Max-Planck-Institut fur Mathematik in den Naturwissenschaften Leipzig Uniformly distributed measures in Euclidean spaces by Bernd Kirchheim and David Preiss Preprint-Nr.: 37 1998

Uniformly Distributed Measures in Euclidean spaces Bernd Kirchheim and David Preiss For every complete metric space X there is, up to a constant multiple, at most one Borel regular measure over X such that 0 < (B(x; r)) = (B(y; r)) < 1 whenever x; y 2 X and r 2 (0; 1); where the symbol B(x; r) stands for the closed ball with center in x and radius r. In fact, the existence of such a measure imposes rather strong regularity conditions on the space X itself (see [2], [7], [10]); in particular from [3] we know that, if X is a bounded subset of R n, then it is contained in a sphere. In the case when X R n, regularity results about X turned out to be the most important tool in the study of densities of measures in R n (see [9], or, for a more detailed but less complete account, [8]). One of our main results considerably improves several results from [9] by showing that any X R n is an analytic variety. Our proof is much simpler than those presented in [9], which enables us to give as an application a simple proof of a remarkable result of Marstrand [6] according to which the existence of non-trivial s-dimensional densities (of a measure in R n ) implies that s is an integer. Our results, however, do not seem to be strong enough to show the main result of [9] that measures having s-dimensional density are s rectiable, because they do not give any information about the behaviour of X at innity. In addition to the analyticity result mentioned above, we also show that X is an algebraic variety provided that it is bounded and obtain more precise results in the special cases n = 1 and n = 2. For various reasons, including the political changes in Europe, this note has not been published for many years, although our main approach has become known and used in the literature. We would like to thank the Max 1

Planck Institute for bringing us together and thus enabling us to recover the manuscript and prepare it for publication. Under a measure over R n we will understand a monotone, -additive mapping of the family of all subsets of R n into [0; 1] such that (A) = inff(b); A B and B Borel setg and that (A [ B) = (A) + (B) whenever dist(a; B) > 0 (i.e., is a Borel regular metric outer measure). The last property is equivalent to the -additivity of on the family of all Borel sets, see [4], 2.3.2(9). The support of the measure over R n is dened by spt() = fx 2 R n ; (B(x; r)) > 0 for every r > 0g: Since in our situation we will always have X = spt() R n, we avoid the use of any special symbol for the space by adopting the following denition. Denition A measure over R n is said to be uniformly distributed if there is a function f : (0; 1)! [0; 1] such that (i) (B(x; r)) = f (r) for every x 2 spt() and r > 0, and (ii) f (r) < 1 for some r > 0. 1 Uniformly distributed measures in R n Lemma 1 Let be a uniformly distributed measure over R n, x 2 R n, and 0 < s < r < 1, then (B(x; r)) 5 n (r=s) n f (s). Proof. Let B(x; r) be a maximal set such that kz 1?z 2 k > s=2 whenever z 1 ; z 2 2 and z 1 6= z 2. Since the balls B(z; s=4), z 2, are mutually disjoint, by comparison of their volumes we infer that (s=4) n card (5r=4) n. Hence card 5 n (r=s) n and (B(x; r)) X z2 (B(z; s=2)) = f (s) card 5 n (r=s) n f (s) since (B(z; s=2)) = 0 in case when B(z; s=2)\spt() = ; and (B(z; s=2)) (B(x; s)) if x 2 B(z; s=2) \ spt(). Corollary 2 Let be a uniformly distributed measure over R n. Then 2

