Let's say for a specific simple decoherence model you end up with the following density matrix for the "system" (I will use the common separation in "system" and "environment" usually used in decoherence literature):
$$\frac{1}{2} \begin{pmatrix}1&r(t)\\ r^*(t)& 1\end{pmatrix}$$
with $r(t)=e^{-2i\omega t}$ ($\omega$ being some parameter of the interaction hamiltonian standing for the strength of the interaction between system and environment).
Usually, when $r(t)\rightarrow 0$, one says that whatever system state has been considered, it is "fragile" to the decoherence process in consideration. Here although, $|r(t)|^2=1$. So what is the important parameter?
Since an arbitrary density matrix can be written as $$\rho=\begin{pmatrix}1-\rho_{00}&\rho_{01}\\ \rho_{01}^*&1-\rho_{00}\end{pmatrix}$$ and a measurement probability is calculated along $p(\lambda)=\text{Tr}(\rho P)$ with $P$ e.g. being $P=|+\rangle\langle+|$ (in this case), one finds that $p(\lambda)=\frac{1}{2}\left(1+2\text{Re}(\rho_{01})\right)$. Therefore, only the real part of $r(t)$ is important for whether an observer experiences the system to act quantum-mechanically (but only relative to the basis $|+\rangle,|-\rangle$.
Therefore one can say that for the considered model, the considered state of the system (before the interaction) is also fragile to this concrete decoherence process.
Is it correct that only the real part of $r(t)$ (see above) matters for whether "the system loses non classical properties" (in regard to a local observer)?
More generally: When using an arbitrary state $|\psi\rangle=a|0\rangle+b|1\rangle$ instead of $|+\rangle$ one arrives for the probability at $|a|^2\rho_{00}+ab^*\rho_{01}^*+a^*b\rho_{01}+|b|^2(1-\rho_{00})$ but from this one can again only say that the probability depends on the real part of $ab^*\rho_{01}$. So that doesn't support the suspicion that only the real part is important, since $Re(x)*Re(y)\neq Re(x*y)$.
