[proofplan]
Decompose $H$ into mutually orthogonal cyclic reducing subspaces for $T$, then apply the cyclic spectral theorem on each summand. On each cyclic model, the spectral projections are multiplication by indicator functions; transporting these projections back to the original summand and taking their orthogonal direct sum gives the global projection-valued measure. The representation of $T$, the [Borel functional calculus](/theorems/2696), and the norm formula are then checked on the direct-sum multiplication model. Finally, uniqueness follows by comparing scalar matrix-coefficient measures: the moments against all continuous functions agree, so the [Riesz representation theorem](/theorems/218) forces equality of the measures and hence equality of the projections.
[/proofplan]
[step:Decompose $H$ into cyclic reducing summands]
By the [C^*(T)-Cyclic Reducing Decomposition For Normal Operators]([citetheorem:8411]), there are an index set $I$ and a mutually orthogonal family $(H_i)_{i\in I}$ of nonzero closed reducing subspaces for $T$ such that
\begin{align*}
H=\bigoplus_{i\in I}H_i.
\end{align*}
For each $i\in I$, define
\begin{align*}
T_i:=T|_{H_i}\in\mathcal L(H_i).
\end{align*}
Then each $T_i$ is normal, because $H_i$ reduces $T$. For each $i\in I$, let $C^*(T_i)\subset\mathcal L(H_i)$ denote the unital $C^*$-subalgebra generated by $T_i$ and $I_{H_i}$. The cited decomposition theorem also asserts the existence of a vector $x_i\in H_i$ such that the closed linear span of $\{A x_i:A\in C^*(T_i)\}$ is $H_i$, so $T_i$ is $C^*(T_i)$-cyclic.
For each $i\in I$, apply the [Cyclic Spectral Theorem]([citetheorem:8410]) to the normal cyclic operator $T_i$. Thus there are a compact set $K_i\subset\mathbb C$, a finite positive Borel measure $\mu_i$ on $K_i$, and a unitary map
\begin{align*}
U_i:H_i\to L^2(K_i,\mu_i)
\end{align*}
such that
\begin{align*}
U_iT_iU_i^{-1}=M_{z,i},
\end{align*}
where
\begin{align*}
M_{z,i}:L^2(K_i,\mu_i)\to L^2(K_i,\mu_i),\qquad g\mapsto z g
\end{align*}
is multiplication by the coordinate function. We take $K_i=\sigma(T_i)$, as allowed by the cyclic theorem.
Since $H_i$ reduces $T$, the spectrum of $T_i$ is contained in $\sigma(T)$. Hence every Borel subset $B\subset\sigma(T)$ has a Borel intersection $B\cap K_i\subset K_i$.
[/step]
[step:Build the spectral projections on each cyclic summand]
For each $i\in I$ and each $B\in\mathcal B(\sigma(T))$, define an operator
\begin{align*}
\widetilde E_i(B):L^2(K_i,\mu_i)\to L^2(K_i,\mu_i),\qquad g\mapsto \mathbb{1}_{B\cap K_i}g
\end{align*}
Then $\widetilde E_i(B)$ is the [orthogonal projection](/theorems/437) onto the closed subspace of functions supported in $B\cap K_i$, and
\begin{align*}
\|\widetilde E_i(B)\|_{\mathcal L(L^2(K_i,\mu_i))}\le 1.
\end{align*}
Transport this projection to $H_i$ by defining
\begin{align*}
E_i(B):=U_i^{-1}\widetilde E_i(B)U_i\in\mathcal L(H_i).
\end{align*}
Because unitary conjugation preserves products, adjoints, and strong limits, the map
\begin{align*}
E_i:\mathcal B(\sigma(T))\to\mathcal L(H_i)
\end{align*}
is a projection-valued measure.
[guided]
Fix an index $i\in I$. The cyclic spectral theorem has converted the operator $T_i$ into the concrete multiplication operator $M_{z,i}$ on $L^2(K_i,\mu_i)$. In this model, the natural spectral projection associated to a Borel set $B\subset\sigma(T)$ is multiplication by the indicator of the part of $B$ visible inside the model space $K_i$.
