Since as tends to infinity, there exists such that the first summand in the last inequality is smaller than for all . Choose an . Then, we can take such that the third summand is smaller than , because it approaches . In addition, once , we can refine so that the second summand becomes smaller than , and we finish the proof.
For a proof of the next lemma, it is enough to adapt the proof found in [107, Theorem 16] for the case of Banach space-valued functions.
Lemma 1.95: .
Now, we are able to prove the next inclusion.
Theorem 1.96: .
Proof. By Lemma 1.95, . Then, following the steps of the proof of Lemma 1.95 and using Lemma 1.94, we obtain the desired result.
For the next result, which says that the indefinite integral of any function of belongs to , we employ a trick based on the fact that if almost everywhere, then and , that is, the indefinite integrals of and coincide. This fact follows by a straightforward adaptation of [108, Theorem 9.10] for Banach space-valued functions (see also [70]). Thus, if we change a function on a set of Lebesgue measure zero, its indefinite integral does not change. Therefore, we consider, for instance, that vanishes at such points.
Lemma 1.97: If , then .
Proof. It is enough to show that every has a neighborhood where is of bounded variation. By hypothesis, given , there exists a gauge on such that for every -fine
