It is obvious that ˜ a$_{k}$$_{+1}$ ≤ ˜ a$_{k}$$_{+2}$ for every integer k. ˜ A D l is given as The second equality in (A.11) follows from ˜ a$_{l}$$_{+1}$ = a$_{l}$ + δ = ˜ a$_{l}$ + δ , and the third equality holds because This implies that Therefore, This means that for any sequence of sets { A$_{k}$ } ∞ k = −∞ with a$_{k}$$_{+1}$ > a$_{k}$ + δ for some integer k , one can always construct a seqeunce of sets { ˜ A$_{k}$ } ∞ k = −∞ with ˜ a$_{k}$$_{+1}$ ≤ ˜ a$_{k}$ + δ for every integer k satisfying This can be intuitively understood by comparing Figure A.3(a) to Figure A.3(b), where the sum of the lower bound on each triangle is equal to ∞ ∑ k = −∞ max $^{{}$µ$_{0}$ ( A$_{k}$ ) − µ$_{1}$ $^{(}$A D k $^{)}$, 0 $^{}}$and ∞ ∑ k = −∞ max $^{{}$µ$_{0}$ ( ˜ A$_{k}$ $^{)}$− µ$_{1}$ ( ˜ A D k $^{)}$, 0 $^{}}$, respectively. Therefore, it is innocuous to assume a$_{k}$$_{+1}$ ≤ a$_{k}$ + δ at the optimum. 65 ∞ ∑ k = −∞ max $^{{}$µ$_{0}$ ( ˜ A$_{k}$ $^{)}$− µ$_{1}$ ( ˜ A D k $^{)}$, 0 } ≥ ∞ ∑ k = −∞ max $^{{}$µ$_{0}$ ( A$_{k}$ ) − µ$_{1}$ $^{(}$A D k $^{)}$, 0 $^{}}$. ∞ ∑ k = −∞ max $^{{}$µ$_{0}$ ( A$_{k}$ ) − µ$_{1}$ $^{(}$A D k $^{)}$, 0 } ≤ ∞ ∑ k = −∞ max $^{{}$µ$_{0}$ ( ˜ A$_{k}$ $^{)}$− µ$_{1}$ ( ˜ A D k $^{)}$, 0 } max $^{{}$µ$_{0}$ ( ˜ A$_{k}$ $^{)}$− µ$_{1}$ ( ˜ A D k $^{)}$, 0 } = max $^{{}$µ$_{0}$ ( A$_{k}$ ) − µ$_{1}$ $^{(}$A D k $^{)}$, 0 } for k ≤ l, max $^{{}$µ$_{0}$ ( ˜ A$_{k}$$_{+1}$ $^{)}$− µ$_{1}$ ( ˜ A D k +1 $^{)}$, 0 } = max $^{{}$µ$_{0}$ ( A$_{k}$ ) − µ$_{1}$ $^{(}$A D k $^{)}$, 0 } for k ≥ l + 1 , A D l = (min { a$_{l}$ + δ, a$_{l}$$_{+1}$ } , ∞ ) = ( a$_{l}$ + δ, ∞ ) . ˜ A D l = (min {$_{˜}$ a$_{l}$ + δ,$_{˜}$ a$_{l}$$_{+1}$ } , ∞ ) (A.11) = ( a$_{l}$ + δ, ∞ ) = A D l (A.12)