How to show $t \mapsto E[Z|\mathscr{F}_t]$ is a.s. borel measurable. 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.
 A: 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.
