Corrigendum to: An enhanced uncertainty principle for the Vaserstein distance (Bulletin of the London Mathematical Society, (2020), 52, 6, (1158-1173), 10.1112/blms.12390)

One of the main results in the paper mentioned in the title is the following. (Formula presented.) Here (Formula presented.) indicates the Vaserstein distance between the positive and negative parts of (Formula presented.). The proof, which we now recall briefly, was based on a recursive selection, through a stopping time argument, of dyadic squares (Formula presented.) where either the mass of (Formula presented.) is irrelevant, called empty, or (Formula presented.) is much larger than (Formula presented.) (or the other way around), called unbalanced. Assume, without loss of generality, that (Formula presented.) is finite and that (Formula presented.) is normalized so that (Formula presented.). Consider now the standard dyadic partition of the unit cube: first split (Formula presented.) into (Formula presented.) subcubes of length (Formula presented.) and then, recursively, split each of the new subcubes into (Formula presented.) “descendants” each with side length half that of the “parent”. A cube (Formula presented.) is balanced if (Formula presented.) where (Formula presented.) indicates the volume and (Formula presented.) On the other hand, (Formula presented.) is empty if (Formula presented.). Recursively, at each generation we set aside the cubes (Formula presented.) such that either: (i)the cube (Formula presented.) itself is empty; (ii)one of the (Formula presented.) direct descendants (Formula presented.) of (Formula presented.) is full (non empty) and unbalanced. In this case, denote by (Formula presented.) the full, unbalanced descendant of (Formula presented.) for which (Formula presented.) is maximal. This gives a decomposition (Formula presented.) where (Formula presented.) and (Formula presented.) indicate respectively the indices of the empty and full selected cubes, and where (Formula presented.) is the remaining set (whatever has not been selected in the process above). We proved (Proposition 2) that (Formula presented.) and that the full cubes (Formula presented.) carry most of the mass, i.e. (Formula presented.), and next we claimed (Remark 1) that the corresponding descendants (Formula presented.), (Formula presented.), still carry a significant part of that mass: 2 (Formula presented.) Benjamin Jaye has kindly pointed out that this last estimate may in principle not be true in general, in that the mass in the full, unbalanced descendants (Formula presented.) may not be comparable to the total mass of (Formula presented.) in (Formula presented.) – this mass may lie predominantly in the balanced descendants of (Formula presented.). He suggested circumventing this difficulty by constructing the cubes (Formula presented.) by means of a continuous stopping time argument together with the Besicovitch covering theorem, replacing our original dyadic-based decomposition. With these new cubes, Propositions 3 and 4 remain valid, hence also the statement of Theorem 1. The dyadic decomposition used to prove Theorem 2 requires analogous modification. The result itself, as well as all other results, remains correct as stated. Details of the modifications to the proof of Theorem 1 are as follows. The function (Formula presented.) is extended by 0 from (Formula presented.) to all of (Formula presented.). The definitions of balanced and unbalanced cubes are modified so that a cube (Formula presented.) is balanced if 3 (Formula presented.) The choice of the factor (Formula presented.) is in accordance with the constant appearing in the Besicovitch covering theorem (see below). A similar adjustment is made to the definition of full and empty cubes, in that a cube (Formula presented.) is empty whenever (Formula presented.) For every (Formula presented.) such that (Formula presented.), there exists (Formula presented.) such that the open cube (Formula presented.) centred at (Formula presented.) and of side length (Formula presented.) is simultaneously balanced and unbalanced in that either (Formula presented.) or (Formula presented.) equals (Formula presented.). This can be achieved by continuity, since for (Formula presented.) very small the cube centred at (Formula presented.) and of side length (Formula presented.) is infinitely unbalanced, while for side length (Formula presented.) it is balanced. Then there must be an intermediate side length (Formula presented.) for which one of the inequalities in (3) is actually an identity. These cubes (Formula presented.) cover (Formula presented.) (up to at most a zero-measure set). According to the Besicovitch covering theorem [1, Theorem 18.1] (see also [2]), one can find (Formula presented.) sequences (Formula presented.), (Formula presented.), such that for each (Formula presented.) the cubes (Formula presented.) are pairwise disjoint and together still cover (Formula presented.) : (Formula presented.) Since (Formula presented.), there is at least one family of cubes (Formula presented.) such that (Formula presented.) From this particular sequence of cubes we select those that are full, and further relabel the cubes themselves as (Formula presented.). These cubes are thus disjoint and carry most of the mass, since (Formula presented.) In conclusion, estimate (2) holds for the cubes (Formula presented.), up to a change of constants. Lemma 1 follows as before with only minor modifications due to extending (Formula presented.) by 0 to all of (Formula presented.). With the revised choice of the side length of the cubes (Formula presented.), both the estimates for the size of the zero set (Proposition 3) and for the Vaserstein distance (Proposition 4) hold for the new (Formula presented.), with the obvious modifications. The proof of Theorem 1 then concludes as before. The proof of Theorem 2 follows similar arguments. One option is just to take in the existing proof the cells (Formula presented.), where the cubes (Formula presented.) are now obtained as explained above. Another option is to use Nash's embedding theorem and consider the (Formula presented.) -dimensional compact manifold (Formula presented.) as isometrically embedded in (Formula presented.). Then the standard Besicovitch theorem in (Formula presented.) yields a similar Besicovitch covering theorem for (Formula presented.) (with constant (Formula presented.) instead of (Formula presented.)).