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I am trying to understand the notion of predictable process. Let $(Ω,F_t,P)$ be a filtered measure space, satisfying the usual condition. Things starts with the predictable $\sigma$-algebra ${\mathcal P}$, which is generated by sets of the form $A\times (a,b]$ with $A\in{\mathcal F}_a$ and $A\times \{0\}$ with $A\in{\mathcal F}_0$.

My question: is it true that $S\in {\mathcal P}$ if and only if $S$ is progressive and $\{\omega|(\omega,t)\in S\}\in{\mathcal F}_{t−}$ for all $t$? In another word, is it true that $X$ is predictable if and only if $X$ is progressive and $X$ is adapted to the filtration ${\mathcal F}_{t−}$?

The only if part is easy but I am not sure about the if part. I feel that $X$ being ${\mathcal F}_{t−}$-measurable seems to be a more "reasonable" definition of "predictable", but maybe I am wrong.

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No, it's not true.

First of all, recall that any adapted càdlàg process $(X_t)_{t \geq 0}$ is progressively measurable. This means that for any such process $(X_t)_{t \geq 0}$ your assertion reads

$(X_t)_{t \geq 0}$ is predictable $\iff$ $X_{t}$ is $\mathcal{F}_{t-}$-measurable for any $t \geq 0$.

Now consider for instance a Poisson process $(X_t)_{t \geq 0}$, and let $(\mathcal{F}_t)_{t \geq 0}$ be its completed canonical filtration. By the very definition of $\mathcal{F}_{t-}$, we know that $X_{t-} = \lim_{s \uparrow t} X_s$ is $\mathcal{F}_{t-}$-measurable. Since $X_t = X_{t-}$ almost surely we find that $X_t$ is $\mathcal{F}_{t-}$-measurable for any $t \geq 0$. However, $(X_t)_{t \geq 0}$ is not predictable. Indeed: If $(X_t)_{t \geq 0}$ was predictable, then

$$M_t := X_t -t \mathbb{E}(X_1), \qquad t \geq 0,$$

would be a predictable martingale which would imply that $(M_t)_{t \geq 0}$ has continuous sample paths (see e.g. here) which is clearly not true; hence $(X_t)_{t \geq 0}$ is not predictable.

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