Lets say we have 2 commuting operators, $\Omega$ and $\Lambda$. The first operator has 3 separate eigenvalues, $\omega_1$, $\omega_2$ and $\omega_3$. The second operator has eigenvalues $\lambda_1$ and $\lambda_2=\lambda_3=\lambda$.
These commute and so they share eigenkets. The state of the system is: $$ |\psi\rangle=\alpha|\omega_1\lambda_1\rangle+\beta|\omega_2\lambda\rangle+\gamma|\omega_3\lambda\rangle $$
I know what happens if we measure $\Lambda$ then $\Omega$. However what happens if we measure $\Omega$ and then $\lambda$?
I will assume $\Omega$ results in one of the last 2 eigenvalues, lets say $\omega_2$ and so the state collapses to:
$$ |\psi\rangle=|\omega_2\lambda\rangle $$ and obviously we measure $\omega_2$. If we now measure $\Lambda$ we will obviously get $\lambda$.
My question is this: Does the state stay the same or does it become a superposition of both eigenkets which have the eigenvalue $\lambda$? If you measure the degenerate operator second does the system stay well defined in the other operator?