Thus we define the bounded operator
\begin{align*}
\widetilde E_i(B): L^2(K_i,\mu_i) \to L^2(K_i,\mu_i), \qquad g \mapsto \mathbb{1}_{B\cap K_i}g.
\end{align*}
This is well-defined because $B\cap K_i$ is a Borel subset of $K_i$. It is bounded because
\begin{align*}
\|\mathbb{1}_{B\cap K_i}g\|_{L^2(K_i,\mu_i)}\le \|g\|_{L^2(K_i,\mu_i)}.
\end{align*}
It is a projection because $\mathbb{1}_{B\cap K_i}^2=\mathbb{1}_{B\cap K_i}$ pointwise $\mu_i$-a.e., and it is self-adjoint because $\mathbb{1}_{B\cap K_i}$ is real-valued. Hence $\widetilde E_i(B)$ is an orthogonal projection.
If $(B_n)_{n=1}^{\infty}$ is a pairwise disjoint sequence of Borel subsets of $\sigma(T)$ and $B=\bigcup_{n=1}^{\infty}B_n$, then for every $g\in L^2(K_i,\mu_i)$,
\begin{align*}
\mathbb{1}_{B\cap K_i}g=\lim_{N\to\infty}\sum_{n=1}^{N}\mathbb{1}_{B_n\cap K_i}g
\end{align*}
in $L^2(K_i,\mu_i)$. Indeed, the squared norm of the difference is
\begin{align*}
\int_{K_i}\left|\mathbb{1}_{B\cap K_i}(z)-\sum_{n=1}^{N}\mathbb{1}_{B_n\cap K_i}(z)\right|^2 |g(z)|^2\,d\mu_i(z),
\end{align*}
and this tends to $0$ by the [dominated convergence theorem](/theorems/4), with dominator $|g|^2\in L^1(K_i,\mu_i)$. Therefore $\widetilde E_i$ is countably additive in the strong operator topology.
Finally, define
\begin{align*}
E_i(B):=U_i^{-1}\widetilde E_i(B)U_i.
\end{align*}
Unitary conjugation preserves orthogonal projections, products, adjoints, and strong convergence. Therefore
\begin{align*}
E_i:\mathcal B(\sigma(T))\to\mathcal L(H_i)
\end{align*}
is a projection-valued measure.
[/guided]
[/step]
[step:Take the orthogonal direct sum of the cyclic projection-valued measures]
For $B\in\mathcal B(\sigma(T))$, define
\begin{align*}
E(B):H\to H,\qquad \bigoplus_{i\in I}x_i\mapsto \bigoplus_{i\in I}E_i(B)x_i
\end{align*}
This operator is well-defined and bounded because each $E_i(B)$ is an orthogonal projection, so
\begin{align*}
\sum_{i\in I}\|E_i(B)x_i\|_{H_i}^2\le \sum_{i\in I}\|x_i\|_{H_i}^2.
\end{align*}
Thus $\|E(B)\|_{\mathcal L(H)}\le 1$.
For each $B$, $E(B)$ is an orthogonal projection because each $E_i(B)$ is an orthogonal projection. Also,
\begin{align*}
E(\sigma(T))=\bigoplus_{i\in I}I_{H_i}=I_H.
\end{align*}
If $(B_n)_{n=1}^{\infty}$ is a pairwise disjoint sequence in $\mathcal B(\sigma(T))$ and $B=\bigcup_{n=1}^{\infty}B_n$, then for $x=\bigoplus_{i\in I}x_i\in H$,
\begin{align*}
E(B)x-\sum_{n=1}^{N}E(B_n)x=\bigoplus_{i\in I}\left(E_i(B)x_i-\sum_{n=1}^{N}E_i(B_n)x_i\right).
