functions implies $a_j=b_\ell$. Hence
\begin{align*}
\sum_{j=1}^{m}a_jE(\Delta_j)=\sum_{j=1}^{m}\sum_{\ell=1}^{n}a_jE(\Gamma_{j\ell})
\end{align*}
and
\begin{align*}
\sum_{\ell=1}^{n}b_\ell E(\Omega_\ell)=\sum_{j=1}^{m}\sum_{\ell=1}^{n}b_\ell E(\Gamma_{j\ell}).
\end{align*}
The summands agree whenever $E(\Gamma_{j\ell})$ is attached to a set on which the two representations differ only on the empty set, and empty-set terms vanish because $E(\varnothing)=0$. Therefore the two operator sums are equal.
[guided]
The only possible ambiguity in the definition is that a [simple function](/page/Simple%20Function) can be written using many different finite partitions. To remove this ambiguity, we compare two representations on their common refinement.
Let
\begin{align*}
s=\sum_{j=1}^{m} a_j\mathbb 1_{\Delta_j}
\end{align*}
and
\begin{align*}
s=\sum_{\ell=1}^{n} b_\ell\mathbb 1_{\Omega_\ell}
\end{align*}
be two representations on finite Borel partitions of $K$. For each pair of indices $j\in\{1,\dots,m\}$ and $\ell\in\{1,\dots,n\}$, define the Borel set
\begin{align*}
\Gamma_{j\ell}:=\Delta_j\cap\Omega_\ell.
\end{align*}
The sets $\Gamma_{j\ell}$ form a common finite refinement of the two partitions. Since $E$ is finitely additive on disjoint Borel sets, we have
\begin{align*}
E(\Delta_j)=\sum_{\ell=1}^{n}E(\Gamma_{j\ell})
\end{align*}
and
\begin{align*}
E(\Omega_\ell)=\sum_{j=1}^{m}E(\Gamma_{j\ell}).
\end{align*}
Therefore
\begin{align*}
\sum_{j=1}^{m}a_jE(\Delta_j)=\sum_{j=1}^{m}\sum_{\ell=1}^{n}a_jE(\Gamma_{j\ell})
\end{align*}
and
\begin{align*}
\sum_{\ell=1}^{n}b_\ell E(\Omega_\ell)=\sum_{j=1}^{m}\sum_{\ell=1}^{n}b_\ell E(\Gamma_{j\ell}).
\end{align*}
Why do the coefficients match on the refined pieces? If $\Gamma_{j\ell}$ is nonempty and $x\in\Gamma_{j\ell}$, then the first representation gives $s(x)=a_j$ while the second gives $s(x)=b_\ell$, so $a_j=b_\ell$. If $\Gamma_{j\ell}=\varnothing$, then $E(\Gamma_{j\ell})=0$. Thus every refined summand agrees, and the operator assigned to $s$ is independent of the representation.
[/guided]
[/step]
[step:Prove the algebraic identities for simple functions]
Let $s,t\in\mathcal S_b(K)$. Choose a finite Borel partition $\Delta_1,\dots,\Delta_m$ of $K$ and scalars $a_1,\dots,a_m,c_1,\dots,c_m\in\mathbb C$ such that
\begin{align*}
s=\sum_{j=1}^{m}a_j\mathbb 1_{\Delta_j}
\end{align*}
and
\begin{align*}
t=\sum_{j=1}^{m}c_j\mathbb 1_{\Delta_j}.
\end{align*}
Such a common partition is obtained by refining any two finite partitions for $s$ and $t$.
For $\alpha,\beta\in\mathbb C$, linearity follows directly:
\begin{align*}
\Phi_E(\alpha s+\beta t)=\sum_{j=1}^{m}(\alpha a_j+\beta c_j)E(\Delta_j).
\end{align*}
Thus
\begin{align*}
\Phi_E(\alpha s+\beta t)=\alpha\Phi_E(s)+\beta\Phi_E(t).
\end{align*}
For multiplication, use the projection-valued measure identity
\begin{align*}
E(\Delta_i)E(\Delta_j)=E(\Delta_i\cap\Delta_j).
