Let $\mathscr{F}$ be a filtration and $\mathscr{F}^+$ be the right continuous version. ($\mathscr{F} = \{\mathscr{F}_t\}_{t\in [0,T]})$
Suppose $X$ is right-continuous $\mathscr{F}$-adapted process.
Claim X is $\mathscr{F}^+$-progressively measurable.
If I do not misunderstood, it follows by discretize time and approximate $X$ as a limit of left-continuous simple functions as follows, \begin{equation} X_n(s,w) = \sum_{k=0}^{2^n} X\left(t_k, w\right)\mathbb{1}\left\{t_{k-1}< s \leq t_k\right\} \end{equation} where $t_k = kT/2^n$. ($n$ implicit).
Each $X_n$ is $(\mathscr{B}([0,T])\times \mathscr{F}_T, \mathscr{B}(\mathbb{R}))$-measurable, hence $X = \lim_n X_n$ is measurable too.
Question I couldn't figure out why we require right-continuity of filtration.
For $B \in \mathscr{B}(\mathbb{R})$, \begin{equation} \{X_n(s,w) \in B\} = \cup_{k=0}^{2^n} \left\{(t_{k-1}, t_k)\times w : X(t_k, w) \in B \right\} \end{equation} Hence $\{X_n \in B\} \in (\mathscr{B}([0,T])\times \cup_k\mathscr{F}_{t_k})$ because $X$ is adapted.