\end{align*}
For each fixed $i$, the $H_i$-norm of the $i$th component tends to $0$ by strong countable additivity of $E_i$. Moreover,
\begin{align*}
\left\|E_i(B)x_i-\sum_{n=1}^{N}E_i(B_n)x_i\right\|_{H_i}\le \|x_i\|_{H_i}.
\end{align*}
Since $\sum_{i\in I}\|x_i\|_{H_i}^2<\infty$, dominated convergence for the counting measure on the countable support of $x$ gives
\begin{align*}
\lim_{N\to\infty}\left\|E(B)x-\sum_{n=1}^{N}E(B_n)x\right\|_H=0.
\end{align*}
Therefore
\begin{align*}
E:\mathcal B(\sigma(T))\to\mathcal L(H)
\end{align*}
is a projection-valued measure.
[/step]
[step:Identify $T$ with the spectral integral of the coordinate function]
Let
\begin{align*}
\zeta:\sigma(T)\to\mathbb C,\qquad z\mapsto z
\end{align*}
be the coordinate function. For each $i\in I$, the construction gives
\begin{align*}
T_i=U_i^{-1}M_{z,i}U_i=\int_{\sigma(T)}\zeta\,dE_i.
\end{align*}
Therefore, for $x=\bigoplus_{i\in I}x_i\in H$,
\begin{align*}
\left(\int_{\sigma(T)}\zeta\,dE\right)x=\bigoplus_{i\in I}\left(\int_{\sigma(T)}\zeta\,dE_i\right)x_i=\bigoplus_{i\in I}T_ix_i=Tx.
\end{align*}
Hence
\begin{align*}
T=\int_{\sigma(T)}z\,dE(z).
\end{align*}
[/step]
[step:Define the Borel functional calculus and prove the algebraic rules]
By the [Basic Construction of Operator Integrals]([citetheorem:8407]) applied to the projection-valued measure $E$ on the compact [metric space](/page/Metric%20Space) $\sigma(T)$, the map
\begin{align*}
\Phi:B_b(\sigma(T))\to\mathcal L(H),\qquad f\mapsto \int_{\sigma(T)}f\,dE
\end{align*}
is a unital $*$-homomorphism. Thus, for bounded Borel functions $f,g:\sigma(T)\to\mathbb C$ and scalars $\alpha,\beta\in\mathbb C$,
\begin{align*}
(\alpha f+\beta g)(T)=\alpha f(T)+\beta g(T),
\end{align*}
\begin{align*}
(fg)(T)=f(T)g(T),
\end{align*}
and
\begin{align*}
\overline f(T)=f(T)^*.
\end{align*}
Also $\mathbb{1}_{\sigma(T)}(T)=I_H$, so the homomorphism is unital.
In the direct-sum model, this calculus is multiplication by the restricted function on each summand: for each bounded Borel function $f:\sigma(T)\to\mathbb C$, let
\begin{align*}
M_{f|_{K_i},i}: L^2(K_i,\mu_i) \to L^2(K_i,\mu_i), \qquad g \mapsto (f|_{K_i})g
\end{align*}
denominate multiplication by $f|_{K_i}$. Then
\begin{align*}
U_i f(T)|_{H_i} U_i^{-1} = M_{f|_{K_i},i}.
\end{align*}
This follows first for simple functions from the definition of $E_i$, and then for bounded Borel functions by bounded convergence in the scalar matrix coefficients supplied by the operator-integral construction.
[/step]
[step:Compute the operator norm from the spectral measure class]
Define the spectral null ideal
\begin{align*}
\mathcal N_E:=\{B\in\mathcal B(\sigma(T)):E(B)=0\}.
\end{align*}
For a bounded Borel function $f:\sigma(T)\to\mathbb C$, define
\begin{align*}
M_E(f):=\inf\{a\ge 0:E(\{z\in\sigma(T):|f(z)|>a\})=0\}.
\end{align*}
This is precisely the essential supremum of $|f|$ with respect to the spectral measure class of $E$.
