[proofplan]
The construction proceeds in two stages: first define $\Phi$ on the dense subspace of simple Borel functions in $L^\infty(P)$, then extend by isometry to the closure $L^\infty(P)$. On simple functions $s = \sum_{i=1}^m \alpha_i \mathbb{1}_{E_i}$ with $\{E_i\}$ a Borel partition of $K$, set $\Phi(s) := \sum_{i=1}^m \alpha_i P(E_i)$. The disjointness of the partition combined with the multiplicativity axiom $P(E)P(F) = P(E \cap F)$ makes $\Phi$ a $*$-homomorphism on simples, and the orthogonality of the projections $P(E_i)$ produces the isometric identity $\|\Phi(s)x\|^2 = \int_K |s|^2 \, dP_{x,x}$. The unique continuous extension to $L^\infty(P)$ inherits all algebraic and metric properties by continuity. Property (iii) follows because the projections $P(E)$ generate the image of $\Phi$ in $\mathcal{L}(H)$.
[/proofplan]
[step:Recall the structural axioms of a resolution of the identity]
A resolution of the identity $P$ of $H$ over $K$ is, by definition, a map $P: \mathcal{B} \to \mathcal{L}(H)$ from the Borel $\sigma$-algebra of $K$ such that:
(R1) Each $P(E)$ is an orthogonal projection.
(R2) $P(\varnothing) = 0$ and $P(K) = I$.
(R3) Multiplicativity: $P(E \cap F) = P(E) P(F)$ for all $E, F \in \mathcal{B}$.
(R4) Disjoint additivity: $P(E \cup F) = P(E) + P(F)$ whenever $E \cap F = \varnothing$.
(R5) For each pair $x, y \in H$, the set function $P_{x,y}: \mathcal{B} \to \mathbb{C}$ defined by $P_{x,y}(E) := (P(E)x, y)_H$ is a regular complex Borel measure on $K$, with total variation $\|P_{x,y}\|_1 \le \|x\|_H \|y\|_H$.
We will use these axioms repeatedly. Two consequences are recorded for later use:
*Disjoint range orthogonality.* If $E \cap F = \varnothing$, then $P(E)P(F) = P(E \cap F) = P(\varnothing) = 0$ by (R3) and (R2). Hence the ranges of $P(E)$ and $P(F)$ are mutually orthogonal subspaces of $H$.
*Self-adjointness.* Each $P(E)$ is an orthogonal projection (R1), hence Hermitian: $P(E)^* = P(E)$.
[/step]
[step:Define the map $\Phi$ on simple functions]
A *simple function* $s: K \to \mathbb{C}$ has the form
\begin{align*}
s = \sum_{i=1}^m \alpha_i \mathbb{1}_{E_i}, \qquad \alpha_i \in \mathbb{C},
\end{align*}
where $\{E_i\}_{i=1}^m$ is a Borel partition of $K$ (i.e.\ $E_i \in \mathcal{B}$, the $E_i$ are pairwise disjoint, and $K = \bigsqcup_i E_i$). Denote the space of simple Borel functions on $K$ by $\mathcal{S}(K)$.
Define
\begin{align*}
\Phi: \mathcal{S}(K) &\to \mathcal{L}(H) \\
s = \sum_{i=1}^m \alpha_i \mathbb{1}_{E_i} &\mapsto \sum_{i=1}^m \alpha_i P(E_i).
\end{align*}
*Well-definedness.* Suppose $s = \sum_{i=1}^m \alpha_i \mathbb{1}_{E_i} = \sum_{j=1}^n \beta_j \mathbb{1}_{F_j}$ are two representations of the same simple function with respect to two Borel partitions $\{E_i\}$ and $\{F_j\}$. Form the common refinement $\{E_i \cap F_j\}_{i,j}$, which is again a Borel partition (after dropping empty intersections). On $E_i \cap F_j$, $s$ takes the value $\alpha_i = \beta_j$ whenever $E_i \cap F_j \ne \varnothing$. By (R4) applied iteratively (a finite disjoint union),
\begin{align*}
P(E_i) = \sum_{j=1}^n P(E_i \cap F_j), \qquad P(F_j) = \sum_{i=1}^m P(E_i \cap F_j).
\end{align*}
Hence
\begin{align*}
\sum_i \alpha_i P(E_i) = \sum_{i,j} \alpha_i P(E_i \cap F_j) = \sum_{i,j} \beta_j P(E_i \cap F_j) = \sum_j \beta_j P(F_j),
\end{align*}
where in the middle step we used $\alpha_i P(E_i \cap F_j) = \beta_j P(E_i \cap F_j)$ — true because $\alpha_i = \beta_j$ when $E_i \cap F_j \ne \varnothing$, and both sides vanish when $E_i \cap F_j = \varnothing$ (since $P(\varnothing) = 0$).
