Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 28, 2003, 415–431

A GENERALISED CONICAL DENSITY THEOREM FOR UNRECTIFIABLE SETS

Andrew Lorent

Scuola Normale Superiore, Piazza dei Cavalieri 7, I-56126 Pisa, Italy Current address: Mathematical Institute 24–29 St Giles, Oxford OX1 3LB, UK; lorent@maths.ox.ac.uk

Abstract. We provide a strengthening of an elementary technique in geometric measure theory. Given an s-set S ⊂ R^{n}, in the language of tangent measures, this technique establishes the existence of tangent measures to the measure H_{⌊S}^{s} on one side of an (n − 1)-plane.

If S is purely unrectifiable or of dimension less than n − 1, our strengthening consists of being able to find tangent measures on one side of an (n − 1)-plane containing a vector of our choice. We give an application of this result to the problem of characterising removable sets for harmonic functions.

1. Introduction

In this note we give a strengthening of a well-known technique in the subfield of geometric measure theory known as “Rectifiability and densities”. Arguments using this technique are usually called “touching point arguments”, they are used to establish conical density theorems of the following type: Given set S, we can (roughly speaking) show that for almost all points x ∈ S, there exists a sequence of radii r_{n} → 0 such that B_{r_{n}}(x) ∩ S lies mostly on one side of an (n − 1)-plane intersecting x. This essentially follows from the existence of large empty sub-balls in any ball centered on any point of S, which in turn follows from basic density estimates. The technique originates from work of Besicovitch [1], and has been used in an essential way in [6], [8] and [12].

For those familiar with the language of tangent measures, these arguments establish the existence of a tangent measure to the measure H_{⌊S}^{s} that exists on one side of an (n − 1)-plane. For sets S that are unrectifiable or of dimension less than n − 1, our strengthening consists of being able to show that B_{r_{n}}(x) ∩ S lies increasingly on one side of an (n − 1)-plane containing a vector in S^{n−1} of our choice.

This result comes from a reworking of the basic idea using Besicovitch–Federer projection theorem instead of density estimates. The proof is intricate, but in our opinion the method is classical. This work arose from the author’s attempts to generalise rectifiability and density theorems outside Euclidean space, [5]. Our main lemma is an n-dimensional version of Lemma 14 from [5]. Touching point arguments have application in a variety of geometrical problems. We hope the results of this note might also be of some utility; by way of example, we apply the theorem to the results of Mattila and Paramonov, to give a strengthening (under more restrictive hypotheses) of one of their theorems related to the problem of characterising removable sets for harmonic functions.

2000 Mathematics Subject Classification: Primary 28A75.

First some notation. Let B_{r}(z) := {y ∈ R^{n} : |y − z| < r}. Let H^{s} denote Hausdorff s-measure. Let P_{V} be the orthogonal projection onto subspace V .

We define the cone X(x, v, α) = {y ∈ R^{n} : α|P_{v^{⊥}}(y − x)| ≤ |P_{⟨v⟩}(y − x)|}; see Figure 1. Let X^{+}(x, v, α) = {y ∈ R^{n} : y ∈ X(x, v, α) and (y − x) · v > 0}.

Theorem 1. Given an integer n ≥ 3 and real numbers s ∈ (0, n − 1] and ϱ > 0, let S ⊂ R^{n} be a set (purely unrectifiable in the case s = n − 1) of positive finite H^{s} measure.

Then for any ψ ∈ S^{n−1} we have that for H^{s} a.e. x ∈ S

(1)  lim inf_{r→0} \frac{H^{s}(B_{r}(x) ∩ S ∩ X^{+}(x, ϕ, ϱ))}{r^{s}} = 0

for some ϕ ∈ ψ^{⊥} ∩ S^{n−1} .

Note that for a set S ⊂ R^{2} that is purely unrectifiable or of dimension s < 1 a much stronger result holds: Given ϕ ∈ S^{1} and ϱ > 0, for H^{s} a.e. x ∈ S equation (1) holds true. This can be seen by simplified versions of the arguments of this paper. However in higher dimensions this stronger result is not true, a counter example is given by taking the cartesian product of a purely unrectifiable set with a interval.

1.1. Application: Removable sets for harmonic functions. A compact set E ⊂ C is said to be removable for bounded analytic functions if and only if the following holds true:

If U ⊂ C is open, E ⊂ U and f: U\E → C is a bounded analytic function, then f has an analytic extension to U.

The long standing problem of geometrically characterising removable sets was resolved for sets of finite H^{1} measure in [2], building on work of [3] and [10]. It is known and relatively easy to show that (see [9]) sets of zero H^{1} measure are removable and sets of Hausdorff dimension greater than 1 are not removable. So the characterisation is needed for sets of dimension 1. The complete characterisation for sets of dimension 1 (i.e. including sets of infinite H^{1} measure) was achieved in [13].

First some notation: An m-rectifiable set S in R^{n} is a set that can be covered by countably many C^{1} submanifolds of dimension m. A purely m-unrectifiable set T is a set with the property; H^{m}(T ∩ M) = 0 for all C^{1} m-dimensional submanifolds M.

It was proved in [2] that if S has finite H^{1} measure and is not removable then S must have a 1-rectifiable subset with positive H^{1} measure. The converse is well known, so sets of finite H^{1} measure are removable if and only if they are purely 1-unrectifiable. Analogous questions can be asked about harmonic functions.

Following [11] a compact subset E of R^{n} is called Lip_{1} -removable for harmonic functions, (abbreviated L_{1}RH) if for each domain D ⊂ R^{n} every locally Lipschitz function f: D → R which is harmonic in D\E is harmonic in D.

