International Congress of Mathematicians

Harmonie Aleasure and "Locally Flat" Domains 703where D denotes Hausdorff distance. Note that this is a significant condition onlyfor 5 < 1. We will always assume Ö < -fi^. We say that 0 is Reifenberg vanishingif, as r —t 0, we can take Ö —¥ 0. For instance, the domain above the graph of a A»function is Reifenberg vanishing. In general, Reifenberg vanishing domains are notlocal graphs; they do not have tangent planes or a "surface measure". This class ofdomains was introduced by Reifenberg [20] in his study of the Plateau problem forminimal surfaces in higher dimensions.In order to state our analogue of Kellogg's theorem in this setting, we needto introduce "multiplicative" analogues of (I)o- A measure p, supported on 90, isdoubling if, Vif ÇC R" +1 , there exists RK > 0 such that, if 0 < r < RK, thenp(B(2r, Q) n 90) < C p(B(r, Q) n 90).Such a p is called asymptotically optimal doubling (see [], []) for details) if it isdoubling andUm . nf p(B(rr,Q)ndn)_ i]m p(B(rr, Q) n 90) _ ^r^oQGOfinif p(B(r,Q)ndü) r^o Qe9nnK p(B(r,Q)ndü)for 0 < r < 1, K CC R". For example, if 0 is of class C 1,a and da denotessurface measure, then a(B(r, Q) n 90) = a n r n + 0(r n+a ), Q £ 90, and hence a isasymptotically optimal doubling. If logfc £ C a , then the same is true for du = kda.Our analog of Kellog's theorem is:Theorem 1. ([15]) If ii is a Reifenberg vanishing domain, then u (OJ°°) is asymptoticallyoptimal doubling.The proof uses the fact that oJ-Reifenberg flat domains are NTA domains ([9],[15]). One then uses the theory of the boundary behavior of harmonic functions onNTA domains ([9]) and comparisons to half-planes, using the Reifenberg vanishingcondition and the maximum principle.To understand a possible converse to Theorem 1, we recall a geometric measuretheory (GAIT) problem, first posed by Besicovitch: let p be a positive Radonmeasure on R" +1 such that, for each Q £ S (S the support of p) and each r > 0,we havep(B(r, Q)) = ar n , a>0 fixed. (B)Then, what can be said about pi Clearly, if dp = dx on R" C R" +1 , then (B)holds. Nevertheless, in 1987, D. Preiss found the following interesting example: letSc be the light cone x\ = x\ + x\ + x\, and dp = daj: c its surface measure. Then psatisfies (B). Aloreover, the general case of (B) is settled by the following remarkabletheorem of Kowalski-Preiss [17].Theorem. ([17]) Let p be a non-zero measure with property (B), and put S =suppM Ç K n+1 . If n = 1,2, then S = R". If n > 3, then either S = R" orS = Sc ® R" -3 , modulo rigid motions.The connection of the Preiss example to our problem comes from the fact([16]) that, if 0 = {x\ < x\ + x\ + x\}, OCR 4 , then dui 00 = daj: c (separation of