I'm going through Revuz and Yor and am stuck at a technicality. Suppose $Z$ is bounded and $A$ is bounded increasing continuous with $A_0 =0$. The goal of the problem is to show $E[ZA_\infty] = E\int_0^\infty E[Z|\mathscr{F}_t] dA_t$. I'm having trouble seeing why $t \longmapsto E[Z|\mathscr{F}_t]$ should be measurable for a fixed $\omega$, so that the integral on the right even makes sense.

I see that $E[Z|\mathscr{F}_t]$ is a UI martingale. So for $t_n \uparrow t$ we have $E[Z|\mathscr{F}_{t_n}] \to E[Z|\mathscr{F}_{t_-}]$ and similarly for $t_n \downarrow t$ we have $E[Z | \mathscr{F}_{t_n}] \to E[Z|\mathscr{F}_{t_+}]$ a.s. and in $L^1$. I don't see how any of this will lead to measurability though.

Edit: Suppose the filtration is right continuous. Then the previous line looks like it means $E[Z|\mathscr{F}_t]$ is right continuous, but I don't think it does. The convergence occurs almost surely, and the almost sure set depends on $t$ and the sequence $t_n \downarrow t$. Since there are uncountably many $t$ and sequences $(t_n)$, I don't see how we can conclude right continuity.

  • $\begingroup$ Any assumptions on right-/left-continuity of the filtration? $\endgroup$ – saz Feb 19 '15 at 6:33
  • $\begingroup$ Nope, it's Ch. IV 1.31 in R&Y if you want to check. $\endgroup$ – nullUser Feb 19 '15 at 7:05
  • $\begingroup$ In this question math.stackexchange.com/q/557257, it was already discussed that the claim does not hold for any filtration $\mathcal{F}_t$. (There, we agreed on the "usual conditions", so this doesn't answer your question.) $\endgroup$ – saz Feb 19 '15 at 9:51

Recall that conditional expectation is only defined up to a set of probability 0, so the first issue is to choose versions of the conditional expectations. Probably Revuz and Yor meant to assume that the filtration is right-continuous, which implies that there is a modification of the conditional expectations that is also right-continuous (e.g., Thm II.2.9 of Revuz and Yor). However, that assumption is not needed. By Thm 2.IV.1 of Doob's book on potential theory (pp. 463-4, which assumes only that the filtration is complete), there is a modification of the conditional expectations that has left and right limits everywhere. Any such function is continuous except on a countable set (for example, see this link: Prove that the number of jump discontinuities is countable for any function); in particular, it is Borel. In fact, it is Reimann-Stieltjes integrable w.r.t. any continuous measure.

Then to solve the problem from there, you can use the solution outlined here: Weird equality of expectations involving stochastic integral.


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