I'm currently looking over the proof(s) of the theorem of Fujisaki, Kallianpur and Kunita regarding the MRT-like characterisation of square integrable random variables measurable with respect to the augmented observation filtration. Unfortunately I'm seeing problems with respect to measurability in the two proofs I'm aware of:

The proof in Fundamentals of Stochastic Filtering (Crisan and Bain) which most of my intuition about the ideas of the proof revolves around: http://books.google.co.jp/books?id=hE3KF5Wf6ecC&pg=PA18&lpg=PA18&dq=path+regularity+for+optional+projection&source=bl&ots=XoOtAHQ6YG&sig=z1b6Zp_P3c7ExNl2DqpNHGHB54M&hl=en&sa=X&ei=asjCUIjYEO2tiQfj-YCwBA&ved=0CCoQ6AEwAA#v=onepage&q=path%20regularity%20for%20optional%20projection&f=false

The original proof: Link

For the first proof, my understanding breaks down around the bottom of page 36 and the start of page 37, especially where they state that $\int_0^t \|\bar{\pi}_s(h)\| ds$ is continuous. However, as I understand this process is only equal to a continuous process with respect to indistinguishability, which is not sufficient for the $T^n$ defined later to be $\mathcal{Y}_t^o$ stopping times.

For the original proof (which I've admittely skimmed), the authors seem to suggest defining $\hat{h}$ to be the optional projection of $h$ with respect to the raw (unaugmented) filtration of the observation process $z$ (which I believe can be done by a theorem of Dellacherie and Meyer though I have not read the proof). Later on we would like to define stopping times $T_n$ such that $T_n$ is the hitting time of $|\int_0^t \hat{h}_s \ ds |$ of the level n. For these to be stopping times with respect to the raw observation filtration, we'd need $\int_0^t \hat{h}_s \ ds $ to be continuous, but this can only be done if we have some path regularity of $\hat{h}_s$. In the case of taking the optional projection with respect to an augmented, right continuous filtration, given certain integrability conditions on $h$ we can choose a cadlag version of $\hat{h}$, but in the unaugmented filtration case, as far as I know, this is not possible.

As it stands, I only feel I can prove the theorem in the case $h_t = h(X_t)$ is surely bounded by a deterministic function $K_t$, in which case we don't need to apply any probabilistic stopping arguments.

As this is quite a fundamental result in the field of stochastic filtering, these issues are quite worrying to me. I'm hoping that someone more experienced in this area could help clear things up. My impression so far is that the literature seems rather fraught with problems of this nature (the proof of Lemma A.24 in the book linked, required for the proof of the same theorem, being another example: https://math.stackexchange.com/questions/168566/stopped-filtration-filtration-generated-by-stopped-process).


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Having looked at the proof in Diffusions, Markov Processes and Martingales Vol. 2 (Rogers and Williams) I believe I have resolved the issue. We define $\mathcal{Y}_t$-stopping times $T_n$ using the cadlag version of $\pi_t(h)$. By continuity of $\int_0^t \pi_s(h) ds$ these are $\mathcal{Y}_{t-}$ stopping times. We then can find $\mathcal{Y}_t^o$-stopping times $S_n$ equal to $T_n$ almost surely. This enough for the proof in the book linked.


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