Hence $\Phi(s)$ depends only on $s$, not on the chosen representation.
[/step]
[step:Verify that $\Phi$ is a $*$-homomorphism on simple functions]
Linearity is direct: if $s = \sum_i \alpha_i \mathbb{1}_{E_i}$ and $t = \sum_i \gamma_i \mathbb{1}_{E_i}$ are written over a common refinement $\{E_i\}$, then $\Phi(s + \lambda t) = \sum_i (\alpha_i + \lambda \gamma_i) P(E_i) = \Phi(s) + \lambda \Phi(t)$.
*Adjoint preservation.* For $s = \sum_i \alpha_i \mathbb{1}_{E_i}$, the conjugate is $\bar{s} = \sum_i \bar{\alpha}_i \mathbb{1}_{E_i}$, and
\begin{align*}
\Phi(s)^* = \left(\sum_i \alpha_i P(E_i)\right)^* = \sum_i \bar{\alpha}_i P(E_i)^* = \sum_i \bar{\alpha}_i P(E_i) = \Phi(\bar{s}),
\end{align*}
using continuity of the involution and the self-adjointness of each $P(E_i)$ from Step 1.
*Multiplicativity.* For $s = \sum_i \alpha_i \mathbb{1}_{E_i}$ and $t = \sum_j \beta_j \mathbb{1}_{F_j}$, write the product
\begin{align*}
st = \sum_{i, j} \alpha_i \beta_j \mathbb{1}_{E_i \cap F_j}.
\end{align*}
The sets $\{E_i \cap F_j\}_{i,j}$ form a Borel partition (after refinement). Hence
\begin{align*}
\Phi(st) = \sum_{i,j} \alpha_i \beta_j P(E_i \cap F_j) \overset{(R3)}{=} \sum_{i,j} \alpha_i \beta_j P(E_i) P(F_j) = \left(\sum_i \alpha_i P(E_i)\right)\left(\sum_j \beta_j P(F_j)\right) = \Phi(s) \Phi(t).
\end{align*}
*Unitality.* The constant function $1 = \mathbb{1}_K$ is simple, and $\Phi(1) = P(K) = I$ by (R2).
Hence $\Phi: \mathcal{S}(K) \to \mathcal{L}(H)$ is a unital $*$-homomorphism.
[/step]
[step:Establish the integration formulae (i) and (ii) on simple functions]
Fix $x, y \in H$ and a simple function $s = \sum_i \alpha_i \mathbb{1}_{E_i}$.
*Formula (i) on simples.* By linearity of the inner product,
\begin{align*}
(\Phi(s)x, y)_H = \sum_i \alpha_i (P(E_i)x, y)_H = \sum_i \alpha_i P_{x,y}(E_i) = \int_K s \, dP_{x,y},
\end{align*}
where the last equality is the definition of the integral of a simple function against the complex measure $P_{x,y}$ from axiom (R5).
*Formula (ii) on simples.* Compute $\|\Phi(s)x\|_H^2 = (\Phi(s)x, \Phi(s)x)_H$. Using $\Phi(s)^* \Phi(s) = \Phi(\bar{s} s) = \Phi(|s|^2)$ from Step 3,
\begin{align*}
\|\Phi(s)x\|_H^2 = (\Phi(s)^* \Phi(s) x, x)_H = (\Phi(|s|^2) x, x)_H = \int_K |s|^2 \, dP_{x,x},
\end{align*}
applying formula (i) with $y = x$ and $|s|^2 = \sum_i |\alpha_i|^2 \mathbb{1}_{E_i}$ (a simple function).
[/step]
[step:Prove $\Phi$ is isometric on simple functions]
We show $\|\Phi(s)\|_{\mathcal{L}(H)} = \|s\|_{L^\infty(P)}$ for every simple $s$.
*Upper bound $\|\Phi(s)\| \le \|s\|_{L^\infty(P)}$.* The norm on $L^\infty(P)$ is the *essential supremum with respect to $P$*: $\|s\|_{L^\infty(P)} = \inf\{c \ge 0 : P(\{|s| > c\}) = 0\}$. For a simple function $s = \sum_i \alpha_i \mathbb{1}_{E_i}$, this is $\max\{|\alpha_i| : P(E_i) \ne 0\}$. Set $M := \|s\|_{L^\infty(P)}$. Then $|s|^2 \le M^2$ $P$-a.e. on $K$. By Step 4 formula (ii),
\begin{align*}
\|\Phi(s)x\|_H^2 = \int_K |s|^2 \, dP_{x,x} \le M^2 \int_K dP_{x,x} = M^2 P_{x,x}(K) = M^2 (Ix, x)_H = M^2 \|x\|_H^2,
\end{align*}
using $P(K) = I$ from (R2). Hence $\|\Phi(s)\| \le M = \|s\|_{L^\infty(P)}$.
