It is well-known that a POVM $\boldsymbol\mu\equiv (\mu_a)_{a\in\Sigma}$ describes outcome probabilities, but not post-measurement outcomes, which in many scenarios exist and are of interest. To actually describe post-measurement outcomes associated to a given measurement, the most natural approach seems to me to be use a channel. As far as I can tell, the most general possible way to write a channel $\Phi$ that corresponds to a measurement $\boldsymbol\mu$ is via the formalism of quantum instruments. That is, to consider a channel of the form $$\Phi(\rho) = \sum_{a\in\Sigma} \Phi_a(\rho)\otimes \mathbb{P}_a, \qquad \mathbb{P}_a\equiv |a\rangle\!\langle a|,$$ with $\Phi_a$ CP trace-non-increasing maps. For $\Phi$ to be compatible with the POVM we need $\operatorname{tr}(\Phi_a(\rho))=\langle \mu_a,\rho\rangle$ for all $\rho$, which should be equivalent to $\Phi_a^\dagger(I)=\mu_a$ with $\Phi_a^\dagger$ adjoint channel of $\Phi_a$.
As discussed in Given a state $\rho$ and operator $0\le \Lambda\le I$, what does $\sqrt\Lambda \rho \sqrt\Lambda$ represent?, and What is the most general way to describe post-measurement states?, it is possible to have channels corresponding to a given POVM that preserve different amounts of coherence. Entanglement-breaking channels would be an extreme case where al of the coherence in the original states is destroyed. At the opposite side of the spectrum, it seems we have "gentle measurements", that is, the channels of the form $$\Phi(\rho) = \sum_{a\in\Sigma} V_a \sqrt{\mu_a} \rho \sqrt{\mu_a}V_a^\dagger \otimes \mathbb{P}_a.$$ for any set of isometries $V_a$.
Is there a way to prove that this choice is indeed the one that optimally preserves the coherence in the input states (assuming this to be the case)? In fact, this question probably boils down to what is a good way to quantify what precisely "preserves the coherence" means here. Given the gentle operator lemma, my guess is that this should mean the channel such that the post-measurement outcomes have minimal (averaged?) trace-distance, but I'm not sure how exactly to do this.