Hello I have a question regarding KAM curves. KAM theory says that if we add a very small non-integrable Hamiltonian perturbation to an integrable Hamiltonian system, then most of the invariant tori are preserved if the perturbation is very small and the measure of the destroyed invariant tori scales with the magnitude of the perturbation. Now, are the KAM curves for perturbed Hamiltonian systems "exact invariants" or "invariants upto all orders". Since the perturbation is non-integrable, I would believe that the KAM curves would only be "invariants upto all orders" and not really be "exact invariants". Is this right? If the above is right, then, in addition to Arnold Diffusion, should not there be an exponentially slow diffusion across the KAM curves also since they are not exact invariants? And if the above statement is wrong, then how does a non-integrable perturbation happen to conserve the integrability of the Hamiltonian in certain regions in phase-space? Thanks, Kushal.