2026-07-072026-07-07http://salesiana.dossiersoluciones.com/handle/123456789/229670We give simple proofs that for a continuous local martingale M_t: 1) \liminf_{ε->0} ε\log Ee^{(1-ε) <M>_\infty /2} < \infty ==> E\exp(M_\infty - <M>_\infty /2) = 1, 2) \liminf_{ε->0} ε\log\sup_{t>=0} Ee^{(1-ε)M_t/2} < \infty ==> E\exp(M_\infty - <M>_\infty /2) = 1 .3 pages, few glitches correctedProbability60H05text