*Lower bound $\|\Phi(s)\| \ge \|s\|_{L^\infty(P)}$.* Pick an index $j$ with $|\alpha_j| = M$ and $P(E_j) \ne 0$. The range $P(E_j)(H)$ is a non-zero closed subspace, so we may choose $x \in P(E_j)(H)$ with $\|x\|_H = 1$. Then $P(E_j)x = x$.
For $i \ne j$, $E_i \cap E_j = \varnothing$, so by Step 1's disjoint-range observation, $P(E_i)P(E_j) = 0$, and hence
\begin{align*}
P(E_i) x = P(E_i) P(E_j) x = 0 \cdot x = 0.
\end{align*}
Therefore
\begin{align*}
\Phi(s) x = \sum_i \alpha_i P(E_i) x = \alpha_j x,
\end{align*}
giving $\|\Phi(s) x\|_H = |\alpha_j| \|x\|_H = M$, so $\|\Phi(s)\| \ge M$.
Combining the two bounds, $\|\Phi(s)\| = \|s\|_{L^\infty(P)}$.
[guided]
We split the isometry into an upper bound and a lower bound, each handled by a different mechanism.
*The upper bound.* The natural way to bound an operator norm is to bound $\|\Phi(s)x\|_H$ by some constant times $\|x\|_H$. Formula (ii) of Step 4 hands us this: $\|\Phi(s)x\|_H^2 = \int_K |s|^2 \, dP_{x,x}$. The integrand is bounded by $M^2 := \|s\|_{L^\infty(P)}^2$ $P$-a.e., and $|s|^2 - M^2 \le 0$ on the set $\{|s| \le M\}$ which has full $P$-measure. Crucially, $P_{x,x}$ is a *positive* measure (because $P_{x,x}(E) = (P(E)x, x)_H = \|P(E)x\|^2 \ge 0$), and "full $P$-measure" implies "full $P_{x,x}$-measure" for every $x$. So we can integrate the inequality $|s|^2 \le M^2$ against $P_{x,x}$ and conclude $\int |s|^2 \, dP_{x,x} \le M^2 P_{x,x}(K) = M^2 \|x\|_H^2$ — using $P(K) = I$ for the total mass.
*The lower bound.* For the reverse inequality we need to *exhibit* a vector $x$ on which $\Phi(s)$ acts with norm close to $M$. The clue is that $\Phi(s) = \sum_i \alpha_i P(E_i)$ is a sum of orthogonal projections weighted by scalars. If we choose $x$ in the range of *exactly one* projection $P(E_j)$ — the one with the largest weight $|\alpha_j| = M$ — then the other projections kill $x$, and $\Phi(s)$ scales $x$ by $\alpha_j$. The disjointness $E_i \cap E_j = \varnothing$ for $i \ne j$ is precisely what produces this clean decomposition: $P(E_i) P(E_j) = P(\varnothing) = 0$. So $P(E_i)x = P(E_i)P(E_j)x = 0$ for $i \ne j$, and $\Phi(s)x = \alpha_j x$. Why does such an $x$ exist? Because $|\alpha_j| = M$ being attained on a set with $P(E_j) \ne 0$ means $P(E_j)(H)$ is a non-zero closed subspace, hence contains a unit vector.
The upper and lower bounds match, so $\|\Phi(s)\| = M = \|s\|_{L^\infty(P)}$. The isometry is what allows us to extend $\Phi$ from simples to all of $L^\infty(P)$ by continuity in the next step.
[/guided]
[/step]
[step:Extend $\Phi$ from simples to $L^\infty(P)$ by density and continuity]
The space $\mathcal{S}(K)$ of simple Borel functions is dense in $L^\infty(P)$ in the $L^\infty(P)$-norm. (This is a standard fact: any bounded Borel function $f$ is a uniform limit of simple Borel functions, by partitioning the range $\{|f| \le \|f\|_\infty\}$ into small Borel pieces $E_i^{(n)} := f^{-1}(B_i^{(n)})$ for shrinking partitions $\{B_i^{(n)}\}$ of the range and forming $s_n := \sum_i c_i^{(n)} \mathbb{1}_{E_i^{(n)}}$ with $c_i^{(n)}$ a value sampled from $B_i^{(n)}$; then $\|f - s_n\|_\infty \le \operatorname{diam}(B_i^{(n)}) \to 0$.)