It would be reasonable to conjecture that E is Lip_{1} -removable if and only if it is purely (n − 1) rectifiable.

The only results on this conjecture are from [11] (Theorem 5.5) where it is proved:

Theorem 2 (Mattila–Paramonov). Let X ⊂ R^{n} be a compact set of finite H^{n−1} measure such that for some constant A > 0,

(2)  H^{n−1}(X ∩ B_{r}(a)) ≤ Ar^{n−1}

for a ∈ R^{n} and r > 0. Suppose that the following holds at H^{n−1} almost all points a ∈ X: For every v ∈ S^{n−1} there is a δ > 0 such that

(3)  lim inf_{r→0} r^{1−n}H^{n−1}(y ∈ X ∩ B_{r}(a) : |(y − a) · v| > δ|y − a|) > 0.

Then X is Lip_{1} -removable.

Given a set S of positive finite H^{n−1} measure, if a ∈ S has the property that

(4)  lim inf_{r→0} r^{1−n}H^{n−1}(y ∈ S ∩ B_{r}(a) : |(y − a) · v| > δ|y − a|) = 0

for any δ > 0 then we say S has a weak tangent V := v^{⊥} at a. Existence of weak tangents is a well-known partial result to rectifiability (see [7] where the term was coined); if the lim inf of (4) is replaced by a lim sup, then this condition holding for almost every point a ∈ S and for any δ > 0 is equivalent to (n − 1)-rectifiability, see [9].

So Theorem 2 says that any set (with density bound (2)) which is not Lip_{1} - removable must have a subset B of positive H^{n−1} measure such that for each a ∈ B there exists v ∈ S^{1} with condition (4) holding for every δ > 0. Informally: a non Lip_{1} removable set must have weak tangents on some subset of positive measure. This result suggests that these sets should be rectifiable.

Suppose the conjecture for harmonic functions is not true for sets with density bound (2) and so we have a purely (n − 1) unrectifiable set X which is not Lip_{1} -removable satisfying (2). Then by directly inserting the statement of Theorem 1 into the proof of Theorem 5.5 of Mattila–Paramonov [11] (instead of their Lemma 5.2) we have the following result:

Theorem 3. Suppose S ⊂ R^{n} is a purely unrectifiable (n − 1)-set which is not Lip_{1} -removable and has the property: For some constant A > 0

(5)  H^{n−1}(S ∩ B_{r}(a)) ≤ Ar^{n−1}

for all a ∈ R^{n}, r > 0.

Then there must exist a subset B ⊂ S of positive H^{n−1} measure such that for any ψ ∈ S^{n−1} we have that for almost all x ∈ B the set S must have a weak tangent V at x with the property that ψ ∈ V .

So if there exists a purely unrectifiable (n − 1)-set S with density bound (5) which is not Lip_{1} removable then it must have a subset B of positive measure such that for almost every point z ∈ B there exists an infinite collection of (n−1) subspaces {V_{1}^{z}, V_{2}^{z}, …} each of which is a weak tangent to S at z.

Acknowledgement. I would like to thank the referee for reading the paper with great care and in particular for pointing out a hypothesis in Theorem 1 was not necessary.

2. Background and notation

First we will need to introduce some more notation. If x ∈ v^{⊥}, v ∈ S^{n−1} , s > 0, let K(x, v, s) := P_{v^{⊥}}^{−1}(B_{s}(x) ∩ v^{⊥}) be the standard definition of a cylinder.

We will also need to define a kind of double cone object, see Figure 1. Given x ∈ R^{n}, v ∈ S^{n−1} , s ∈ (0, 1) and r > 0 define:

Ψ^{u}(x, v, s, r) := {z ∈ B_{r}(x) : |P_{v^{⊥}}(z − x)| ≤ s|P_{v}(z − (x + rv))|},

Ψ^{d}(x, v, s, r) := {z ∈ B_{r}(x) : |P_{v^{⊥}}(z − x)| ≤ s|P_{v}(z − (x − rv))|},

and Ψ(x, v, s, r) = Ψ^{u}(x, v, s, r) ∩ Ψ^{d}(x, v, s, r). Let A(x, α, β) := B_{β}(x)\\overline{B_{α}(x)}.

3. Proof

We will first prove that Theorem 1 holds true for a.e. ψ ∈ S^{n−1} (in the sense of H^{n−1} measure on S^{n−1} ), and then show how this implies the result for every ψ ∈ S^{n−1} .

Suppose the statement of Theorem 1 was false for some set Y ⊂ S^{n−1} of positive measure. Then there must exist ψ ∈ Y for which the following two statements hold true:

– We can take some subset S_{0} ⊂ S such that for some ϱ > 0 and some λ_{1} > 0 we have for all ϕ ∈ ψ^{⊥} ∩ S^{n−1}

lim inf_{r→0} \frac{H^{s}(B_{r}(x) ∩ X^{+}(x, ϕ, ϱ) ∩ S)}{r^{s}} > 2λ_{1},

for all x ∈ S_{0} .

– By the fact that S is unrectifiable or of dimension less than n − 1, from the Besicovitch–Federer projection theorem ([9, Theorem 18.1]) (or in the latter case by an elementary result, ([9, Theorem 7.5])) we have

(6)  L^{2}(P_{ψ^{⊥}}(S)) = 0.

