We know that predictable $\implies$ optional $\implies$ progressively measurable. Source
Predictable processes have obvious/simple counterparts in discrete time.
Do optional processes and progressively measurable processes also have counterparts (of any kind) in discrete time?
Are optional processes in discrete time just predictable processes? (Since optional times and stopping times are the same thing in discrete time -- also why do they call them stopping times instead of predictable times?)
Are progressively measurable processes in discrete time just predictable processes (since the power set of the real numbers is a $\sigma-$algebra)?
"Progressive" and "optional" are refinements of "adapted" involving some degree of joint measurability of $(\omega,t)\mapsto X_t(\omega)$. The three notions coalesce in discrete time. Consider, for example, the notion "optional" in a discrete-time setting. So let $(\Omega,(\mathcal F_n)_{n\ge 0},\mathcal F,\Bbb P)$ be a filtered probability space. The optional $\sigma$-field $\mathcal O$ is the $\sigma$-field on $\Omega\times\{0,1,2,\ldots\}$ generated by the sets of the form $[T,\infty):=\{(\omega,n)\in\Omega\times\{0,1,\ldots\}:n\ge T(\omega)\}$, as $T$ varies over the $(\mathcal F_n)$ stopping times. A process $X=(X_n)_{n\ge 0}$ is optional provided $(\omega,n)\mapsto X_n(\omega)$ is $\mathcal O$-measurable.
If $X$ is optional then $X_T1_{\{T<\infty\}}$ is $\mathcal F_T$-measurable for each stopping time $T$; in particular, $X_n$ is $\mathcal F_n$-measurable for each $n\in\{0,1,\ldots\}$. Therefore $X$ is adapted.
Suppose, conversely, that $X$ is adapted and $B$ is a Borel subset of the real line. Define stopping times $T_m$, $m=0,1,\ldots$ by $$ T_m=\cases{m,&if $X_m\in B$;\cr \infty,&otherwise.\cr} $$ (These are stopping times because $X$ is adapted.) The graph $[T_m]:=\{(\omega,n): T_m(\omega)=n\}$ of $T_m$ is optional (i.e., an element of $\mathcal O$---just notice that $[T_m]=[T_m,\infty)\setminus[T_m+1,\infty)$) and so $$ X^{-1}(B)=\cup_{m=0}^\infty[T_m] $$ is an element of $\mathcal O$. It follows that $X$ is optional.
"Progressive" can be handled similarly.