Pauli projectors define a finite measurement instrument

Each ${\mathcal{I}}_b$ is complex-linear because matrix multiplication distributes over finite sums and scalars: $P_b(aX+cY)P_b=a{P}_{b}X{P}_b+c{P}_{b}Y{P}_b$.

Fix $r\ge 1$, take an orthonormal basis $(e_j{)}_{j<r}$ of the ancilla, and write $X=[X_{ij}]$ for a nonnegative operator on $H\otimes {\mathbb{C}}^r$. Self-adjointness of $X$ by [F6] gives $X_{ij}^{†}=X_{ji}$: for $\xi ,\eta \in H$, $⟨X_{ij}\xi ,\eta ⟩=⟨X(\xi \otimes e_j),\eta \otimes e_i⟩=⟨\xi \otimes e_j,X(\eta \otimes e_i)⟩=⟨\xi ,X_{ji}\eta ⟩$ by the product pairing [F5]. Consequently the blocks $P_b{X}_{ij}{P}_b$ satisfy the same adjoint relation, so ${\mathcal{I}}_b^{(r)}(X)$ is self-adjoint.

For every operator $X$, cyclicity [F7] and idempotence [F2] give $\mathrm{Tr}({\mathcal{I}}_0(X))+\mathrm{Tr}({\mathcal{I}}_1(X))=\mathrm{Tr}(X{P}_0)+\mathrm{Tr}(X{P}_1)=\mathrm{Tr}(X(P_0+P_1))=\mathrm{Tr}(X)$. For a density operator $\rho$, the same calculation gives $\mathrm{Tr}({\mathcal{I}}_b(\rho ))=\mathrm{Tr}(\rho P_b)$, the branch trace.

Write $v={\sum }_{j<r}{v}_j\otimes e_j$ and $w={\sum }_{j<r}{P}_b{v}_j\otimes e_j$. By [F4] and the product pairing [F5], $⟨{\mathcal{I}}_b^{(r)}(X)v,v⟩={\sum }_{i,j<r}⟨P_b{X}_{ij}{P}_b{v}_j,v_i⟩={\sum }_{i,j<r}⟨X_{ij}{P}_b{v}_j,P_b{v}_i⟩=⟨Xw,w⟩\ge 0$. Thus the ampliation preserves nonnegative operators. Since $r$ was arbitrary, ${\mathcal{I}}_b$ is completely positive by [F3].

The two maps are completely positive by step 2.1 and their total output trace equals the input trace by step 1.3. They therefore form a finite instrument by [F8].

By the branch-weight clause of [F8], the outcome weight on $\rho$ is $\mathrm{Tr}({\mathcal{I}}_b(\rho ))$. Step 1.3 identifies this with $\mathrm{Tr}(\rho P_b)$.