stopped filtration = filtration generated by stopped process?

Solution 1:

I've been worried about the same thing. Here's what I came up with:

Assume that $X$ is progressively measurable (e.g. cadlag), then the inclusion $\sigma (X_s^T : s \le t)\subset\mathcal{X}_T^0$ trivial. Without this condition, this inclusion won't hold in general, I believe. For the other direction:

$A \in \mathcal{X}_T^0 \Leftrightarrow \mathbb{1}_A$ is $\mathcal{X}_T^0$ measurable $\Leftrightarrow \mathbb{1}_A = Y_T$ for some $\mathcal{X}_t^0$-optional process $Y_t$.

Thus it suffices to show that for all optional processes $Y_t$, $Y_T$ is $\sigma (X_s^T: s \ge 0)$ measurable.

Now, the optional processes are generated by stochastic intervals of the form $[\sigma,\infty)$, so using a functional class argument it is enough to show that $\{\sigma \le T\} \in \sigma (X_s^T : s \ge 0)$ for all $\mathcal{X}_t^0$-stopping times $\sigma$, $T$. Do this via discretisation of $\sigma$, $T$ and taking limits. (Similarly we can prove that a $\mathcal{X}_t^0$-stopping time $T$ is a $\sigma (X_s^T : s \ge 0)$ stopping time, I believe.)

EDIT: I spoke too soon, this discretisation argument doesn't work unless the filtration is right continuous.... I have no idea how to proceed.

EDIT2: It appears that the proof is IV.100 of Probabilities and Potential (Dellacherie and Meyer)　though I do not currently have this to hand.