# 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.

• Any assumptions on right-/left-continuity of the filtration? – saz Feb 19 '15 at 6:33
• Nope, it's Ch. IV 1.31 in R&Y if you want to check. – nullUser Feb 19 '15 at 7:05
• 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.) – saz Feb 19 '15 at 9:51