Purely unrectifiable sets ~~~~~~~~~~~~~~~~~~~~~~~~~ .. index:: Purely unrectifiable set .. prf:definition:: Purely unrectifiable set :label: purely-unrectifiable A :math:\mathcal H^n-measurable set :math:S\subset \mathbb R^m is purely :math:n-unrectifiable if, for all :math:n-rectifiable :math:E\subset \mathbb R^n, :math:\mathcal H^n(S\cap E)=0. .. prf:lemma:: The four corner Cantor set :math:K\subset \mathbb R^2 is purely 1-unrectifiable. .. prf:proof:: Observe that the coordinate projections of :math:K have :math:\mathcal L^1 measure zero (see :prf:ref:cantor-projections). If there existed a rectifiable :math:\gamma\subset \mathbb R^2 with :math:\mathcal H^1(\gamma\cap K)>0, then :math:\gamma\cap K is a 1-rectifiable set of positive measure and hence, by :prf:ref:projection-of-rectifiable-cor, one of the coordinate projections must have positive measure, a contradiction. For a second proof see :prf:ref:cantor-pu. .. prf:lemma:: :label: rect-decomp Let :math:A\subset \mathbb R^m be :math:\mathcal H^m-measurable with :math:\mathcal H^m(A)<\infty. There exists a decomposition :math:A=E\cup S with :math:E :math:n-rectifiable and :math:S purely :math:n-unrectifiable. .. prf:proof:: Let .. math:: t = \sup\{\mathcal H^n(E): E\subset A, n\text{-rectifiable}\}. Since :math:\mathcal H^n(A)<\infty, :math:t<\infty. Let :math:E_i\subset A be :math:n-rectifiable with :math:\mathcal H^n(E_i)\to t. Then :math:E=\bigcup_{i\in \mathbb N}E_i is :math:n-rectifiable and is contained in :math:A. Therefore .. math:: t\geq \mathcal H^n(E)\geq \mathcal H^n(E_i)\to t and so :math:\mathcal H^n(E)=t. Then :math:S=A\setminus E is purely :math:n-unrectifiable. Indeed, if :math:f\colon \mathbb R^n\to \mathbb R^m is Lipschitz, :math:E'=(A\setminus E)\cap f(\mathbb R^n), is :math:\mathcal H^m-measurable and so .. math:: t\geq \mathcal H^n(E\cup E')=\mathcal H^n(E)+\mathcal H^n(E')= t +\mathcal H^n(E'). We now state a very important theorem on the structure of purely unrectifiable sets. It requires the notion of a natural measure :math:\gamma_{n,m} on :math:G(n,m) that is invariant under the action of :math:SO(m). There are several ways to construct this measure. The simplest is to consider :math:G(n,m) as a (compact) metric space equipped with the metric .. math:: d(V,W)=\|\pi_V-\pi_W\|. Then :math:\gamma_{n,m} is given by (a scalar multiple of) :math:\mathcal H^{n(m-n)}. We will not discuss the specific details of this measure. When :math:n=1, we may identify :math:G(1,m) with :math:\mathbb S^{m-1}. In this case, :math:\gamma_{1,m} is simply :math:\mathcal H^{m-1}. .. index:: Besicovitch-Federer projection theorem .. prf:theorem:: Besicovitch--Federer projection theorem Let :math:S\subset \mathbb R^m be purely :math:n-unrectifiable with :math:\mathcal H^n(S)<\infty. Then, for :math:\gamma_{n,m}-a.e. :math:V\in G(n,m), .. math:: \mathcal L^n(\pi_V(S))=0. Conversely, if :math:E\subset \mathbb R^m is purely :math:n-unrectifiable with :math:\mathcal H^n(E)>0, for :math:\gamma_{n,m}-a.e. :math:V\in G(n,m), .. math:: \mathcal