August 15th, 2009, 2:28 pm
This is just what I thought, I'm not sure if it's right. When you have a weak solution to a diffusion, you have a measure on the space of continuous functions, which is equivalent to specifying all finite dimensional distributions of the process. But if a stopping time takes on uncountably many values, then the finite dimensional distributions don't give you enough information. In other words, if you have a strong markov process and you take a modification of it, the new process need not be strong markov, so a priori weak solutions don't tell you anything about the strong markov property.When you do have uniqueness, you want to show that under (say) has the same law as X under , where T is some stopping time. But KS show in one of the preceding lemmas that the measure induced by the first process solves the martingale problem for the second process. By uniqueness, they conclude that those two measures must be the same. I don't really have an intuitive feel for this part though.