1. Then, b(c) ? [0, x]. By (B), b is strictly increasing;= G � B(c) = G � B(c) ? G;Conversely, consider c ? G
2. As b is strictly increasing, c = c for every c ? C. By Lemma 3.1, G is surjective and G (resp., b) is the function inverse of b (resp;As G is strictly increasing, it is injective
3. If G is continuous and strictly increasing, then so is b by Lemma 3.1. Hence, g 1 is continuous. The converse is proved as follows. (a) Suppose G is discontinuous at ? ? (0, ? 1 ) with G(?) = c. Then, G(?-) = c < c = G(?);As b = ? on