\end{align*}
Since the partition sets are pairwise disjoint, $E(\Delta_i\cap\Delta_j)=0$ for $i\ne j$, while $E(\Delta_j\cap\Delta_j)=E(\Delta_j)$. Hence
\begin{align*}
\Phi_E(s)\Phi_E(t)=\sum_{i=1}^{m}\sum_{j=1}^{m}a_i c_j E(\Delta_i)E(\Delta_j).
\end{align*}
Therefore
\begin{align*}
\Phi_E(s)\Phi_E(t)=\sum_{j=1}^{m}a_jc_jE(\Delta_j)=\Phi_E(st).
\end{align*}
Since each $E(\Delta_j)$ is an [orthogonal projection](/theorems/437), $E(\Delta_j)^*=E(\Delta_j)$. Hence
\begin{align*}
\Phi_E(s)^*=\left(\sum_{j=1}^{m}a_jE(\Delta_j)\right)^*.
\end{align*}
Using conjugate-linearity of the Hilbert-space adjoint,
\begin{align*}
\Phi_E(s)^*=\sum_{j=1}^{m}\overline{a_j}E(\Delta_j)=\Phi_E(\overline{s}).
\end{align*}
Finally, the constant function $1:K\to\mathbb C$ satisfies $1=\mathbb 1_K$, so normalization of the projection-valued measure gives
\begin{align*}
\Phi_E(1)=E(K)=I.
\end{align*}
[/step]
[step:Establish the simple-function norm estimate]
Let $s\in\mathcal S_b(K)$. Choose a finite Borel partition $\Delta_1,\dots,\Delta_m$ of $K$ and scalars $a_1,\dots,a_m\in\mathbb C$ such that
\begin{align*}
s=\sum_{j=1}^{m}a_j\mathbb 1_{\Delta_j}.
\end{align*}
For $x\in H$, the vectors $E(\Delta_1)x,\dots,E(\Delta_m)x$ are pairwise orthogonal because, for $i\ne j$,
\begin{align*}
(E(\Delta_i)x,E(\Delta_j)x)_H=(E(\Delta_j)E(\Delta_i)x,x)_H=0.
\end{align*}
Therefore the Pythagorean identity gives
\begin{align*}
\|\Phi_E(s)x\|_H^2=\sum_{j=1}^{m}|a_j|^2\|E(\Delta_j)x\|_H^2.
\end{align*}
Since $|a_j|\le \|s\|_\infty$ for every $j$,
\begin{align*}
\|\Phi_E(s)x\|_H^2\le \|s\|_\infty^2\sum_{j=1}^{m}\|E(\Delta_j)x\|_H^2.
\end{align*}
Because the sets $\Delta_j$ partition $K$, finite additivity and $E(K)=I$ imply
\begin{align*}
\sum_{j=1}^{m}E(\Delta_j)x=x.
\end{align*}
Again using orthogonality,
\begin{align*}
\sum_{j=1}^{m}\|E(\Delta_j)x\|_H^2=\|x\|_H^2.
\end{align*}
Thus
\begin{align*}
\|\Phi_E(s)x\|_H\le \|s\|_\infty\|x\|_H
\end{align*}
for every $x\in H$. Taking the supremum over all $x\in H$ with $\|x\|_H\le 1$ gives
\begin{align*}
\|\Phi_E(s)\|_{\mathcal L(H)}\le \|s\|_\infty.
\end{align*}
[guided]
This estimate is the main reason the construction extends from simple functions to all bounded Borel functions. We want a bound in operator norm controlled only by the scalar supremum norm.
Write the simple function on a finite Borel partition of $K$:
\begin{align*}
s=\sum_{j=1}^{m}a_j\mathbb 1_{\Delta_j}.
\end{align*}
The corresponding operator is
\begin{align*}
\Phi_E(s)=\sum_{j=1}^{m}a_jE(\Delta_j).