Now we can take a closed subset S_{1} ⊂ S_{0} and some small r_{0} > 0 such that

(7)  \frac{H^{s}(B_{r}(x) ∩ X^{+}(x, ϕ, ϱ) ∩ S)}{r^{s}} > λ_{1}

for all ϕ ∈ ψ^{⊥} ∩ S^{n−1} , x ∈ S_{1} , r ∈ (0, r_{0}). And (using [9, Theorem 6.2] for the upper bound on the density)

(8)  λ_{1} < \frac{H^{s}(B_{r}(x) ∩ S)}{r^{s}} < 2

for all x ∈ S_{1} , r ∈ (0, r_{0}).

First we prove the following.

Lemma 1. Given an integer n ≥ 3 and a real number s ∈ (0, n) suppose we have a set S ⊂ R^{n} of positive finite H^{s} measure and a closed subset S_{1} ⊂ S with the following two properties:

Firstly, there exists a vector ψ ∈ S^{n−1} and a number r_{0} > 0 such that for some small λ_{1} > 0 and ϱ > 0 we have

\frac{H^{s}(B_{r}(x) ∩ X^{+}(x, ϕ, ϱ) ∩ S)}{r^{s}} > λ_{1}

for all ϕ ∈ ψ^{⊥} ∩ S^{n−1} and all x ∈ S_{1} , r ∈ (0, r_{0}).

Secondly, we have

(9)  λ_{1} < \frac{H^{s}(B_{r}(x) ∩ S)}{r^{s}} < 2

for all x ∈ S_{1} , r ∈ (0, r_{0}).

Then we can find constants κ_{ϱ}^{λ_{1}} > 0 and ϑ_{ϱ}^{λ_{1}} > 0 such that the following statement holds true: Suppose x ∈ S_{1} and d ∈ (0, r_{0}) is such that

(10)  \frac{H^{s}(B_{4d}(x)\S_{1})}{d^{s}} ≤ ε

then for all z ∈ (ψ^{⊥} + x) ∩ K(x, ψ, κ_{ϱ}^{λ_{1}}d) we have that

K(z, ψ, ε^{1/s}ϑ_{ϱ}^{λ_{1}}d) ∩ S_{1} ∩ B_{2d}(x) ≠ ∅.

Proof. Firstly to simplify the expressions, we let μ := H_{⌊S}^{s}. Let r ∈ (0, r_{0}). We start by showing the following:

If a ∈ ∂K(x, ψ, r) let W_{a} denote the (n − 1)-dimensional tangent plane of the boundary of the cylinder K(x, ψ, r) at point a. Let n_{a} ∈ W_{a}^{⊥} ∩ S^{n−1} be the unit normal pointing “inwards” towards the center of cylinder K(x, ψ, r).

Step 1: We will show that we have

(11)  X^{+}(a, n_{a}, ϱ) ∩ B_{2rϱ/(1+ϱ)}(a) ⊂ K(x, ψ, r).

To see this, it makes the calculations a lot easier if we change our orthonormal basis. Let ε_{1} = n_{a} and ε_{2} = ψ and complete this to get an orthonormal set of vectors {ε_{j}} where {ε_{2}, …, ε_{n}} span W_{a}. Let c = a + rε_{1} . See Figure 2. Now in our new basis (using point c) the definition of the cylinder is as follows

K(x, ε_{2}, r) = {z ∈ R^{n} : ((z − c) · ε_{1})^{2} + ∑_{j=3}^{n}((z − c) · ε_{j})^{2} < r^{2}}.

And in our basis we have that

X^{+}(a, ε_{1}, ϱ) = \{z ∈ R^{n} : (z − a) · ε_{1} ≥ ϱ\sqrt{(∑_{k=2}^{n}((z − a) · ε_{k})^{2})} \}.

So if z ∈ X^{+}(a, ε_{1}, ϱ) ∩ B_{2rϱ/(1+ϱ)}(a) we have that (z − a) · ε_{1} ∈ (0, 2rϱ/(1 + ϱ)) and so as ϱ is small (z − a) · ε_{1} ∈ (0, r) and we have

(12) \begin{aligned}((z − c) · ε_{1})^{2} + ∑_{k=3}^{n}((z − c) · ε_{k})^{2} & = ((c − a) · ε_{1} − (z − a) · ε_{1})^{2} \\ & + ∑_{k=3}^{n}((z − c) · ε_{k})^{2} \\ & = (r − (z − a) · ε_{1})^{2} + ∑_{k=3}^{n}((z − c) · ε_{k})^{2} \\ & ≤ (r − ϱ\sqrt{(∑_{k=2}^{n}((z − a) · ε_{k})^{2})})^{2} \\ & + ∑_{k=3}^{n}((z − c) · ε_{k})^{2} \\ & = r^{2} − 2rϱ\sqrt{(∑_{k=2}^{n}((z − a) · ε_{k})^{2})} \\ & + ϱ^{2}(∑_{k=2}^{n}((z − a) · ε_{k})^{2}) \\ & + ∑_{k=3}^{n}((z − c) · ε_{k})^{2}. \end{aligned}

Now as a = c − rε_{1} we know that (c − a) · ε_{k} = 0 for k ∈ {2, 3, …, n} so

\begin{aligned}∑_{k=3}^{n}((z − a) · ε_{k})^{2} & = ∑_{k=3}^{n}((z − c) · ε_{k} + (c − a) · ε_{k})^{2} \\ & = ∑_{k=3}^{n}((z − c) · ε_{k})^{2}. \end{aligned}

