Put $p=\langle\psi|M_1^2|\psi\rangle\geq1-\varepsilon$. Since $M_1$ is a <positive contraction>, the successful <post-measurement state> is the pure state represented by $|\psi'\rangle=M_1|\psi\rangle/\sqrt p$. Its squared <quantum fidelity> with the input is
$$
F(\rho,\rho')^2=|\langle\psi|\psi'\rangle|^2
=\frac{\langle\psi|M_1|\psi\rangle^2}{p}.
$$
The <finite-dimensional spectral theorem> and $0\leq M_1\leq I$ give $M_1^2\leq M_1$, so $\langle M_1\rangle\geq p$. Therefore the <pure-state gentle measurement bound> is
$$
\boxed{F(\rho,\rho')^2\geq p\geq1-\varepsilon}.
$$
The positive-operator assumption matters: an arbitrary unitary measurement operator can have success probability one while rotating the state to an orthogonal vector.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 323 / 3 / iii / Solution

{title2=$0\leq M\leq I$}

A positive contraction is a <positive operator> whose <operator norm> is at most one. Its <spectrum> lies in $[0,1]$, so the <spectral theorem for normal operators> gives $M^2\leq M$. Positive contractions are the effects in a <POVM> and occur in the <pure-state gentle measurement bound>.

Positive contraction

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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 Incoming links: Pure-state gentle measurement bound