Consistent maximally mixed local densities

Each ${\rho }_{\mathrm{\Lambda }}$ is positive because it is a positive scalar multiple of the identity. Its trace is $q^{-\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}{\mathrm{Tr}}_{\mathrm{\Lambda }}(I_{\mathrm{\Lambda }})=1$ by [F1], including $\mathrm{\Lambda }=\mathrm{\varnothing }$ where $q^0=1$.

Put $\mathrm{\Delta }={\mathrm{\Lambda }}^{\mathrm{\prime }}\setminus \mathrm{\Lambda }$ and $r=\mathrm{\mid }\mathrm{\Delta }\mathrm{\mid }$. For every $A\in {\mathcal{A}}_{\mathrm{\Lambda }}$, the defining partial-trace pairing gives ${\mathrm{Tr}}_{\mathrm{\Lambda }}\text{\hspace{0.17em}⁣}(A{\mathrm{Tr}}_{\mathrm{\Delta }}(I_{\mathrm{\Lambda }}\otimes I_{\mathrm{\Delta }}))={\mathrm{Tr}}_{{\mathrm{\Lambda }}^{\mathrm{\prime }}}((A\otimes I_{\mathrm{\Delta }})(I_{\mathrm{\Lambda }}\otimes I_{\mathrm{\Delta }}))=q^r{\mathrm{Tr}}_{\mathrm{\Lambda }}(A),$ where the last equality follows by summing the diagonal matrix coefficients over a configuration basis of ${\mathcal{H}}_{\mathrm{\Delta }}$, which has $q^r$ elements. The trace pairing is nondegenerate: if $\mathrm{Tr}(AT)=0$ for every matrix $A$, taking $A=E_{ji}$ makes each entry $T_{ij}$ zero. Hence ${\mathrm{Tr}}_{\mathrm{\Delta }}(I_{\mathrm{\Lambda }}\otimes I_{\mathrm{\Delta }})=q^r{I}_{\mathrm{\Lambda }}$. Since $\mathrm{\mid }{\mathrm{\Lambda }}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }+r$, ${\mathrm{Tr}}_{\mathrm{\Delta }}({\rho }_{{\mathrm{\Lambda }}^{\mathrm{\prime }}})=q^{-\mathrm{\mid }{\mathrm{\Lambda }}^{\mathrm{\prime }}\mathrm{\mid }}{q}^r{I}_{\mathrm{\Lambda }}=q^{-\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}{I}_{\mathrm{\Lambda }}={\rho }_{\mathrm{\Lambda }}\mathrm{.}$ Thus the family is consistent for every inclusion, including equality of regions ($r=0$).

By [F3], this family extends to a unique algebraic state $\omega$ on $\mathcal{A}$, and its local restriction is $\omega (A)={\mathrm{Tr}}_{\mathrm{\Lambda }}({\rho }_{\mathrm{\Lambda }}A)=q^{-\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}{\mathrm{Tr}}_{\mathrm{\Lambda }}(A)$ for $A\in {\mathcal{A}}_{\mathrm{\Lambda }}$.

If $A\in {\mathcal{A}}_{\mathrm{\Lambda }}$ and $B\in {\mathcal{A}}_{\mathrm{\Delta }}$ for disjoint finite regions, choose orthonormal configuration bases $(e_i)$ and $(f_j)$ for their local Hilbert spaces. The tensor basis gives ${\mathrm{Tr}}_{\mathrm{\Lambda }\cup \mathrm{\Delta }}(A\otimes B)={\sum }_{i,j}⟨A{e}_i,e_i⟩⟨B{f}_j,f_j⟩={\mathrm{Tr}}_{\mathrm{\Lambda }}(A){\mathrm{Tr}}_{\mathrm{\Delta }}(B)\mathrm{.}$ Substitution into step 2.1 gives $\omega (AB)=\omega (A)\omega (B)$ on these disjoint local factors; this product formula follows from the displayed density family and is not an assumption about arbitrary states.