So putting this into (12) we get that

(13)  \begin{aligned}((z − c) · ε_{1})^{2} + ∑_{k=3}^{n}((z − c) · ε_{k})^{2} & ≤ r^{2} − 2rϱ\sqrt{(∑_{k=2}^{n}((z − a) · ε_{k})^{2})} \\ & + (1 + ϱ^{2})(∑_{k=2}^{n}((z − a) · ε_{k})^{2}). \end{aligned}

Thus if

(14)  2rϱ\sqrt{(∑_{k=2}^{n}((z − a) · ε_{k})^{2})} ≥ (1 + ϱ^{2})(∑_{k=2}^{n}((z − a) · ε_{k})^{2})

then z ∈ K(x, ε_{2}, r). Now (14) is equivalent to

(15)  2rϱ ≥ (1 + ϱ^{2})\sqrt{(∑_{k=2}^{n}((z − a) · ε_{k})^{2})}

and as z ∈ B_{2rϱ/(1+ϱ^{2})}(a), (15) obviously holds and so (14) also holds and z ∈ K(x, ψ, r); we have shown Step 1.

Step 2: Now suppose a ∈ ∂K(x, ψ, r) and n_{a} ∈ S^{n−1} ∩ ψ^{⊥} such that

X^{+}(a, n_{a}, ϱ) ∩ B_{rϱ/(1+ϱ^{2})}(a) ⊂ K(x, ψ, r)

and we have that

(16)  μ(X^{+}(a, n_{a}, ϱ) ∩ B_{rϱ/(1+ϱ^{2})}(a) ∩ S_{1}) ≥ \frac{λ_{1}}{2} (\frac{rϱ}{(1 + ϱ^{2})})^{s}.

We will show there exists some number ζ_{ϱ}^{λ_{1}} > 0 with the property that

(17)  B_{rϱ/(1+ϱ^{2})}(a) ∩ X^{+}(a, n_{a}, ϱ) ∩ K(x, ψ, (1 − ζ_{ϱ}^{λ_{1}})r) ∩ S_{1} ≠ ∅.

Now note that for any β > 0, if z ∈ X^{+}(a, n_{a}, ϱ) ∩ B_{β}(a) then X^{+}(a, n_{a}, ϱ) ∩ B_{β}(a) ⊂ B_{2β}(z). So if X^{+}(a, n_{a}, ϱ) ∩ B_{β}(a) ∩ S_{1} ≠ ∅ then by (9)

μ(X^{+}(a, n_{a}, ϱ) ∩ B_{β}(a)) ≤ 2(2β)^{s}

hence there exists some small number σ_{λ_{1}} ∈ (0, \frac{1}{8}ϱ) such that

μ(A(a, σ_{λ_{1}}r, \frac{rϱ}{(1 + ϱ^{2})}) ∩ X^{+}(a, n_{a}, ϱ) ∩ S_{1}) ≥ \frac{λ_{1}}{4} (\frac{rϱ}{(1 + ϱ^{2})})^{s}.

We are going to use this and (13) to show Step 2. Firstly let

g(p) = −2rϱp + (1 + ϱ^{2})p^{2},

the two zeros of this quadratic are at 0 and 2rϱ/(1 + ϱ^{2}). Since σ_{λ_{1}} is smaller than \frac{1}{8}ϱ its easy to see that g(σ_{λ_{1}}r) = ((1 + ϱ^{2})σ_{λ_{1}}^{2} − 2ϱσ_{λ_{1}})r^{2} and g(rϱ/(1 + ϱ^{2})) = −(rϱ)^{2}/(1 + ϱ^{2}) are both less than zero.

If we let

ζ_{ϱ}^{λ_{1}} = \frac{1}{2} min\{−\frac{g(σ_{λ_{1}}r)}{r^{2}}, −\frac{g(rϱ/(1 + ϱ^{2}))}{r^{2}}\}

then we have that for q ∈ (σ_{λ_{1}}r, rϱ/(1 + ϱ^{2}))

g(q) ≤ −2ζ_{ϱ}^{λ_{1}}r^{2}.

Let z ∈ A(a, σ_{λ_{1}}r, rϱ/(1 + ϱ^{2}))∩X^{+}(a, n_{a}, ϱ)∩S_{1} . Now in our notation inequality (13) becomes

((z − c) · ε_{1})^{2} + ∑_{j=3}^{n}((z − c) · ε_{j})^{2} ≤ r^{2} + g(\sqrt{∑_{k=2}^{n}((z − a) · ε_{k})^{2}} )
≤ (1 − 2ζ_{ϱ}^{λ_{1}})r^{2}
≤ (1 − ζ_{ϱ}^{λ_{1}})^{2}r^{2}

so z ∈ K(x, ψ, (1 − ζ_{ϱ}^{λ_{1}})r) and this establishes Step 2.

Let ξ_{ϱ} = ϱ/(1 + ϱ^{2}) and let κ_{ϱ}^{λ_{1}} = ζ_{ϱ}^{λ_{1}}/ξ_{ϱ}. Given x ∈ S_{1} let d > 0 such that for some small ε > 0

(18)  \frac{μ(B_{4d}(x)\S_{1})}{d^{s}} ≤ ε.

Let z ∈ ψ^{⊥} + x be such that Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ⊂ B_{2d}(x).

Define

H(z, ψ^{⊥}, α) := ⋃_{h∈(−α,α)} z + hψ + ψ^{⊥}.

Let ϕ_{k} = ∑_{j=0}^{k} ξ_{ϱ}κ_{ϱ}^{λ_{1}}d(1 − ζ_{ϱ}^{λ_{1}})^{j} where we let ϕ_{−1} = 0. Define

(19)  Γ_{k} := H(z, ψ^{⊥}, ϕ_{k}) ∩ K(z, ψ, (1 − ζ_{ϱ}^{λ_{1}})^{k}κ_{ϱ}^{λ_{1}}d).