Since $f(T)$ is the orthogonal direct sum of the multiplication operators $M_{f|_{K_i},i}$, we have
\begin{align*}
\|f(T)\|_{\mathcal L(H)}=\sup_{i\in I}\|M_{f|_{K_i},i}\|_{\mathcal L(L^2(K_i,\mu_i))}.
\end{align*}
For each $i$, the norm of a bounded multiplication operator equals the $\mu_i$-essential supremum of the multiplier. Hence
\begin{align*}
\|f(T)\|_{\mathcal L(H)}=\sup_{i\in I}\operatorname{ess\,sup}_{z\in K_i,\mu_i}|f(z)|.
\end{align*}
By construction, $E(B)=0$ iff $E_i(B)=0$ for every $i\in I$, which is equivalent to
\begin{align*}
\mu_i(B\cap K_i)=0
\end{align*}
for every $i\in I$. Therefore the last supremum is exactly $M_E(f)$. Thus
\begin{align*}
\|f(T)\|_{\mathcal L(H)}=\operatorname{ess\,sup}_{z\in\sigma(T)}|f(z)|.
\end{align*}
[/step]
[step:Prove uniqueness by comparing scalar spectral measures]
Let
\begin{align*}
F:\mathcal B(\sigma(T))\to\mathcal L(H)
\end{align*}
be another projection-valued measure such that
\begin{align*}
T=\int_{\sigma(T)}z\,dF(z).
\end{align*}
For $h,k\in H$, define finite complex measures
\begin{align*}
E_{h,k}(B):=(E(B)h,k)_H
\end{align*}
and
\begin{align*}
F_{h,k}(B):=(F(B)h,k)_H
\end{align*}
for $B\in\mathcal B(\sigma(T))$. These are finite complex Borel measures by the [Countable Additivity in Matrix Coefficients]([citetheorem:8404]).
Apply the [Basic Construction of Operator Integrals]([citetheorem:8407]) to both projection-valued measures $E$ and $F$ on the compact metric space $\sigma(T)$. Thus the associated operator-integral maps on $B_b(\sigma(T))$ are unital $*$-homomorphisms.
Let $p:\sigma(T)\to\mathbb C$ be any polynomial in the two coordinate functions $z$ and $\overline z$. Since both operator-integral maps send $z$ to $T$, the $*$-homomorphism property implies that both send $\overline z$ to $T^*$. Therefore
\begin{align*}
\int_{\sigma(T)}p(z,\overline z)\,dE_{h,k}(z)=\int_{\sigma(T)}p(z,\overline z)\,dF_{h,k}(z).
\end{align*}
The complex [Stone-Weierstrass theorem](/theorems/886) applies to the unital self-adjoint algebra of polynomials in $z$ and $\overline z$, which separates points of the [compact space](/page/Compact%20Space) $\sigma(T)$. Hence these polynomials are uniformly dense in $C(\sigma(T))$. Since $E_{h,k}$ and $F_{h,k}$ are finite measures, uniform approximation gives
\begin{align*}
\int_{\sigma(T)}g\,dE_{h,k}=\int_{\sigma(T)}g\,dF_{h,k}
\end{align*}
for every $g\in C(\sigma(T))$.
By the [Riesz representation theorem](/theorems/221) for finite regular complex Borel measures on compact Hausdorff spaces, equality of integrals against all functions in $C(\sigma(T))$ implies
\begin{align*}
E_{h,k}=F_{h,k}.
\end{align*}
Thus, for every $B\in\mathcal B(\sigma(T))$ and every $h,k\in H$,
\begin{align*}
(E(B)h,k)_H=(F(B)h,k)_H.
\end{align*}
Since the [inner product](/page/Inner%20Product) separates vectors in $H$, this implies $E(B)h=F(B)h$ for every $h\in H$. Hence $E(B)=F(B)$ for every Borel set $B\subset\sigma(T)$, so $E=F$.
This proves existence, the functional calculus properties, the norm formula, and uniqueness of the spectral measure.
[/step]