(i) f (r) < 1 for every r > 0, (ii) spt() is bounded if and only if (R n ) < 1, (iii) R R n exp(?skz? xk 2 ) d(z) = R 1 0 (B(x;p?(ln r)=s)) dr converges for all s > 0 and x 2 R n, and (iv) R R n exp(?skz?xk 2 ) d(z) = R R n exp(?skz?yk 2 ) d(z) whenever s > 0 and x; y 2 spt(). Proof. Since f (s) < 1 for some s > 0, (i) follows from Lemma 1. If spt() is unbounded then there is an innite set spt() such that kz 1? z 2 k > 2 whenever z 1 ; z 2 2 and z 1 6= z 2. Since (B(z; 1)) = f (1) > 0 for every z 2, we have (R n ) = 1. The formula in (iii) follows from Fubini's theorem (see, e.g., [8], Theorem 1.15), and the niteness statement follows by using Lemma 1 to estimate 1 0 (B(x; p?(ln r)=s)) dr 5 n 1 0 (?(ln r)=s) n=2 f (1) dr < 1: Finally, (iv) follows from the formula in (iii), since for x 2 spt(), 1 does not depend on x. 0 (B(x;?(ln r)=s)) dr = 1 0 f ( p?(ln r)=s) dr Theorem 3 Let be a uniformly distributed measure over R n and let x 0 2 spt(). For every x 2 R n and s > 0 let Then F (x; s) = R n [exp(?skz? xk 2 )? exp(?skz? x 0 k 2 )] d(z) : (i) F (x; s) is well-dened and nite for any x 2 R n and any s > 0; moreover, its denition is independent of the choice of x 0 2 spt(). (ii) x 2 spt() if and only if F (x; s) = 0 for every s > 0, and (iii) if x =2 spt() then there is an s x 2 (0; 1) such that F (x; s) < 0 for every s s x. 3

Proof. Statement (i) follows from Corollary 2, (iii) and (iv). If x 2 spt() and s > 0 then F (x; s) = 0 according to Corollary 2, (iv). If x =2 spt(), let " 2 (0; 1) be such that B(x; ") \ spt() = ;. Using Lemma 1, we estimate R n exp(?skz? xk 2 ) d(z) = 1X 1X k=1 k=1 (10) n 1 X k=1 B(x;(k+1)")nB(x;k") exp(?skz? xk 2 ) d(z) exp(?sk 2 " 2 )(B(x; (k + 1)")) exp(?sk 2 " 2 )(k + 1) n f ("=2) : Hence lim s!1 R R Rn exp(?skz? xk 2 ) d(z) R n exp(?skz? x 0 k 2 ) d(z) lim s!1 lim s!1 (10) n = 0: R R n exp(?skz? xk 2 ) d(z) exp(?s" 2 =4)f ("=2) 1X k=1 exp(?sk 2 (" 2? " 2 =4))(k + 1) n Consequently, there is an s x 2 (0; 1) such that for every s s x R n exp(?skz? xk 2 ) d(z) < 1 2 R n exp(?skz? x 0 k 2 ) d(z) ; which proves (iii) as well as the remaining implication in (ii). Theorem 4 Let be a uniformly distributed measure over R n. Then (i) spt() is an analytic variety, and (ii) there is an integer m 2 f0; 1; : : : ; ng, a constant c 2 (0; 1) and an open subset G of R n such that (1) G \ spt() is an m-dimensional analytic submanifold of R n, (2) R n n G is the union of countably many analytic submanifolds of R n of dimensions less than m, (R n n G) = H m (R n n G) = 0. (3) (A) = ch m (A \ G \ spt()) = ch m (A \ spt()) for every subset A R n. (H m denotes the m-dimensional Hausdor measure.) 4