424 Andrew Lorent

Figure 3.

Step 3: We will show

(20)  Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ⊂ ⋃_{j=0}^{∞} Γ_{j}.

See Figure 3.

We argue inductively. Firstly recall that ϕ_{0} = ξ_{ϱ}κ_{ϱ}^{λ_{1}}d and note that it is obvious that

Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ∩ H(z, ψ^{⊥}, ξ_{ϱ}κ_{ϱ}^{λ_{1}}d) ⊂ Γ_{0}.

Suppose we have that

(21)  Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ∩ H(z, ψ^{⊥}, ϕ_{k}) ⊂ ⋃_{j=0}^{k} Γ_{j}.

Let ϑ_{k+1} be the radius of the two congruent spheres given by ∂Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ∩ ∂H(z, ψ^{⊥}, ϕ_{k}). So by definition of Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) we know

(22)  \begin{aligned}ϑ_{k+1} & = κ_{ϱ}^{λ_{1}}d − κ_{ϱ}^{λ_{1}}ϕ_{k} \\ & = κ_{ϱ}^{λ_{1}}d − κ_{ϱ}^{λ_{1}}(∑_{j=0}^{k} ξ_{ϱ}κ_{ϱ}^{λ_{1}}d(1 − ζ_{ϱ}^{λ_{1}})^{j}) \\ & = κ_{ϱ}^{λ_{1}}d − ξ_{ϱ}(κ_{ϱ}^{λ_{1}})^{2}d(∑_{j=0}^{k}(1 − ζ_{ϱ}^{λ_{1}})^{j}). \end{aligned}

And note

∑_{j=0}^{k}(1 − ζ_{ϱ}^{λ_{1}})^{j} = \frac{1}{ζ_{ϱ}^{λ_{1}}} (1 − (1 − ζ_{ϱ}^{λ_{1}})^{k+1}).

Recall κ_{ϱ}^{λ_{1}} = ζ_{ϱ}^{λ_{1}}/ξ_{ϱ} so putting the above expression into (22) we get

ϑ_{k+1} = κ_{ϱ}^{λ_{1}}d(1 − ζ_{ϱ}^{λ_{1}})^{k+1}

which is exactly the width of the cylinder Γ_{k+1} .

Now ϑ_{k+1} is the biggest radius of the spheres given by ψ^{⊥} slices of

∂Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ∩ (H(z, ψ^{⊥}, ϕ_{k+1})\H(z, ψ^{⊥}, ϕ_{k})).

So we know that Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d)∩(H(z, ψ^{⊥}, ϕ_{k+1})\H(z, ψ^{⊥}, ϕ_{k})) is “thin” enough to fit into Γ_{k+1} . It is also, by definition, “short” enough to fit into Γ_{k+1} . So by inductive assumption (21) we have

Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ∩ H(z, ψ^{⊥}, ϕ_{k+1}) ⊂ ⋃_{j=0}^{k+1} Γ_{j}.

This establishes Step 3.

Step 4: Recall z ∈ ψ^{⊥} + x such that Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ⊂ B_{2d}(x).

Suppose Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d)∩S_{1} ≠ ∅. Let W := {k : Γ_{k}∩Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d)∩S_{1} ≠ ∅}. If W is finite we define k_{1} := max{k : k ∈ W }, otherwise we define k_{1} to be any positive integer such that λ_{1}(1 − ζ_{ϱ}^{λ_{1}})^{s(k_{1}+1)}(ξ_{ϱ}κ_{ϱ}^{λ_{1}}d)^{s} < 2εd^{s}.

In the case where W is finite we will show k_{1} is sufficiently big so that

λ_{1}(1 − ζ_{ϱ}^{λ_{1}})^{s(k_{1}+1)}(ξ_{ϱ}κ_{ϱ}^{λ_{1}}d)^{s} < 2εd^{s}.

Suppose not and

(23)  λ_{1}(1 − ζ_{ϱ}^{λ_{1}})^{s(k_{1}+1)}(ξ_{ϱ}κ_{ϱ}^{λ_{1}}d)^{s} ≥ 2εd^{s}.

We know that Γ_{k_{1}+1} ∩ Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ∩ S_{1} = ∅. Let y_{1} ∈ Γ_{k_{1}} ∩ Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ∩ S_{1} , let h = |P_{ψ^{⊥}}(y_{1} − z)| so we know that

(24)  h ∈ ((1 − ζ_{ϱ}^{λ_{1}})^{k_{1}+1}κ_{ϱ}^{λ_{1}}d, (1 − ζ_{ϱ}^{λ_{1}})^{k_{1}}κ_{ϱ}^{λ_{1}}d).

Now y_{1} ∈ Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ⊂ B_{2d}(x) and as hξ_{ϱ} ≤ ξ_{ϱ}(1−ζ_{ϱ}^{λ_{1}})^{k_{1}}κ_{ϱ}^{λ_{1}}d = ζ_{ϱ}^{λ_{1}}(1−ζ_{ϱ}^{λ_{1}})d so as in Step 1, letting n_{y_{1}} be the “inwards” pointing unit normal of ∂K(z, ψ, h) at point y_{1} , we have

(25)  X^{+}(y_{1}, n_{y_{1}}, ϱ) ∩ B_{hξ_{ϱ}}(y_{1}) ⊂ B_{d}(y_{1}) ⊂ B_{4d}(x).

