“…(t, y)∇t = 1 (t, y)∇t + (t, y)∇t,(23) we use(21) and(22) and the fact that, ifA ∩ B = ∅, then χ[A] • χ[B] = 0 to simplify the integrand, ∞ i=1 χ[z i , z i−1 ] |t − t i | y k , z i−1 ] |t − t k | y χ[z i−k+1 , z i−k ] |t − t i−k+1 | y = ∞ i=1 χ[z i , z i−1 ] |t − t i | 2y a.e.,and so (23) may be written as (t, y)∇t = l , z l−1 ]|t −…”