Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 24 4 b Solution Created 2026-10-03 Updated 2026-10-07
For each nonnegative rational , equality of the two versions gives . Intersect these countably many full-probability events with the full-probability event on which both paths are càdlàg. Call the resulting event ; then .
Fix and any real . Choose rational numbers decreasing to . Right continuity givesThe same event works for every , because the argument is pathwise after is fixed. 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.