Since $\Phi: \mathcal{S}(K) \to \mathcal{L}(H)$ is linear and isometric (Step 5), and $\mathcal{L}(H)$ is complete, $\Phi$ extends uniquely to a bounded linear isometric map
\begin{align*}
\Phi: L^\infty(P) \to \mathcal{L}(H).
\end{align*}
Call this extension $\Phi$ as well.
*Algebraic properties persist.* The unital $*$-homomorphism identities $\Phi(1) = I$, $\Phi(\bar{f}) = \Phi(f)^*$, $\Phi(fg) = \Phi(f) \Phi(g)$ hold on $\mathcal{S}(K)$ (Step 3); by continuity of $\Phi$, the involution $A \mapsto A^*$ on $\mathcal{L}(H)$, and operator multiplication (jointly continuous on bounded sets), they extend to all of $L^\infty(P)$.
*Property (i) on $L^\infty(P)$.* Fix $x, y \in H$. The map $f \mapsto (\Phi(f)x, y)_H$ is a bounded linear functional on $L^\infty(P)$ (composition of $\Phi$ with the bounded inner-product evaluation). The map $f \mapsto \int_K f \, dP_{x,y}$ is also a bounded linear functional on $L^\infty(P)$, by axiom (R5) which gives $|\int_K f \, dP_{x,y}| \le \|f\|_\infty \|P_{x,y}\|_1 \le \|f\|_\infty \|x\|_H \|y\|_H$. The two functionals agree on simples by Step 4. Two bounded linear functionals that agree on a dense subspace agree everywhere. Hence formula (i) holds on $L^\infty(P)$.
*Property (ii) on $L^\infty(P)$.* For any $f \in L^\infty(P)$, $|f|^2 \in L^\infty(P)$, and applying property (i) to $|f|^2$ with $y = x$:
\begin{align*}
\|\Phi(f)x\|_H^2 = (\Phi(f)^*\Phi(f)x, x)_H = (\Phi(\bar{f} f)x, x)_H = (\Phi(|f|^2)x, x)_H = \int_K |f|^2 \, dP_{x,x}.
\end{align*}
Uniqueness of $\Phi$ as an isometric, unital $*$-homomorphism $L^\infty(P) \to \mathcal{L}(H)$ satisfying (i): the value of $\Phi$ on simples is determined by (i) (specialised to characteristic functions: $(\Phi(\mathbb{1}_E)x, y)_H = P_{x,y}(E) = (P(E)x, y)_H$, so $\Phi(\mathbb{1}_E) = P(E)$, and linearity extends this to all simples), and continuity then determines $\Phi$ on $L^\infty(P)$ by density.
[/step]
[step:Establish property (iii) — the commutant characterisation]
We show that for $S \in \mathcal{L}(H)$,
\begin{align*}
S \Phi(f) = \Phi(f) S \text{ for all } f \in L^\infty(P) \iff S P(E) = P(E) S \text{ for all } E \in \mathcal{B}.
\end{align*}
*$(\Rightarrow)$.* Apply the assumption to the characteristic function $f = \mathbb{1}_E \in L^\infty(P)$. By the construction of $\Phi$ on simples, $\Phi(\mathbb{1}_E) = P(E)$ (Step 2 applied to the partition $\{E, K \setminus E\}$ with $\alpha_1 = 1, \alpha_2 = 0$). Hence $SP(E) = P(E)S$ for every $E \in \mathcal{B}$.
*$(\Leftarrow)$.* Assume $S$ commutes with every $P(E)$. Then for any simple $s = \sum_i \alpha_i \mathbb{1}_{E_i}$,
\begin{align*}
S \Phi(s) = S \sum_i \alpha_i P(E_i) = \sum_i \alpha_i S P(E_i) = \sum_i \alpha_i P(E_i) S = \Phi(s) S.
\end{align*}
The set $\{T \in \mathcal{L}(H) : ST = TS\}$ — the *commutant* of $S$ — is closed in the operator norm, since the maps $T \mapsto ST$ and $T \mapsto TS$ are continuous linear (operator multiplication on the left and right by a fixed $S$ has operator norm $\|S\|$). The commutant therefore contains the closure of $\Phi(\mathcal{S}(K))$ in $\mathcal{L}(H)$, which by isometry of $\Phi$ (Step 5) equals $\Phi(L^\infty(P))$. Hence $S$ commutes with $\Phi(f)$ for every $f \in L^\infty(P)$.
This completes the proof of (iii) and of the theorem.
[/step]
