Theorem: Residual obstruction with an identity coarse instrument

Record agreement does not imply residual-source agreement, even when $T=\mathrm{id}$ and the coarse instrument has one outcome with map ${\mathcal{I}}_{\ast }=\mathrm{id}$. There is a qubit input for which the recorded residual outputs have trace distance $1/2$.

Proof. Take the computational projectors $P_0,P_1$, aggregate both outcomes to $\ast$, and take $T=\mathrm{id}$. Both record effects are $I$, so the record laws agree on every input.

For $\rho =|+⟩⟨+|$,

Their difference has residual block $X/2$, where $X=|0⟩⟨1|+|1⟩⟨0|$. Its eigenvalues are $1/2,-1/2$. Hence

The residual observable $X$ has expectations $1$ and $0$, respectively.