Proof. (i) For a direct proof, let H(x) = R 1 1 exp(?s)f 2 (x; s) ds, where F is the function dened in Theorem 3. We easily see that H is a well-dened function on R n and, using Theorem 3 (iii) and (iv), that spt() = fx 2 R n ; H(x) = 0g. Moreover, H is an analytic function, since the function ~H taking at (x 1 ; : : : ; x n ) 2 C n the value 1 1 e?s "R n (exp(?s nx i=1 (z i? x i ) 2 ))? exp(?s nx i=1 (z i? x 0 i ) 2 )) d(z)# 2 ds is well-dened and holomorphic on the whole complex n-space C n. An alternative proof is obtained by noting that, for each s > 0, the set fx : F (x; s) = 0g is an analytic variety, and that the intersection of (an arbitrary number of) analytic varieties is again an analytic variety. (ii) Whenever U R n is an open set and U \ spt() is a nonempty k- dimensional analytic submanifold of R n, then the Besicovitch Dierentiation Theorem (see [4], 2.9.7) implies that the limit f(z) = lim r&0 (B(z; r)=( + H k spt())(b(z; r)) exists at ( + H k (A) = A spt())-almost every z 2 U and f(z) d( + H k spt())(z) for any Borel set A U : (*) (Here H k spt()(m) = H k (M \ spt()) for any set M.) Then clearly f(z) = lim r&0 f (r)=(f (r) + (k)r k ) for any z 2 U \ spt() where (k) denotes the volume of the k-dimensional unit ball. Thus c = lim f (r)=((k)r k ) r&0 exists; moreover c 6= 0 since otherwise (*) would imply (U) = 0 and c 6= 1 since otherwise (*) would imply H k (U \ spt()) = 0. Consequently, f = c=(c + 1) at ( + H k spt())-almost every point of U \ spt(), from which, using (*) once more, we infer that (A) = c H k spt()(a) for every Borel set A U. Since the statement (ii) is trivial if is the zero measure, we may assume that spt() 6= ;. Let G be a maximal open subset of R n such that G \ spt() is an analytic submanifold of R n. Since G \ spt() 6= ; (see [4], 3.4.8), we 5

infer from the above analysis that there is an integer m 2 f0; 1; : : : ; ng such that 0 < c = lim r&0 f (r)=((m)r m ) < 1. Since this integer m is determined uniquely, we conclude that G \ spt() is an m-dimensional analytic submanifold of R n. Therefore, according to [4], 3.4.8, R n n G is a countable union of analytic submanifolds of R n, all of dimension less than m. Hence H m (R n n G) = 0 and, since is absolutely continuous with respect to H m (see [4], 2.10.19(1)), (R n n G) = 0. This proves (2). The statement (3) is an easy corollary of (2) and of the analysis at the beginning of the proof of (ii). Theorem 5 Let be a uniformly distributed measure over R n with bounded support and let x 0 2 spt(). Then x 2 spt() if and only if P k (x) = for every natural number k 1. R n hz? x; z? xi k? hz? x 0 ; z? x 0 i k d(z) = 0 Proof. Because spt() is a compact set, F (x; s) = 1X k=1 (?s) k k! R n hz? x; z? xi k? hz? x 0 ; z? x 0 i k d(z) and hence F (x; s) = 0 for every s > 0 if and only if P k (x) = 0 for every k = 1; 2; : : :. Corollary 6 Let be a uniformly distributed measure over R n with bounded support. Then this support is an algebraic variety. Proof. According to Theorem 6, spt() = T 1 k=0 fx 2 Rn ; P k (x) = 0g. But each P k (x) is a polynomial in the coordinates of the point x. Hence Hilbert's theorem for polynomials over a Noetherian ring (see [1], chap.iii, x2.10, Corollary 2) implies that there are natural numbers k 1 ; : : : ; k l such that spt() = T l j=1 fx 2 Rn ; P k j(x) = 0g. Hence the support of the measure is the zero-set, for example, of the polynomial P (x) = P l j=1 P 2 kj (x). By considering the condition P 1 (x) = 0 we obtain the following result which is contained already in [3]. Proposition 7 Let be a uniformly distributed measure over R n with bounded support. Then this support is contained in some (n? 1)-dimensional sphere. 6

