= Solution
For each nonnegative rational $q$, equality of the two versions gives $\mathbb P(X_q=Y_q)=1$. Intersect these countably many full-probability events with the full-probability event on which both paths are <càdlàg>. Call the resulting event $A$; then $\mathbb P(A)=1$.
Fix $\omega\in A$ and any real $t\geq0$. Choose rational numbers $q_n>t$ decreasing to $t$. <Right continuity> gives
$$
X_t(\omega)=\lim_nX_{q_n}(\omega)
=\lim_nY_{q_n}(\omega)=Y_t(\omega).
$$
The same event $A$ works for every $t$, because the argument is pathwise after $\omega$ is fixed. \b[The <stochastic processes> are therefore <indistinguishable>.] This proves that <càdlàg versions are indistinguishable>; the left limits are not needed for this implication, since <right continuity> alone suffices.
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