The definition of an adapted process $X$ is that $X_i$ be $(\mathcal{F_i}, \Sigma)$-measuriable where $\mathcal{F.} = (\mathcal{F_i})_{i \in S}$ is a filtration of the sigma algebra $\mathcal{F}$ (probability space) and $\Sigma$ is part of the measurable space $(S, \Sigma)$. (and there are other required ones, but I will skip those parts.)
And predictable processes are the ones that $X_t$ is measurable with respect to $F_{t-}$.
It seems that adapted left-continuous processes cannot be predictable processes – after all, if they are adapated, $X_t$ must be $F_t$ measurable, while the definition of predictable processes say that $X_t$ must be $F_{t-}$-measurable.
Why are these seemingly nonsenses working?
Best Answer
If $X_t$ is $\mathcal{F}_t$-measurable, this does not imply that $X_t$ is not $\mathcal{F}_{t-}$-measurable. Actually, left-continuous adapted processes are predictable:
Let $t \geq 0$. The information about the path $[0,t) \ni s \mapsto X(s,\omega)$ already determines the value at time $t$, since $$X(t,\omega) = \lim_{s \uparrow t} X(s,\omega)$$ by the left-continuity of $X$. This means (intuitively) that you can predict $X_t$ by the knowledge of $(X_s)_{s <t}$. Since $X_s$ is $\mathcal{F}_{t-}$-measurable for all $s<t$, this implies the $\mathcal{F}_{t-}$-measurability of $X_t$.