 Solution

Let $\mathcal{P}$ be the predictable sigma-algebra on $\Omega \times [0,\infty )$, generated by $A\times \{0\}$ with $A\in {\mathcal{F}}_0$ and $A\times (s,u]$ with $A\in {\mathcal{F}}_s$. A stochastic process $H$ is previsible precisely when $(\omega ,t)\mapsto H_t(\omega )$ is $\mathcal{P}$-measurable.

Every left-continuous adapted process is previsible. Indeed, on intervals $(k{2}^{-n},(k+1)2^{-n}]$, put $H_t^{(n)}=H_{k{2}^{-n}}$, and put $H_0^{(n)}=H_0$. Each approximation is a predictable process, since its coefficient is ${\mathcal{F}}_{k{2}^{-n}}$-measurable; left continuity gives $H_t^{(n)}\to H_t$ for every $t$, so measurability passes to the limit.

An adapted càdlàg process need not be previsible. For a concrete example, let $\xi$ take values $0,1$ with equal probabilities, let ${\mathcal{F}}_t$ be trivial for $t<1$ and ${\mathcal{F}}_t=\sigma (\xi )$ for $t\ge 1$, and set $H_t=\xi 1_{[1,\infty )}(t)$. This is an adapted càdlàg process, but $H_1=\xi$ is not ${\mathcal{F}}_{1-}$-measurable, where ${\mathcal{F}}_{1-}=\sigma ({\bigcup }_{s<1}{\mathcal{F}}_s)$. Sections at time $1$ of the generators of $\mathcal{P}$ all belong to ${\mathcal{F}}_{1-}$, so every predictable process has an ${\mathcal{F}}_{1-}$-measurable value there; this proves the counterexample directly.