By Balakrishnan A.V.

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N, the Riemann–Stieltjes sum SRS (f, dh; τ ) converges to SYS (f, dµ; J, T ). 24. For f , µ, h, J and T as above, the sum SYS (f, dµ; J, T ) can be approximated arbitrary closely by Riemann–Stieltjes sums SRS (f, dh; τ ) based on tagged partitions τ of subintervals of J such that all tags of T are tags of τ . Recall that the class of all Y -valued additive and upper continuous interval functions on a nonempty interval J is denoted by AI(J; Y ). Now we can show that in deﬁning the Kolmogorov integral with respect to an additive upper continuous interval function it is enough to take Young interval partitions.

T2m−2 := sm−1 , t2m−1 ∈ Asm−1 ∩ Asm ∩ (sm−1 , sm ), and t2m := sm . Thus (b) holds with n = 2m + 1. (b) ⇒ (c). Given ǫ > 0, choose a Young interval partition {(ti−1 , ti )}ni=1 of J as in (b). Deﬁne a step function fǫ on J by fǫ (t) := f (si ) with si ∈ (ti−1 , ti ) if t ∈ (ti−1 , ti ) for some i ∈ {1, . . , n} and fǫ (ti ) := f (ti ) for i ∈ {0, . . , n} if ti ∈ J. Then fǫ (t) − f (t) < ǫ for each t ∈ J. Since ǫ is arbitrary, (c) follows. (c) ⇒ (a). Given ǫ > 0, choose a step function fǫ such that fǫ (t)−f (t) < ǫ for each t ∈ J.

Let f : S → X be Bochner µ-integrable. 26) A exists and does not depend on the choice of {fk } satisfying the definition. 3 The Reﬁnement Young–Stieltjes and Kolmogorov Integrals 39 Proof. 31. Taking a subsequence, we can assume that gk dµ < 4−k for all k. Then there is a set B ∈ S with µ(B) = 0 such that for s ∈ / B, gk (s) → 0, and so fk (s) → f (s). The set of ﬁnite rational linear combinations of elements in the union of ranges of all fk is countable, so its closure is a separable subspace of X, in which f (s) takes values for s ∈ / B.