“…As for the equivalence, we still haveQ ∼P whenever (λ ,Q,P) is a dual optimizer (see [17], Theorem 2.7). Also, by an exhaustion argument, there exists a maximal solution (λ ,Q,P) in the sense that if (λ , Q, P) is another dual optimizer, then P P (hence Q Q ) and λ dQ/dP =λ dQ/dP, P-a.s., where the density dQ/dP is defined P-a.s. in the sense of Lebesgue decomposition.…”