Now by the fact that y_{1} ∈ S_{1} (recall (7)) we know that

μ(B_{hξ_{ϱ}}(y_{1}) ∩ X^{+}(y_{1}, n_{y_{1}}, ϱ)) ≥ λ_{1}(ξ_{ϱ}h)^{s}.

By using (23), (24) we know λ_{1}(ξ_{ϱ}h)^{s} ≥ 2εd^{s} and so from (25) and the density estimate (10) we know that we have

μ(B_{hξ_{ϱ}}(y_{1}) ∩ X^{+}(y_{1}, n_{y_{1}}, ϱ) ∩ S_{1}) ≥ λ_{1}(ξ_{ϱ}h)^{s} − εd^{s} ≥ \frac{1}{2}λ_{1}(ξ_{ϱ}h)^{s}

and so by equations (16) and (17) (recall ξ_{ϱ} = ϱ/(1 + ϱ^{2})) we can pick y_{2} ∈ X^{+}(y_{1}, n_{y_{1}}, ϱ) ∩ B_{hξ_{ϱ}}(y_{1}) ∩ S_{1} such that

(26)  y_{2} ∈ K(z, ψ, (1 − ζ_{ϱ}^{λ_{1}})h) ⊂ K(z, ψ, (1 − ζ_{ϱ}^{λ_{1}})^{k_{1}+1}κ_{ϱ}^{λ_{1}}d).

Suppose y_{1} ∈ Ψ^{u}(z, ψ, κ_{ϱ}^{λ_{1}}, d). This means

h = |P_{ψ^{⊥}}(y_{1} − z)| ≤ κ_{ϱ}^{λ_{1}}|P_{⟨ψ⟩}(y_{1} − (z + dψ))|.

So from (26)

|P_{ψ^{⊥}}(y_{2} − z)| ≤ (1 − ζ_{ϱ}^{λ_{1}})h ≤ (1 − ζ_{ϱ}^{λ_{1}})κ_{ϱ}^{λ_{1}} |P_{⟨ψ⟩}(y_{1} − (z + dψ))|.

And since y_{2} ∈ B_{hξ_{ϱ}}(y_{1})

|P_{⟨ψ⟩}(y_{2} − (z + dψ))| ≥ |P_{⟨ψ⟩}(y_{1} − (z + dψ))| − hξ_{ϱ}
≥ |P_{⟨ψ⟩}(y_{1} − (z + dψ))| − ξ_{ϱ}κ_{ϱ}^{λ_{1}} |P_{⟨ψ⟩}(y_{1} − (z + dψ))|
= |P_{⟨ψ⟩}(y_{1} − (z + dψ))|(1 − ζ_{ϱ}^{λ_{1}}).

Putting these two together we have

|P_{ψ^{⊥}}(y_{2} − z)| ≤ κ_{ϱ}^{λ_{1}} |P_{⟨ψ⟩}(y_{2} − (z + dψ))|.

So y_{2} ∈ Ψ^{u}(z, ψ, κ_{ϱ}^{λ_{1}}, d) and similarly

y_{1} ∈ Ψ^{d}(z, ψ, κ_{ϱ}^{λ_{1}}, d) implies y_{2} ∈ Ψ^{d}(z, ψ, κ_{ϱ}^{λ_{1}}, d)

so we have y_{2} ∈ Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d).

Now we want to show that y_{2} ∈ (⋃_{j=k_{1}+1}^{∞} Γ_{j}). We know from (20) that y_{2} ∈ ⋃_{j=1}^{∞} Γ_{j}. However we also know from (26) that |P_{ψ_{⊥}}(y_{2} − z)| is sufficiently small (recall the definition of Γ_{k_{1}+1} , see (19)) so that we must have y_{2} ∈ (⋃_{j=k_{1}+1}^{∞} Γ_{j}) and hence

y_{2} ∈ Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ∩ ( ⋃_{j=k_{1}+1}^{∞} Γ_{j}) ∩ S_{1},

contradicting the maximality of k_{1} . So we have established Step 4, and so we know (recall (23)) that

(27)  λ_{1}(1 − ζ_{ϱ}^{λ_{1}})^{s(k_{1}+1)}(ξ_{ϱ}κ_{ϱ}^{λ_{1}}d)^{s} < 2εd^{s}.

Let θ = (1 − ζ_{ϱ}^{λ_{1}})^{k_{1}}κ_{ϱ}^{λ_{1}}d and recall that this is the width of cylinder Γ_{k_{1}} . So

2εd^{s} > λ_{1}(1 − ζ_{ϱ}^{λ_{1}})^{s(k_{1}+1)}(ξ_{ϱ}κ_{ϱ}^{λ_{1}}d)^{s} = λ_{1}(1 − ζ_{ϱ}^{λ_{1}})^{s}ξ_{ϱ}^{s}θ^{s}.

And so

θ < \frac{(2ε)^{1/s}d}{λ_{1}^{1/s}(1 − ζ_{ϱ}^{λ_{1}})ξ_{ϱ}} .

To summarise what we have proved: we have shown that for any z ∈ ψ^{⊥} + x such that Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ⊂ B_{2d}(x), if we know that

Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ∩ S_{1} ≠ ∅,

then

(28)  K(z, ψ, \frac{(2ε)^{1/s}d}{λ_{1}^{1/s}(1 − ζ_{ϱ}^{λ_{1}})ξ_{ϱ}} ) ∩ B_{2d}(x) ∩ S_{1} ≠ ∅.

