the formula does not depend on the chosen orientation of $\tau$. We verify the trace property: $$\begin{aligned}
&\sum_{\tau\in \mathcal{T}_p}\langle ST\tau,\tau\rangle_\mu=\sum_{\tau\in \mathcal{T}_p}\langle T\tau,S^*\tau\rangle_\mu\\
=&\int_{{\mathcal{SC}_*}} \sum_{\tau\in \mathcal{T}_p} \langle T_K\tau([K,x]),S_K^*\tau([K,x])\rangle d\mu([K,x])\\
=&\int_{{\mathcal{SC}_*}}\sum_{\sigma\in K(p)}\sum_{\tau\in \mathcal{T}_p}\langle T_K\tau([K,x]),\sigma\rangle\langle\sigma,S_K^*\tau([K,x])\rangle d\mu([K,x])\\
=&\int_{{\mathcal{SC}_*}}\sum_{y\in V(K)}\underbrace{\sum_{\substack{\sigma\in (K,x)_{{\mathbb N}}(p)\\ y\in\sigma}}\sum_{\tau\in \mathcal{T}_p}\frac{\langle T_K\tau([K,x]),\sigma\rangle\langle S_K\sigma,\tau([K,x])\rangle}{p+1}}_{f([K,x,y])}d\mu([K,x]) \end{aligned}$$ Now we use unimodularity (Definition \[definition unimodular\]): $$\begin{aligned}
=&\int_{{\mathcal{SC}_*}}\sum_{y\in V(K)} f([K,y,x])d\mu([K,x])\\
=&\int_{{\mathcal{SC}_*}}\sum_{y\in V(K)}\sum_{\substack{\sigma\in (K,y)_{{\mathbb N}}(p)\\ x\in\sigma}}\sum_{\tau\in \mathcal{T}_p}\frac{\langle T_K\tau([K,y]),\sigma\rangle\langle S_K\sigma,\tau([K,y])\rangle}{p+1}d\mu([K,x]) \\
=&\int_{{\mathcal{SC}_*}}\sum_{y\in V(K)}\sum_{\substack{\sigma\in (K,y)_{{\mathbb N}}(p)\\ x\in\sigma}}\sum_{\tau\in \mathcal{T}_p}\frac{\langle S_K\sigma,\tau([K,y])\rangle\langle \tau([K,y]),T_K^*\sigma\rangle}{p+1}d\mu([K,x]) \\
=&\int_{{\mathcal{SC}_*}}\sum_{\substack{\sigma\in (K,x)_{{\mathbb N}}(p)\\ 0\in\sigma}}\langle S_K\sigma,T_K^*\sigma\rangle d\mu([K,x]) \\
=&\int_{{\mathcal{SC}_*}}\sum_{\tau\in\mathcal{T}}\langle S_K\tau([K,x]),T_K^*\tau([K,x])\rangle d\mu([K,x])=\sum_{\tau\in\mathcal{T}_p}\langle TS\tau,\tau\rangle_\mu\end{aligned}$$

The trace ${\operatorname{tr}}_p^\mu$ is normal, faithful and semi-finite. If the expected $p$-degree at the root $\mathbb{E}_\mu(\deg_p)$ is finite, then ${\operatorname{tr}}_p^\mu$ is a finite trace.

Faithfulness follows from Equation (\[equation-vonneumann algebra on hilbert field\]), Definition \[def-vonneumannalgebra\]. We check semi-finiteness. Let $A\in{\operatorname{Alg}}(\mu)$ be positive. We define the operators $$A_{n,[K,x]}\coloneqq\frac{A_{[K,x]}}{\max\{\deg_p(x)-n,1\}}.$$ We will show that $A_n$ weakly converges to $A$ but first, we verify that ${\operatorname{tr}}_p^\mu(A_n)<\infty$. $$\begin{aligned}
{\operatorname{tr}}_p^\mu(A_n)&=\int_{{\mathcal{SC}_*}}\sum_{\tau\in\mathcal{T}_p}\frac{\langle A \tau,\tau\rangle}{(p+1)\max\{\deg_p(x)-n,1\}}d\mu([K,x])\\
&\le\int_{{\mathcal{SC}_*}}\frac{\|A\|\deg_p(x)}{(p+1)\max\{\deg_p(x)-n,1\}}d\mu([K,x])\\
&\le \frac{\|A\|(n+1)}{p+1}\end{aligned}$$ We check weak convergence. Let $\sigma,\theta\in C_p^{(2)}({\mathcal{SC}_*},\mu)$: $$\begin{aligned}
\langle A_n \sigma,\theta\rangle_\mu=&\int_{{\mathcal{SC}_*}}\langle A_n\sigma,\theta\rangle d\mu=\int_{{\mathcal{SC}_*}}\frac{\langle A\sigma,\theta\rangle}{\max\{\deg_p(x)-n,1\}}d\mu\\
=&\int_{\substack{[K,x]\in{\mathcal{SC}_*}\\ \deg_p(x)\le n}}\langle A\sigma,\theta\rangle d\mu + \int_{\substack{[K,x]\in{\mathcal{SC}_*}\\ \deg_p(x)> n}}\frac{\langle A\sigma,\theta\rangle}{\deg_p(x)-n}d\mu \\
\xlongrightarrow{n\to\infty}&\int_{{\mathcal{SC}_*}}\langle A\sigma,\theta\rangle d\mu=\langle A\sigma,\theta\rangle_\mu\end{aligned}$$ In the last step we used the fact that the elements of ${\mathcal{SC}_*}$ are locally finite, hence the right summand 