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Update: This question has been moved to Mathoverflow. Please answer it there.

There is a theorem as follows:

Theorem. Let $\mathcal{F}_t$ be a filtration which is right-continuous and complete. Assume $M_t$ is a submartingale adapted to $\mathcal{F}_t$ such that $t \mapsto \mathbb{E}M_t$ is right-continuous (which is always true of martingales on right-continuous filtrations). Then there is a RCLL modification of $M_t$.

Question. If I change "right-continuous" and "RCLL" to "continuous", is this still true? In other words, if the filtration is continuous and the map $t\mapsto \mathbb{E}M_t$ is continuous, can I get the stronger conclusion that there is a continuous (not just RCLL) modification?

If it is true, is there a reference (or obvious proof)? If it is false: Is there a nice counterexample? Are there known conditions on the filtration that would guarantee the continuous modification?

I think I have a proof for martingales, but since I cannot find this written anywhere, I am worried I might be mistaken. Also, I know it is true for martingales on the augmented filtration of Brownian motion, but that proof goes through the Martingale Representation Theorem (I believe) and seems like that is overkill (again making me worried I am missing something).

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  • $\begingroup$ I'm fairly sure this is true (as to conditions, see earlier answer). And I wouldn't worry about Martingale representation theorem. In the big scheme of things, this isn't too big a hammer at all. However, I haven't done this is in a long time, so can't proide an instant proof. People use RCLL/CadLag and rc only to use the results for both processes with and w/o jumps, when you obviously couldn't conclude generally that the modficiation is continous. $\endgroup$ – gnometorule Mar 5 '13 at 5:18
  • $\begingroup$ I guess you shall look into Kolmogorov Continuity Theorem and search for the realted counterexamples. $\endgroup$ – Ilya Mar 5 '13 at 8:31

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