Given a filtered probability space $( \Omega, \mathcal{F}, (\mathcal{F}_t)_t, \mathbb{P})$ and an $\mathcal{F} \otimes \mathcal{B}(\mathbb{R}_+)$-measurable bounded process $X: \Omega \times \mathbb{R}_+ \rightarrow \mathbb{R}$, we know by the predictable projection theorem that there exists a process $^pX$ (unique up to evanescence) such that
- $^pX$ is predictable
- $(^pX)_T = \mathbb{E}(X_T | \mathcal{F}_{T-})$ on $\{T<\infty\}$ for every predictable stopping time $T$
Taking for granted this result (I am not sure whether of not it requires any assumption on the filtration), my question is about the predictable projection of a stopped process.
In particular, take $X$ measurable as in the theorem and consider the stopped process $X^S$, where $S$ is any optional time. How can we express $^p(X^S)$ in terms of $^pX$?
As far as I understand (Jacod & Shiryaev, page 23), it is the case that
\begin{equation} ^p(X^S) = (^pX) \mathbb{I}_{[\![0, S]\!]} + X_S \mathbb{I}_{]\!]S, \infty ]\!]} \tag{*} \end{equation}
but I cannot prove either of the defining properties.
With regard to (1), I was hoping to show that both terms on the rhs of (*) are predictable.
$(^pX) \mathbb{I}_{[\![0, S]\!]}$ is predictable, because it is the product of two predictable processes.
The problem is that I do not see why $X_S \mathbb{I}_{]\!]S, \infty ]\!]}$ needs to be predictable. In order to have $Y \mathbb{I}_{]\!]S, \infty ]\!]}$ predictable if $S$ is optional I need $Y$ to be $\mathcal{F}_S$-measurable, but $X_S$ is not necessarily $\mathcal{F}_S$-measurable.
What is even more puzzling is that - if I interpret Jacod & Shiryaev correctly - we can deduct from (*) that $X_S \mathbb{I}_{]\!]S, \infty ]\!]}$ is predictable (since it is the difference of two predictable processes) for every bounded measurable process $X$ and optional time $S$.
Would not $X(\omega, t) = \mathbb{I}_A(\omega)$, with, say, $A \in \mathcal{F} \setminus \mathcal{F}_1$ and $S(\omega) = \frac12$ be a trivial counterexample to this (in this example $X_S \mathbb{I}_{]\!]S, \infty ]\!]}$ would not even be adapted)?
With regard to (2), I am even more clueless.
Apologies for the macroscopic error that I am somewhere certainly making.
