= 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.
\b[Every left-continuous <adapted process> is <previsible>.] Indeed, on intervals $(k2^{-n},(k+1)2^{-n}]$, put $H^{(n)}_t=H_{k2^{-n}}$, and put $H^{(n)}_0=H_0$. Each approximation is a <predictable process>, since its coefficient is $\mathcal F_{k2^{-n}}$-measurable; left continuity gives $H^{(n)}_t\to H_t$ for every $t$, so measurability passes to the limit.
\b[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 $t\geq1$, and set $H_t=\xi\mathbf1_{[1,\infty)}(t)$. 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.
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