the unordered $p$-simplices.

Since we will need it later, we recall the spectral theorem for self-adjoint operators.

\[spectral theorem\] For every self-adjoint operator $T$ on a Hilbert space $\mathcal{H}$ there exist a unique projection valued measure $E_T\colon\mathrm{Bor}({{\mathbb R}})\to \mathcal{P}(\mathcal{H})$ such that for all bounded Borel functions $f$ on ${{\mathbb R}}$ $$f(T)=\int_{{\mathbb R}}f(\lambda)dE_T(\lambda).$$

The following proposition is from Dykema [@dykema Proposition 4.2], it also applies to unbounded operators:

\[prop-dykema\] Let $T=\int_{X}^\oplus T(\xi)d\eta(\xi)$ be a self-adjoint, decomposable operator on the direct integral Hilbert space $\mathcal{H}=\int_X^\oplus \mathcal{H}(\xi)d\eta(\xi)$. For every Borel subset $B\subset {{\mathbb R}}$ we have $$E_T(B)=\int_{X}^\oplus E_{T(\xi)}(B)d\eta(\xi).$$ Moreover, for every Borel measurable function $f\colon{{\mathbb R}}\to{{\mathbb R}}$, we have $$f(T)=\int_{X}^\oplus f(T(\xi))d\eta(\xi).$$

By applying the spectral theorem, we can define the *spectral measure* of a self-adjoint operator $T$ on $C_p^{(2)}({\mathcal{SC}_*},\mu)$ as $$\begin{aligned}
&\nu_T\colon \mathrm{Bor}({{\mathbb R}})\to [0,\infty];\\
&\nu_T(B)={\operatorname{tr}}_p^\mu(E_T(B))=\sum_{\tau\in \mathcal{T}_p}\langle E_T(B)\tau,\tau\rangle_\mu/(p+1).\end{aligned}$$ Note that $\nu_T$ has compact support if $T$ is a bounded operator and further, if $T$ is decomposable, this definition is (by Proposition \[prop-dykema\]) equivalent to $$\begin{aligned}
\nu_T(B)&=\sum_{\tau\in \mathcal{T}_p}\int_{{\mathcal{SC}_*}}\langle E_{T_{[K,x]}}(B)\tau,\tau\rangle d\mu/(p+1)\\
&=\int_{{\mathcal{SC}_*}}\sum_{\substack{\sigma\in K(p)\\ x\in\sigma}}\langle E_{T_{[K,x]}}(B)\sigma,\sigma\rangle/(p+1) d\mu \\
&=:\int_{{\mathcal{SC}_*}} \nu_{T_{[K,x]}}(B)d\mu\end{aligned}$$ Observe that by Theorem \[spectral theorem\] we also have that: $$\begin{aligned}
{\operatorname{tr}}_p^\mu(f(T))&=\sum_{\tau\in\mathcal{T}_p}\langle f(T)\tau,\tau\rangle_\mu/(p+1)\\
&=\sum_{\tau\in\mathcal{T}_p}\int_{{\mathbb R}}f(\lambda)d\langle E_T(\lambda) \tau,\tau\rangle/(p+1)\\
&=\int_{{\mathbb R}}f(\lambda)d\nu_T(\lambda).\end{aligned}$$

We return to the Laplace operator. The 