= Pure-state gentle measurement bound
If $M$ is a <positive contraction> and $p=\langle\psi|M^2|\psi\rangle>0$, the normalized successful <post-measurement state> is $M|\psi\rangle/\sqrt p$. Since $M^2\leq M$,
$$
F\left(|\psi\rangle\langle\psi|,\frac{M|\psi\rangle\langle\psi|M}{p}\right)^2
=\frac{\langle\psi|M|\psi\rangle^2}{p}\geq p.
$$
A likely outcome therefore preserves a pure input well in <quantum fidelity> when its measurement operator is positive. Positivity excludes a hidden unitary rotation that could otherwise change the state even for an outcome of probability one.
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