As we know that for any z ∈ (ψ^{⊥}+x)∩K(x, ψ, κ_{ϱ}^{λ_{1}}d), (recall this is the hypothesis on point z in the statement of Lemma 1), we have

x ∈ Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d)

and thus Ψ(z, ψ, κ_{ϱ}^{λ_{1}}, d) ∩ S_{1} ≠ ∅ and so by (28) for

ϑ_{ϱ}^{λ_{1}} = \frac{2^{1/s}}{λ_{1}^{1/s}(1 − ζ_{ϱ}^{λ_{1}})ξ_{ϱ}}

we have K(z, ψ, ε^{1/s}ϑ_{ϱ}^{λ_{1}}d) ∩ S_{1} ∩ B_{2d}(x) ≠ ∅ which gives us the conclusion of the lemma. Thus we have completed the proof.

Proof of Theorem 1 continued

Let ε > 0 be some very small number. Again we simplify expressions by letting μ := H_{⌊S}^{s}. Let x ∈ S_{1} be a density point of S_{1} , let d ∈ (0, r_{0}) such that

(29)  \frac{μ(B_{16d}(x)\S_{1})}{d^{s}} ≤ ε.

We can partition [2d, 4d] into M := [2/4ε^{1/s}ϑ_{ϱ}^{λ_{1}}] subintervals. Call them I_{1} , I_{2}, …, I_{M} and let c_{k} be the center of interval I_{k}. Now for k ∈ {1, 2, …, M} we know the following is true: For all z ∈ (ψ^{⊥} + x) ∩ K(x, ψ, κ_{ϱ}^{λ_{1}} \frac{1}{2}c_{k}) by Lemma 1 we have

(30)  K(z, ψ, ε^{1/s}ϑ_{ϱ}^{λ_{1}} \frac{1}{2}c_{k}) ∩ S_{1} ∩ B_{c_{k}}(x) ≠ ∅.

Now fix k ∈ {1, 2, …, M}. From (30) we have that for every z ∈ (ψ^{⊥} + x) ∩ K(x, ψ, κ_{ϱ}^{λ_{1}}d) we have K(z, ψ, 2ε^{1/s}ϑ_{ϱ}^{λ_{1}}d) ∩ S_{1} ∩ B_{c_{k}}(x) ≠ ∅.

Let E_{k} = (ψ^{⊥} + x)\P_{ψ^{⊥}}^{−1}(P_{ψ^{⊥}}(S_{1} ∩ \overline{B_{c_{k}}(x)})).

Now recall since S_{1} is purely unrectifiable or of Hausdorff dimension < n − 1 by our choice of ψ (recall we are first proving the theorem for a.e. ψ ∈ S^{n−1} , see (5)) we have

L^{n−1}(E_{k}) ≥ L^{n−1}(B_{1}(0))(κ_{ϱ}^{λ_{1}}d)^{n−1}.

Since S_{1} is closed, E_{k} is open. Thus for each z ∈ E_{k} we can find

(31)  r_{z} ∈ (0, 2ε^{1/s}ϑ_{ϱ}^{λ_{1}}d)

such that – K(z, ψ, r_{z}) ∩ B_{c_{k}}(x) ∩ (ψ^{⊥} + x) ∩ S_{1} = ∅,

– ∂K(z, ψ, r_{z}) ∩ B_{c_{k}}(x) ∩ (ψ^{⊥} + x) ∩ S_{1} ≠ ∅.

Now by the 5r covering theorem ([9, Theorem 2.1]) from the covering of E_{k} given by {B_{r_{z}}(z) : z ∈ E_{k}} we can find a disjoint collection of balls

{B_{r_{m}^{k}}(z_{m}^{k}) : z_{m}^{k} ∈ E_{k}, m = 1, 2, …, N_{k}}

which are centered on points of E_{k} and are such that

(32)  ∑_{m=1}^{N_{k}}(r_{m}^{k})^{n−1} > \frac{(κ_{ϱ}^{λ_{1}}d)^{n−1}}{5^{n−1}} .

Now each m ∈ {1, 2, …, N_{k}} we can pick a point

x_{m}^{k} ∈ ∂K(z_{m}^{k}, ψ, r_{m}^{k}) ∩ \overline{B_{c_{k}}(x)} ∩ S_{1}.

Figure 4.

The essential point is that by the fact that the points {x_{m}^{k} : m = 1, 2, …, N_{k}} satisfy inequality (7) in every direction perpendicular to ψ, unless the points {x_{m}^{k} : m = 1, …, N_{k}} are very near the surface of the ball B_{c_{k}}(x), there will be some fraction of the set S_{1} in the interior of the cylinders {K(z_{m}^{k}, ψ, r_{m}^{k}), m = 1, …, N_{k}} of Hausdorff measure approximately ∑_{k=1}^{N_{k}}(r_{m}^{k})^{s}.

Now since s ∈ (0, n − 1], by the well-known inequality ([4, Theorem 19])

(33)  ( ∑_{m=1}^{N_{k}}(r_{m}^{k})^{s})^{1/s} ≥ ( ∑_{m=1}^{N_{k}}(r_{m}^{k})^{n−1})^{1/(n−1)} ≥ \frac{κ_{ϱ}^{λ_{1}}d}{5}

so we see this can not happen too often, see Figure 4. Proceeding formally, for each k ∈ {1, 2, …, M}, m ∈ {1, 2, …, N_{k}} let n_{x_{m}^{k}} denote the inner normal to the boundary of the cylinder K(z_{m}^{k}, ψ, r_{m}^{k}) at point x_{m}^{k}. Again let us fix k ∈ {1, 2, …, M}. We know that for each m ∈ {1, …, N_{k}}

(34)  μ(X^{+}(x_{m}^{k}, n_{x_{m}^{k}}, ϱ) ∩ B_{ξ_{ϱ}r_{m}^{k}}(x_{m}^{k})) ≥ λ_{1}(ξ_{ϱ}r_{m}^{k})^{s}.