L^n(\pi_V(E))>0. .. prf:remark:: The converse statement is given by :prf:ref:projection-of-rectifiable-cor. We will prove the projection theorem for :math:n=1 and :math:m=2, which was proved by Besicovitch. First we prove some preliminary geometric properties of purely unrectifiable sets. .. prf:lemma:: :label: basic-rect-crit Let :math:E\subset \mathbb R^m, :math:V\in G(m-n,m) and :math:0 3h(x)/4 - h(x)/2\\ &\geq \|\pi_{V^\perp}(x-z)\|/s. That is, :math:z belongs to the first set in :eq:cover-by-cones. By :eq:cover-by-cones and :eq:hyp-cone-density, .. math:: \mathcal H^1(S\cap C_x) \leq 2\lambda (2h(x)s)^n. We apply :prf:ref:5r to the balls .. math:: \pi_{V^\perp}(B(x,sh(x)/20)) with :math:x\in S. This gives countably many :math:x_i\in S, for which these balls are disjoint, and .. math:: S \subset \bigcup_{i\in\mathbb N}C_{x_i}. Therefore .. math:: \mathcal H^n(S) &\leq \sum_{i\in\mathbb N} \mathcal H^n(C_{x_i})\\ &\leq 2\lambda 2^n \sum_{i\in\mathbb N} (sh(x_i))^n\\ &= 2 \lambda 2^n 20^n \sum_{i\in\mathbb N} \left(\frac{sh(x_i)}{20}\right)^n. But, the :math:\pi_{V^\perp}(B(x_i,sh(x_i)/20)) are disjoint subsets of :math:B(\pi_{V^\perp}(a),\delta/2)\subset V^\perp and so the final sum is bounded above by :math:(\delta/2)^n. .. prf:corollary:: :label: pu-density-cones If :math:S\subset \mathbb R^m is purely :math:n-unrectifiable with :math:\mathcal H^n(S)<\infty then for every :math:V\in G(m-n,m), every :math:00, there exists an :math:R>0 such that the set :math:S'' of those :math:x\in S' with .. math:: \sup_{0__ implies :math:\mathcal H^n(S'')=0, a contradiction. For the remainder of the course, we will prove the Besicovitch projection theorem. .. prf:theorem:: Besicovitch projection theorem :label: besicovitch Let :math:S\subset \mathbb R^2 be purely 1-unrectifiable with :math:\mathcal H^1(S)<\infty. Then for :math:\mathcal H^1-a.e. :math:e \in \mathbb S^1, .. math:: \mathcal L^1(\pi_e(S))=0. From :prf:ref:pu-density-cones, we see that a purely unrectifiable set has many radiating out of almost every point in all directions at almost every point. We now precisely describe two ways in which this can occur. .. prf:definition:: Let :math:S\subset \mathbb R^2 and :math:x\in S. For :math:e\in \mathbb S^1 let :math:l_e(x) be the half line :math:x+[0,\infty)e and for :math:I\subset \mathbb S^1, let :math:C(I,x) be the cone :math:\bigcup_{e\in I}l_e(x). For :math:r>0 let :math:H_x(r) be those :math:e\in \mathbb S^1 for which .. math:: |K \cap l_e(x)\cap B(x,r)| \geq 2. That is, :math:S\cap l_e(x)\cap B(x,r) contains *another* point of :math:K. Also let :math:H_x=\bigcap_{r>0} H_x(r), the directions that contain other points of :math:S arbitrarily close to :math:x. For :math:e\in\mathbb S^1, we let :math:H_e be those :math:x\in S for which :math:e\in H_x. For :math:R,M,\epsilon>0 let :math:D_x(R,M,\epsilon) be those :math:e\in \mathbb S^1 for which there exists :math:00} D_x(R,M,\epsilon). For :math:e\in\mathbb S^1, we let :math:D_e be those :math:x\in S for which :math:e\in