Another Stick-Breaking Game: 11089

DiS) > 0. Suppose that 0 nr for all n in T. Then for n in T we have |5 H [nr/2, ?]| > nr/2 and so nrbn/2 oo through values of T, the right-hand side of this inequality converges to 0, and (3) follows. (3) => (2): Let nx in? for / in N, and let S = UieNini, in?]. Then (2) is satisfied because