Predictability of countably valued accessible stopping times on complete and cadlag filtrations

Question

The following question is motivated by this part of the proof of Lemma 2 on page 107 of the book Stochastic integration and differential equations of Philip Protter.

Lemma 2. Let $T$ be a totally inaccessible stopping time. For $\delta > 0$, let $R(\delta) = \sup_{t \leq v} P(t \leq T \leq t + \delta \vert \mathcal{F}_{t})$. Then $R(\delta) \to 0 $ in probability as $\delta \to 0$.

Proof of Lemma 2. Let $a >0$ and $S_{n}(\delta) = \inf \lbrace t \in D_{n}: P(t \leq T \leq t + \delta \vert \mathcal{F}_{t}) > a \rbrace \wedge v.$ First we assume that $S_{n}(\delta)$ is less than $T$. Since $S_{n}$ is countably valued, it is accessible, and since $T$ is totally inaccessible, $P(S_{n}(\delta) = T)=0$. Suppose that $\Gamma \subset \lbrace T< t \rbrace,$ and also $\Gamma \in \mathcal{F}_{t}$. Then

\begin{align} E\left[ E\left[ 1_{\lbrace t \leq T \leq t+ \delta \rbrace}\vert \mathcal{F}_{t}\right] 1_{\Gamma}\right] = E \left[ 1_{\lbrace t \leq T \leq t+ \delta \rbrace} 1_{\Gamma} \right] = 0 \end{align}

$\cdots$

Is every countably accessible stopping time (in a complete and cadlag filtration) a predictable stopping time?

I ask this because what I want to do is to prove that $S_{n}(δ)$ is predictable (I suppose using the fact that it is accessible and countably valued), and therefore I can use the hypothesis that T is totally inaccessible to prove $P(S_{n}(δ)=T)=0$

Definition. A stopping time $T$ is totally inaccessible if for every predictable stopping time $S$,

\begin{align} P\lbrace w: T(w) = S(w) < \infty \rbrace = 0 \end{align}

Any reference or hint will be welcome.

Answer 1

No, not all accessible stopping times are predictable. However, predictable stopping times are accessible.

A stopping time is accessible whenever its graph is contained in a countable union of graphs of predictable stopping times. Put it another way, a stopping time is accessible whenever it is always equal to one of a countable number of predictable stopping times (up to sets of measure zero).

Answer 2

I might be wrong, but I think one can easily construct a counter-example if the underlying process $X_t$ has jumps at predictable times, or branching points (which, in a sense, are jumps at predictable times in the Ray topology).

Consider the natural (right-continuous) filtration of a process $X_t$ which is identically equal to $0$ with probability $\tfrac{1}{2}$, and equal to $\mathbb{1}_{[1, \infty)}(t)$ also with probability $\tfrac{1}{2}$. Then set $\tau = 1$ in the former case and $\tau = 2$ in the latter one. This way $\tau$ is a stopping time with two values, and hence an accessible time, but clearly not a predictable time.