D_x. The main step in proving :prf:ref:besicovitch is the following. .. prf:proposition:: :label: besicovitch-main Let :math:S\subset \mathbb R^2 be purely 1-unrectifiable with :math:\mathcal H^1(S)<\infty. For :math:\mathcal H^1-a.e. :math:x\in S, :math:\mathcal H^1(\mathbb S\setminus H_x\cup D_x)=0. Before proving :prf:ref:besicovitch-main, we will demonstrate how it is used to prove :prf:ref:besicovitch. .. prf:lemma:: Special case of the coarea formula :label: coarea For any :math:e\in \mathbb S^1 and any compact :math:K\subset \mathbb R^2, .. math:: \int_{\mathbb R}\operatorname{card}(K\cap l_e(t)) \, \mathrm{d}t \leq \mathcal H^1(K). In particular, if :math:\mathcal H^1(K)<\infty then for any :math:e\in \mathbb S^1, :math:\mathcal L^1(\pi_{e^\perp}(H_e))=0. .. prf:proof:: Since :math:K is compact, .. math:: f(t) = \operatorname{card}(K\cap l_e(t)) is a Borel function. Indeed, if, for :math:\delta>0, .. math:: f_\delta(t) = \max\{n \in \mathbb N: \exists x_1,\ldots,x_n \in K \cap l_e(t) \text{ with } \|x_i-x_j\|\geq \delta\ \forall 1\leq i\neq j\leq n\} then :math:f_\delta monotonically increases to :math:f as :math:\delta\to 0. Since :math:K is compact, the :math:f_\delta are lower semi-continuous. Therefore, by the monotone convergence theorem, it suffices to bound the integral of each :math:f_\delta. Fix :math:\delta>0 and cover :math:K by sets :math:E_1,E_2,\ldots with :math:\operatorname{diam}E_i <\delta such that .. math:: \sum_{i\in\mathbb N} \operatorname{diam}E_i \leq \mathcal H^1(K) +\delta. Note that .. math:: f_\delta(t) \leq \operatorname{card}(\{i: E_i\cap l_e(t) \neq\emptyset\}). Therefore .. math:: \int_\mathbb Rf_\delta \, \mathrm{d}\mathcal L^1 &\leq \int_\mathbb R\sum_{i\in\mathbb N} \chi_{\{(i,t):E_i\cap l_e(t) \neq\emptyset\}}\\ &= \sum_{i\in\mathbb N} \int_\mathbb R\chi_{\{(i,t):E_i\cap l_e(t) \neq\emptyset\}}\\ &\leq \sum_{i\in\mathbb N} \operatorname{diam}E_i\\ &\leq \mathcal H^1(K) +\delta, as required. .. prf:lemma:: :label: density-projection Let :math:S\subset \mathbb R^2 be :math:\mathcal H^1-measurable with :math:\mathcal H^1(S)<\infty. Then for any :math:e\in \mathbb S^1, :math:\mathcal L^1(\pi_{e^\perp}(D_e))=0. .. prf:proof:: Fix :math:e\in\mathbb S^1. For any :math:M\in \mathbb N and :math:t \in \pi_{e^\perp}(D_e) there exists an :math:x\in D_e, :math:r_x>0 and an interval :math:e\in I_x\subset \mathbb S^1 with :math:\operatorname{diam}I < 1/10 such that .. math:: \mathcal H^1(S\cap C(x,I)\cap B(x,r)) \geq M \mathcal H^1(rI). Apply :prf:ref:5r to the intervals :math:J_x=\pi_{e^\perp}(C(x,I_x)\cap B(x,r_x)) to obtain a disjoint collection :math:J_{x_1},J_{x_2},\ldots \subset \mathbb R such that .. math:: D_e \subset \bigcup_{i\in\mathbb N} 5J_{x_i}. Therefore .. math:: \mathcal L^1(D_e) &\leq \sum_{i\in\mathbb N} 5\mathcal L^1(J_{x_i})\\ &\leq 5\sum_{i\in \mathbb N} \mathcal H^1(rI_{x_i})\\ &\leq \frac{5}{M} \sum_{i\in \mathbb N} \mathcal H^1(S\cap C(x_i,I_{x_i}) \cap B(x_i,r_{x_i}))\\ &\leq \frac{5}{M} \mathcal H^1(S), where the final inequality