\end{align*}
Fix $x\in H$. Since $E$ is projection-valued, each $E(\Delta_j)$ is an orthogonal projection, and since the sets $\Delta_i$ and $\Delta_j$ are disjoint for $i\ne j$, the product identity for a projection-valued measure gives
\begin{align*}
E(\Delta_i)E(\Delta_j)=E(\Delta_i\cap\Delta_j)=E(\varnothing)=0.
\end{align*}
Thus the vectors $E(\Delta_j)x$ are pairwise orthogonal. Consequently,
\begin{align*}
\|\Phi_E(s)x\|_H^2=\left\|\sum_{j=1}^{m}a_jE(\Delta_j)x\right\|_H^2
\end{align*}
equals
\begin{align*}
\sum_{j=1}^{m}|a_j|^2\|E(\Delta_j)x\|_H^2.
\end{align*}
Because $\|s\|_\infty$ is the supremum of the absolute values of $s$, every coefficient appearing on a nonempty partition set satisfies $|a_j|\le \|s\|_\infty$, and empty-set coefficients do not affect the operator. Hence
\begin{align*}
\|\Phi_E(s)x\|_H^2\le \|s\|_\infty^2\sum_{j=1}^{m}\|E(\Delta_j)x\|_H^2.
\end{align*}
The remaining sum is exactly $\|x\|_H^2$. Indeed, the partition covers $K$, so finite additivity and normalization give
\begin{align*}
\sum_{j=1}^{m}E(\Delta_j)=E(K)=I.
\end{align*}
Since the ranges of the projections $E(\Delta_j)$ are mutually orthogonal,
\begin{align*}
\sum_{j=1}^{m}\|E(\Delta_j)x\|_H^2=\left\|\sum_{j=1}^{m}E(\Delta_j)x\right\|_H^2=\|x\|_H^2.
\end{align*}
Combining the two estimates yields
\begin{align*}
\|\Phi_E(s)x\|_H\le \|s\|_\infty\|x\|_H.
\end{align*}
Taking the supremum over the closed unit ball of $H$ proves
\begin{align*}
\|\Phi_E(s)\|_{\mathcal L(H)}\le \|s\|_\infty.
\end{align*}
[/guided]
[/step]
[step:Extend the integral to bounded Borel functions by uniform simple approximation]
Let $f\in B_b(K)$. There exists a sequence $(s_n)_{n=1}^{\infty}$ in $\mathcal S_b(K)$ such that
\begin{align*}
\|s_n-f\|_\infty\to 0.
\end{align*}
For example, approximate the bounded real and imaginary parts of $f$ by finite-valued Borel step functions on intervals of mesh tending to $0$.
The simple-function estimate gives, for all $n,m\in\mathbb N$,
\begin{align*}
\|\Phi_E(s_n)-\Phi_E(s_m)\|_{\mathcal L(H)}=\|\Phi_E(s_n-s_m)\|_{\mathcal L(H)}\le \|s_n-s_m\|_\infty.
\end{align*}
Since $(s_n)_{n=1}^{\infty}$ converges uniformly, it is Cauchy in $\|\cdot\|_\infty$, so $(\Phi_E(s_n))_{n=1}^{\infty}$ is Cauchy in $\mathcal L(H)$. The space $\mathcal L(H)$ is complete in the operator norm, hence the limit
\begin{align*}
\Phi_E(f):=\lim_{n\to\infty}\Phi_E(s_n)
\end{align*}
exists in $\mathcal L(H)$.
This limit is independent of the approximating sequence. If $(t_n)_{n=1}^{\infty}$ is another sequence in $\mathcal S_b(K)$ with $\|t_n-f\|_\infty\to 0$, then
\begin{align*}
\|\Phi_E(s_n)-\Phi_E(t_n)\|_{\mathcal L(H)}\le \|s_n-t_n\|_\infty.
\end{align*}
Also,
\begin{align*}
\|s_n-t_n\|_\infty\le \|s_n-f\|_\infty+\|f-t_n\|_\infty\to 0.
\end{align*}
Therefore both sequences have the same operator-norm limit. This proves that $\Phi_E(f)$ is well defined for every $f\in B_b(K)$.
