Step 1: Gelfand Transform

Consider the commutative $C^*$-algebra $\mathcal{A}$ generated by $T$ and $I$. By the Gelfand representation theorem, $\mathcal{A}$ is isometrically $*$-isomorphic to $C(X)$ for some compact Hausdorff space $X$.

In fact, $X$ can be identified with the spectrum $\sigma(T)$ equipped with the relative topology from $\mathbb{C}$.

Step 2: Riesz Representation

For each $x \in H$, define a linear functional $\phi_x : \mathcal{A} \to \mathbb{C}$ by $\phi_x(S) = \langle Sx, x \rangle$.

This functional is positive ($\phi_x(S^*S) \geq 0$) and bounded, so by the Riesz representation theorem, there exists a unique finite Borel measure $\mu_x$ on $\sigma(T)$ such that $\langle Sx, x \rangle = \int_{\sigma(T)} \hat{S}(\lambda) \, d\mu_x(\lambda)$

where $\hat{S}$ is the Gelfand transform of $S$.

Step 3: Spectral Projections

For each Borel set $B \subset \sigma(T)$, define $E(B)$ as the unique operator satisfying $\langle E(B)x, y \rangle = \int_B \, d\mu_{x,y}(\lambda)$

where $\mu_{x,y}(\lambda) = \frac{1}{4}\sum_{k=0}^3 i^k \mu_{x + i^k y}(\lambda)$ (polarization).

Step 4: Verification

We verify that $E$ is a projection-valued measure:

1. $E(\emptyset) = 0$ and $E(\sigma(T)) = I$ by construction

2. $E(B)$ is a projection: $E(B)^2 = E(B) = E(B)^*$ follows from properties of $\mu_x$

3. $\sigma$-additivity: If $B_n$ are disjoint, then $E\left(\bigcup_{n=1}^\infty B_n\right) = \sum_{n=1}^\infty E(B_n)$ in the strong operator topology, using monotone convergence for measures

Step 5: Spectral Integral

For any bounded Borel function $f : \sigma(T) \to \mathbb{C}$, define $f(T) = \int_{\sigma(T)} f(\lambda) \, dE(\lambda)$

For the identity function $id(\lambda) = \lambda$, we have $\int_{\sigma(T)} \lambda \, dE(\lambda) = T$

This completes the construction.