follows from the disjointness of the sets :math:C(x_i,I_{x_i})\cap B(x_i,r_i), :math:i\in\mathbb N. Since this is true for all :math:M\in\mathbb N, :math:\mathcal L^1(D_e)=0. Proof of :prf:ref:besicovitch using :prf:ref:besicovitch-main. .. prf:proof:: By the inner regularity of :math:\mathcal H^1, it suffices to prove the result for compact :math:S. By definition, we have .. math:: \{(x,e)\in S\times \mathbb S^1: e\not\in H_x\cup D_x\}=\{(x,e): x\not\in H_e \cup D_e\}. :prf:ref:besicovitch-main implies that the left hand expression has :math:\mathcal H^1\times \mathcal H^1-measure zero and so Fubini’s theorem implies that, for :math:\mathcal H^1-a.e. :math:e\in\mathbb S^1, :math:\mathcal H^1(S\setminus H_e\cup D_e)=0. Therefore, by :prf:ref:density-projection and :prf:ref:coarea, :math:\pi_{e^\perp}(S)=0 for :math:\mathcal H^1-a.e. :math:e\in \mathbb S^1.  Proof of :prf:ref:besicovitch-main. .. prf:proof:: Fix :math:R,M,\epsilon>0 and :math:x\in S which satisfies the conclusion of :prf:ref:pu-density-cones. That is, .. math:: :label: cor-high-dens-cone \Theta^{*,1}(S\cap C(x,I),x) \geq c_0 \mathcal H^1(I) for every interval :math:I\subset \mathbb S^1. It suffices to show that :math:\mathcal H^1(\mathbb S^1\setminus H_x(R)\cup D_x(R,M,\epsilon)) =0. In fact, we will show that, for *any* :math:e\in \mathbb S^1, .. math:: \Theta^{*,1}(H_x(R),e)>0 \quad \text{or} \quad \Theta_*^1(D_x(R,M,\epsilon),e)>0, from which the result follows by the Lebesgue density theorem. To this end, fix :math:e\in \mathbb S^1 with :math:\Theta^{*,1}(H_x(R),e)=0. Then for all sufficiently small intervals :math:I with :math:e\in I \subset \mathbb S^1, .. math:: :label: high-dense-ball \mathcal H^1(H_x(R)\cap I) < c_0 \mathcal H^1(I)/4M. Fix such an :math:I. By :eq:cor-high-dens-cone, there exists :math:r0 as required. By :eq:small-density-h we may cover :math:H_x(r)\cap I by disjoint intervals :math:I_1,I_2,\ldots with .. math:: \sum_{i\in \mathbb N} \mathcal H^1(I_i)< c_0\mathcal H^1(I)/4M (indeed, the disjointness of the intervals is shown in :prf:ref:hd-disjoint). By the definition of :math:H_x(r), we know that .. math:: :label: contained-in-Hx S\cap C(x,I) \cap B(x,r) \subset \bigcup_{i\in\mathbb N} S\cap C(x,I_i)\cap B(x,r). Let :math:\mathcal G be those :math:i\in\mathbb N with .. math:: :label: high-density-cone \frac{\mathcal H^1(S\cap C(x,I_i)\cap B(x,r))}{r} \geq M\mathcal H^1(I_i). Note that by :eq:high-dense-use and :eq:contained-in-Hx, .. math:: \sum_{i\in \mathcal G} \frac{\mathcal H^1(S\cap C(x,I_i)\cap B(x,r))}{r} &\geq \frac{\mathcal H^1(S\cap C(x,I)\cap B(x,r))}{r} - \frac{c_0 \mathcal H^1(I)}{4}\\ &\geq \frac{3c_0 \mathcal H^1(I)}{4}. That is, the cones associated to :math:G:=\bigcup_{i\in\mathcal G}I_i cover a large proportion of :math:S\cap C(x,I) \cap B(x,r). Moreover, :math:G\subset D_x(R,M,\epsilon), because if :math:\xi\in G then :math:\xi \in I_i for some :math:i\in \mathcal G which satisfies the definition of :math:D_x(R,M,\epsilon). Thus, to show :eq:besicovitch-goal, it would be