= Solution
Relative to a <filtration> $(\mathcal F_t)$, <progressive measurability> means that, for every $T<\infty$, the map
$$
[0,T]\times\Omega\longrightarrow\mathbb R,\qquad (t,\omega)\longmapsto X_t(\omega)
$$
is measurable for the <product sigma-algebra> $\mathcal B([0,T])\otimes\mathcal F_T$ and the <Borel sigma-algebra> on $\mathbb R$.
Fix $T>0$ and divide $[0,T]$ into $2^m$ equal subintervals with mesh $h_m=T/2^m$. Define
$$
X^{(m)}_t=
\sum_{k=1}^{2^m}X_{kh_m}\mathbf1_{[(k-1)h_m,kh_m)}(t)
+X_T\mathbf1_{\{T\}}(t).
$$
Since $X$ is <adapted>, every <random variable> $X_{kh_m}$ is $\mathcal F_{kh_m}$-measurable, hence $\mathcal F_T$-measurable. Each approximation is consequently $\mathcal B([0,T])\otimes\mathcal F_T$-measurable. For $t<T$, its sampling time lies strictly to the right of $t$, tends to $t$, and never exceeds $T$. <Right continuity> implies $X^{(m)}_t\to X_t$; at $T$ equality is exact. Thus $X$ is the pointwise limit of measurable functions on this product space. As $T$ was arbitrary, \b[$X$ is <progressively measurable>]. This is the theorem that <right-continuous adapted processes are progressively measurable>.
The right-endpoint approximations need not themselves be adapted at their intermediate times. What the proof requires is their joint measurability with respect to the single terminal sigma-algebra $\mathcal F_T$. The proof uses the pathwise <càdlàg> convention. If path regularity is assumed only <almost surely>, under a completed filtration setting the <stochastic process> to zero on its common exceptional null event gives an <indistinguishable> <progressively measurable> version. Arbitrary values on that null event need not make the original <stochastic process> <progressively measurable>: even with a complete filtration, a null sample point may be assigned a non-Borel time function. This is why <almost sure path regularity does not ensure progressive measurability>.
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