[/step]
[step:Pass linearity, adjoints, and the unit relation to bounded Borel functions]
Let $f,g\in B_b(K)$ and let $\alpha,\beta\in\mathbb C$. Choose sequences $(s_n)_{n=1}^{\infty}$ and $(t_n)_{n=1}^{\infty}$ in $\mathcal S_b(K)$ such that
\begin{align*}
\|s_n-f\|_\infty\to 0
\end{align*}
and
\begin{align*}
\|t_n-g\|_\infty\to 0.
\end{align*}
Then $\alpha s_n+\beta t_n\to \alpha f+\beta g$ uniformly. By the definition of $\Phi_E$ and the simple-function linearity,
\begin{align*}
\Phi_E(\alpha f+\beta g)=\lim_{n\to\infty}\Phi_E(\alpha s_n+\beta t_n).
\end{align*}
Thus
\begin{align*}
\Phi_E(\alpha f+\beta g)=\lim_{n\to\infty}\left(\alpha\Phi_E(s_n)+\beta\Phi_E(t_n)\right).
\end{align*}
Norm-continuity of addition and scalar multiplication in $\mathcal L(H)$ gives
\begin{align*}
\Phi_E(\alpha f+\beta g)=\alpha\Phi_E(f)+\beta\Phi_E(g).
\end{align*}
Since $\overline{s_n}\to\overline f$ uniformly, the simple-function adjoint identity gives
\begin{align*}
\Phi_E(\overline f)=\lim_{n\to\infty}\Phi_E(\overline{s_n})=\lim_{n\to\infty}\Phi_E(s_n)^*.
\end{align*}
The adjoint map $A\mapsto A^*$ on $\mathcal L(H)$ is isometric, hence norm-continuous. Therefore
\begin{align*}
\Phi_E(\overline f)=\Phi_E(f)^*.
\end{align*}
The constant function $1:K\to\mathbb C$ is already simple, so the simple-function identity gives
\begin{align*}
\Phi_E(1)=E(K)=I.
\end{align*}
[/step]
[step:Pass multiplicativity and the norm estimate to bounded Borel functions]
Let $f,g\in B_b(K)$. Choose sequences $(s_n)_{n=1}^{\infty}$ and $(t_n)_{n=1}^{\infty}$ in $\mathcal S_b(K)$ such that
\begin{align*}
\|s_n-f\|_\infty\to 0
\end{align*}
and
\begin{align*}
\|t_n-g\|_\infty\to 0.
\end{align*}
Since $f$ and $g$ are bounded and the approximating sequences converge uniformly, the products $s_nt_n$ converge uniformly to $fg$. Indeed,
\begin{align*}
\|s_nt_n-fg\|_\infty\le \|s_n\|_\infty\|t_n-g\|_\infty+\|g\|_\infty\|s_n-f\|_\infty,
\end{align*}
and the sequence $(\|s_n\|_\infty)_{n=1}^{\infty}$ is bounded because $s_n\to f$ uniformly. Therefore, using the definition of $\Phi_E$ and the simple-function multiplicativity,
\begin{align*}
\Phi_E(fg)=\lim_{n\to\infty}\Phi_E(s_nt_n)=\lim_{n\to\infty}\Phi_E(s_n)\Phi_E(t_n).
\end{align*}
Multiplication in $\mathcal L(H)$ is norm-continuous, so
\begin{align*}
\Phi_E(fg)=\Phi_E(f)\Phi_E(g).
\end{align*}
Finally, applying the simple-function norm estimate to $s_n$ gives
\begin{align*}
\|\Phi_E(s_n)\|_{\mathcal L(H)}\le \|s_n\|_\infty.
\end{align*}
Taking limits in $\mathbb R$ and using operator-norm convergence $\Phi_E(s_n)\to\Phi_E(f)$ and [uniform convergence](/page/Uniform%20Convergence) $s_n\to f$, we obtain
\begin{align*}
\|\Phi_E(f)\|_{\mathcal L(H)}\le \|f\|_\infty.
\end{align*}
Together with the previous steps, this proves that $\Phi_E:B_b(K)\to\mathcal L(H)$ is a unital $*$-homomorphism satisfying the stated norm estimate.
[/step]