enough to bound :math:\mathcal H^1(G) from below by a multiple of :math:\mathcal H^1(I). But this is not necessarily true: the intervals that form :math:G could be extremely thin compared to :math:I. To accommodate this, we enlarge the intervals :math:I_i with :math:i\in\mathcal G as follows. For each :math:i\in \mathcal G enlarge :math:I_i until :eq:high-density-cone becomes an equality or until :math:I_i intersects another :math:I_j. If the first possibility occurs then we still have :math:I_i\subset D_x(R,M,\epsilon). If the second possibility occurs then we merge the two intervals; since both sides of :eq:high-density-cone are linear in :math:I, and the boundary of each :math:C(x,I_i) contains no points of :math:S, :eq:high-density-cone remains true after the merge. By the same reasoning, both sides of :eq:high-density-cone are continuous under expanding :math:I and consequently one of the two possibilities must occur. This results in a disjoint collection of intervals :math:\tilde I_i for which :eq:high-density-cone is an equality. Moreover, .. math:: G= \bigcup_{i\in \mathcal G}I_i \subset \bigcup \tilde I_i and, by construction, each :math:\tilde I_i \in D_x(R, M, \epsilon). Therefore .. math:: \mathcal H^1(D_x(r_0,\epsilon,M) \cap I) &\geq \sum \mathcal H^1(\tilde I_i)\\ &= \sum \frac{\mathcal H^1(S\cap B(x,r) \cap C(x,\tilde I_i))}{rM}\\ &\geq \sum_{i\in\mathcal G}\frac{\mathcal H^1(S\cap B(x,r) \cap C(x,I_i))}{rM}\\ &\geq \frac{3c_0\mathcal H^1(I)}{4M}. Exercises ^^^^^^^^^ .. prf:example:: :label: cantor-projections Prove that the coordinate projections of the four corner Cantor set :math:K\subset \mathbb R^2 have Lebesgue measure zero. .. prf:example:: :label: cantor-pu A second proof that the four corner Cantor set is purely 1 unrectifiable. Suppose that :math:f\colon \mathbb R\to \mathbb R^2 satisfies :math:\mathcal H^1(f(\mathbb R)\cap K)>0. #. Prove that there exists :math:x\in \mathbb R that is a density point of :math:f^{-1}(K) such that :math:f'(x)\neq 0. #. Therefore, for sufficiently small :math:r, :math:f(x-r,x+r) is approximated by a line segment of length :math:2rf'(x) that is mostly contained in :math:K. Derive a contradiction. .. prf:example:: Prove that the decomposition given in :prf:ref:rect-decomp is unique up to :math:\mathcal H^n-null sets. .. prf:example:: Think about :prf:ref:pu-density-cones and :prf:ref:besicovitch-main in regard to the four corner Cantor set. .. prf:example:: Let :math:K\subset \mathbb R^2 be compact. Show that .. math:: \{(x,e): e \in H_x\} \quad \text{and} \quad \{(x,e): e \in D_x\} are Borel subsets of :math:K\times \mathbb S^1. .. prf:example:: :label: hd-disjoint Show that in the definitions of :math:\mathcal L^n and :math:\mathcal H^n we may suppose the covering intervals, respectively sets, may be chosen to be disjoint. .. prf:example:: Open problem For :math:k\geq 2, does there exist a compact purely 1-unrectifiable :math:S\subset \mathbb R^2 with :math:\mathcal H^1(S)>0 that intersects every line in at most :math:k` points?