Let I_{1}^{k} = {m ∈ {1, 2, …, N_{k}} : X^{+}(x_{m}^{k}, n_{x_{m}^{k}}, ϱ) ∩ B_{ξ_{ϱ}r_{m}^{k}}(x_{m}^{k}) ⊂ B_{c_{k}}(x)}. Let I_{2}^{k} = {1, 2, …, N_{k}}\I_{1}^{k}. Now firstly for any m ∈ I_{1}^{k} as

X^{+}(x_{m}^{k}, n_{x_{m}^{k}}, ϱ) ∩ B_{ξ_{ϱ}r_{m}^{k}}(x_{m}^{k}) ⊂ K(z_{m}^{k}, ψ, r_{m}^{k}) ∩ B_{c_{k}}(x)

and so X^{+}(x_{m}^{k}, n_{x_{m}^{k}}, ϱ) ∩ B_{ξ_{ϱ}r_{m}^{k}}(x_{m}^{k}) ⊂ B_{c_{k}}(x)\S_{1} , thus we have

μ( ⋃_{m∈I_{1}^{k}} X^{+}(x_{m}^{k}, n_{x_{m}^{k}}, ϱ) ∩ B_{ξ_{ϱ}r_{m}^{k}}(x_{m}^{k})) ≤ εd^{s}.

So

∑_{m∈I_{1}^{k}}(r_{m}^{k})^{s} ≤ \frac{εd^{s}}{λ_{1}ξ_{ϱ}^{s}} .

Thus given sufficient smallness of ε, from (33) we have

(35)  ∑_{m∈I_{2}^{k}}(r_{m}^{k})^{s} ≥ \frac{1}{5^{s}} (κ_{ϱ}^{λ_{1}}d)^{s} − \frac{εd^{s}}{λ_{1}ξ_{ϱ}^{s}} ≥ \frac{1}{2 × 5^{s}} (κ_{ϱ}^{λ_{1}}d)^{s}.

However by definition of I_{2}^{k} we know that for k = 1, 2, …, M (recall (31))

{B_{ξ_{ϱ}r_{m}^{k}}(x_{m}^{k}), m ∈ I_{2}^{k}} ⊂ N_{2ε^{1/s}ϑ_{ϱ}^{λ_{1}} d}(∂B_{c_{k}}(x)).

So using the density estimate of subset S_{1} (i.e. (8)) and (35)

μ(B_{5d}(x)) ≥ μ( ⋃_{k=1}^{M} N_{2ε^{1/s}ϑ_{ϱ}^{λ_{1}} d}(∂B_{c_{k}}(x)))

≥ ∑_{k=1}^{M} ∑_{m∈I_{2}^{k}} μ(B_{ξ_{ϱ}r_{m}^{k}}(x_{m}^{k}))

≥ \frac{λ_{1}ξ_{ϱ}^{s}}{2} (\frac{M}{5^{s}} )(κ_{ϱ}^{λ_{1}}d)^{s}

and since M can be made as large as we like by reducing ε (recall that M = [2/4ε^{1/s}ϑ_{ϱ}^{λ_{1}}]) we end up contradicting the density bound (8) and thus Theorem 1 is proved for almost every ψ ∈ S^{n−1} .

Now for arbitrary ψ ∈ S^{n−1} we can argue in the following way: Suppose the conclusion of Theorem 1 is false for ψ. Then as before we must have some subset S_{1} ⊂ S and some ϱ > 0, λ_{1} > 0 such that for all ϕ ∈ ψ^{⊥} ∩ S^{n−1}

lim inf_{r→0} \frac{H^{s}(B_{r}(x) ∩ X^{+}(x, ϕ, ϱ) ∩ S)}{r^{s}} > λ_{1},

for all x ∈ S_{1} .

Now by elementary geometry we can see that there exists a small number a(ϱ) > 0 such that any ψ̃ ∈ S^{n−1} for which |ψ̃ − ψ| < a(ϱ) has the following property:

For any ϕ̃ ∈ S^{n−1} ∩ ψ̃^{⊥} there exists ϕ ∈ S^{n−1} ∩ ψ^{⊥} such that

(36)  X^{+}(z, ϕ, ϱ) ⊂ X^{+}(z, ϕ̃, \frac{1}{2}ϱ)

for any z ∈ R^{n}.

A generalised conical density theorem for unrectifiable sets 431

Since we have proved Theorem 1 for almost every ψ̃ ∈ S^{n−1} we can certainly find a ψ̃ ∈ S^{n−1} for which it is true and |ψ̃−ψ| < a(ϱ). Then for H^{n} a.e. x_{0} ∈ S_{1} we have that for some ϕ̃ ∈ S^{n−1} ∩ ψ̃^{⊥}:

lim inf_{r→0} \frac{H^{s}(B_{r}(x_{0}) ∩ X^{+}(x_{0}, ϕ̃, \frac{1}{2}ϱ) ∩ S)}{r^{s}} = 0.

However by (36) this implies that

lim inf_{r→0} \frac{H^{s}(B_{r}(x_{0}) ∩ X^{+}(x_{0}, ϕ, ϱ) ∩ S)}{r^{s}} = 0,

which contradicts the definition of S_{1} so we are done.

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Received 14 January 2003