Proof. P 1 (x) = 0 means R n [hx; xi? 2hz; xi] d(z) = R n [hx 0 ; x 0 i? 2hz; x 0 i] d(z): We denote by c = R R n z d(z)=(r n ) the barycentr of the measure. Then hx; xi? 2hc; xi = hx 0 ; x 0 i? 2hc; x 0 i and consequently kx? ck 2 = kx 0? ck 2 for any x 2 spt(). Remark If spt() is a discrete, bounded set, then it is clearly nite. Let k denote its cardinality. Then spt() = fx 2 R n ; P m (x) = 0 for every m kg: This can be shown using the Newton identities for the sums of the m-th powers. 2 Uniformly distributed measures in R and R 2 Now we investigate uniformly distributed measures on the straight line and on the plane. Proposition 8 The support of a uniformly distributed measure over the straight line R is either the whole R or a discrete set. Proof. It suces to assume that ; 6= spt() 6= R and to show that spt() contains an isolated point. Let x 0 =2 spt(). Then there exists some x 1 2 spt() such that jx 0? x 1 j = minfjx 0? xj ; x 2 spt()g. We assume x 0 < x 1 ; the case x 0 > x 1 is similar. If x 1 is an accumulation point of spt() then there are x 2 ; x 3 2 spt() satisfying 0 < 2(x 2?x 1 ) < x 3?x 1 < x 1?x 0. Hence we can choose an " 2 (x 3? x 2 ; x 3? x 1 ). Because spt() \ [x 0 ; x 1 ) = ;, we obtain (B(x 1 ; ")) = ([x 1 ; x 1 + "]) ([x 1 ; x 2 + "]) (B(x 2 ; ")). That implies ((x 1 + "; x 2 + ")) = 0 but x 3 is an interior point of (x 1 + "; x 2 + "). This contradiction shows that x 1 must be an isolated point of spt() and hence that spt() is a discrete set. A simple combinatorial proof of the following lemma and of its corollary may be left to the reader. 7

Lemma 9 Let be an uniformly distributed measure over R with discrete and unbounded support. Then there are positive reals a; b and a bijection :! R mapping the integers onto spt() such that a (i + 1)? (i) = b for even i for odd i Corollary 10 There are three kinds of nonzero uniformly distributed measures over R. The rst kind are multiples of the Lebesgue measure by positive constants. The second kind are the unbounded and discrete uniformly distributed measures described in Lemma 9. The last kind are bounded and discrete uniformly distributed measures; by Proposition 8, such measures may be supported either by a single point or by a pair of points. The following proposition completely characterizes those uniformly distributed measures over the plane that have a bounded support. A number of examples of such measures with unbounded support can be also constructed, but a general description has not been found yet. Proposition 11 There are three kinds of nonzero uniformly distributed measures over R 2 with bounded support. (i) The support of the measure is a circle. (ii) The support is formed by the vertices of a regular polygon. (iii) The support is formed by the vertices of two regular n-gons having the same center and radius. Proof. Because of Proposition 7, we may assume that spt() fx 2 R 2 ; kx? ck = rg. Let : R! R 2 be dened by (t) = c + (r cos t; r sin t) and let g : R! R be such that g(t) = ((0; t]) for t 2 [0; 2) and g(t + 2) = g(t) + (R 2 ) for every t 2 R. Then g is the distribution function of some uniformly distributed measure over R and x 2 spt() if and only if (x) 2 spt(). The statement now follows from Corollary 10. Remark Clearly, a measure over R n is uniformly distributed provided it is locally nite and fullls the following homogeneity condition: For every pair x; y 2 spt() there is an isometry : R n! R n such that (x) = y and # =, where ( # )(A) = (?1 (A)). 8

