# Measurability issues in the proof of Fujisaki, Kallianpur and Kunita for stochastic filtering

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

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.
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.