The measures described in Corollary 10 and Proposition 11 are homogeneous in this sense. On the other hand, the three dimensional measure concentrated on the set f(x 1 ; x 2 ; x 3 ; x 4 ) 2 R 4 ; x 2 4 = x 2 1 + x 2 2 + x 2 3g is uniformly distributed (see [5] or [9]) but not homogeneous. It is unknown whether there are non-homogeneous uniformly distributed measures in R 2 or R 3, or in any R n with bounded support or with discrete support. 3 An application: Proof of Marstrand's Theorem Using our results we will present here an alternative proof of one of the main results in [6] which roughly says that the \density dimension" is always an integer. Theorem 12 Let be a metric outer measure over R n, n 1, s a nonnegative real number and let the sets B A R n satisfy (i) (B) > 0, and (ii) lim r&0 r?s (B(x; r)\a) exists and is positive and nite for any x 2 B. Then s is an integer. Proof. Using standard regularization methods, the measurability of both x! lim sup r&0 r?s (B(x; r) \ A) and x! lim inf r&0 r?s (B(x; r) \ A), and (ii) we may in addition assume that ~ = A is Borel regular (i.e., a measure) and locally nite in some open superset U of A and that B is a compact set on which the functions x! k s ~(B(x; 1 )) uniformly converge, as k tends to innity, to a continuous function d : B! (0; 1). The k Besicovitch Dierentiation theorem ([4], 2.9.7) ensures the existence of an x 0 2 B with lim r&0 ~(B(x 0 ; r) n B)=~(B(x 0 ; r)) = 0. For k 1 we dene ~ k (M) = k s ~(x 0 + 1M) and B k k = k (B? x 0 ). Since lim k!1 ~ k (B(0; i)) = d(x 0 )i s for any integer i 0, we infer from well-known weak compactness results for measures, see e.g., [8], Theorem 1.23 or [9], Proposition 1.12, that there is a sequence k i % 1 of integers and a Radon measure over R n such that (C) lim sup i!1 ~ k i(c) for any compact set C and (G) lim inf i!1 ~ k i(g) for any open set G. We show that is a nonzero 9

uniformly distributed measure with f (r) = r s d(x 0 ). Then it easily follows that s = m for the integer m from Theorem 4 (ii). This will nish our proof. Since (B(0; 1)) d(x 0 ) > 0, we have ; 6= spt(). Let x 2 spt(), R > 0 and x a positive " < R. Then 0 < (B(x; "=2)) lim inf i!1 ~ k i(b(x; ")). According to the choice of x 0 we have lim i!1 ~ k i(b(0; kxk+2")nb k i) = 0 and therefore for any i suciently large we may nd a point y i 2 B(x; ") \ B k i. Consequently, we can estimate (B(x; R)) lim sup ~ k i(b(x; R)) lim sup ~ k i(b(y i ; R? ")) i!1 i!1 = lim sup k s i ~(B(x 0 + y i =k i ; (R? ")=k i )) d(x 0 )(R? ") s i!1 since all (x 0 + y i =k i ) belong to B and the \average density" uniformly converges on B. Similarly, one estimates that the measure of the interior of the ball B(x; R + ") is d(x 0 )(R + 2") s and, tending with " to zero, we infer that (B(x; R)) = d(x 0 )R s as required. References [1] N.Bourbaki, Elements de mathematique, Fascicule XXVIII, Algebre Commmutative, Hermann, Paris 1961. [2] J.P.R.Christensen, On some measures analogous to Haar measures, Mathematica Scandinavica, 26(1970), 103 106. [3] J.P.R.Christensen, Uniform measures and spherical harmonics, Mathematica Scandinavica, 26(1970), 293 302. [4] H.Federer, Geometric measure theory, Grundlehren der math. Wissenschaften 153, Springer 1969. [5] O.Kowalski and D.Preiss, Besicovitch-type properties of measures and submanifolds, Jour. reine, angew. Math., 379(1987), 115 151. [6] J.M.Marstrand, The (; s)-regular subsets of n space, Trans. Amer. Math. Soc., 113(1964), 369 392. [7] P.Mattila, Dierentiation of measures in uniform spaces, in Lecture Notes in Mathematics 794, Springer 1979. 10

[8] P.Mattila, Geometry of sets and measures in Euclidean Spaces, Cambridge University Press 1995. [9] D.Preiss, Geometry of measures in R n : Distribution, rectiability and densities, Annals Math., 125(1987), 537{643. [10] M.Studeny, Dierentiations of measures in metric spaces, Thesis, Prague 1981 Max-Planck-Institute fur Mathematik in den Naturwissenschaften, Inselstrae 22-26, D-04103 Leipzig, Germany 1 and Department of Mathematics, University College London, Gower Street, London WC1E 6BT, UK 1 On leave from UAM MFF UK, Mlynska dolina, 842 15 Bratislava, Slovakia 11