AN EXAMPLE PERTAINING TO THE FAILURE OF THE BESICOVITCH–FEDERER STRUCTURE THEOREM IN HILBERT SPACE

Abstract: We give an example, in the infinite dimensional separable Hilbert space, of a purely unrectifiable Borel set with finite nonzero one dimensional Hausdorff measure, whose projection is nonnegligible in a set of directions which is not Aronszajn null. 2010 Mathematics Subject Classification: 28A75, 28A80, 53C65.


Foreword
We let 0 < m < n be integers.The Grassmannian G(R n , m) is equipped with an O(n) invariant probability Borel measure γ n,m [11, 2.7.16(6)].Given W ∈ G(R n , m) we let P W denote the orthogonal projection onto W .The Besicovitch-Federer Theorem referred to in the title states the following: If S ⊆ R n is Borel measurable and H m (S) < ∞ then the following are equivalent: Here H m is the m dimensional Hausdorff measure in R n .The m rectifiable subsets of R n are defined to be those of the form M = f (A) for some bounded A ⊆ R m and Lipschitz f : A → R n .It follows from Theorems of H. Rademacher [11, 3.1.6],N. Luzin [11, 2.3.5], and H. Whitney [11, 3.1.14],that condition (A) is equivalent to (A ) H m (S ∩ M ) = 0 for every m rectifiable subset M ⊆ R n .

3.3.14] and
The Besicovitch-Federer Theorem has played a distinguished role in the development of the Geometric Calculus of Variations, of which the Plateau problem is a paradigm.The original proof of the Closure Theorem for integral currents due to H. Federer and W. H. Fleming [13] relies upon the Structure Theorem, and so does one more recent proof, perhaps (unfortunately) less known, due to W. H. Fleming [14], see also [12].Other proofs of the Closure Theorem have avoided the Structure Theorem, see the techniques set forth in [1] and [21], as well as [24], whether these have been designed to this end or not.Even more recent versions of the Closure Theorem, when the ambient space R n is replaced with either a Banach space or a complete metric space, rely upon the fact that 0 dimensional slices of integral currents are of bounded variation -an observation that goes back at least to H. Federer's [11, 5.3.5(1)].
However it has been so far unknown whether a version of the Structure Theorem holds when the ambient space R n is replaced with a separable infinite dimensional Banach space, for instance the simplest one, 2 .Complications in stating the problem soon arise: Even though orthogonal projections P W onto W ∈ G( 2 , m) make sense in the Hilbert setting, one needs to face the nonexistence of an invariant probability measure on the infinite dimensional Grassmannian G( 2 , m).Nevertheless stating condition (B) above does not require the existence of such measure, but merely the existence of a distinguished invariant σ ideal of null sets in G( 2 , m) (see also the forthcoming [4]).This puts us in a better position as we explain hereunder.
From now on we shall consider the case m = 1 only.The projective space G( 2 , 1) is the usual quotient of the unit sphere S 2 .In fact the map allows us to push-forward any σ ideal from 2 to G( 2 , 1), as one would do with a measure.Replacing temporarily 2 by R n in this construction, and recalling that the O(n) invariant measure γ n,1 is a normalized quotient of H n−1 S n−1 , we infer from integration in polar coordinates that γ n,1 (E) = 0 if and only if L n (ψ −1 (E)) = 0. Accordingly, a sought for σ ideal of null sets in G( 2 , 1) can be obtained by a ψ push-forward of some σ ideal of null sets in 2 generalizing the Lebesgue null sets of R n .
There are several such choices.In this paper we consider Aronszajn null sets in 2 (see Subsection 1.4 for the definition), which are known to be equivalent to cube null and Gaussian null sets [5], but are not equivalent to Haar null sets, see e.g. the instructive monograph [2], which also discusses the relevance of Aronszajn null sets for instance to the almost everywhere Gâteaux differentiability of Lipschitz functions 2 → R.
Those S ⊆ R n verifying condition (A ) above are termed purely (H m , m) unrectifiable.The definition makes sense in any ambient metric space, in particular in 2 .Possibly the simplest and most classical example of a purely (H 1 , 1) unrectifiable subset of R2 is the self-similar four corners Cantor set C illustrated in Figure 1.This paper contributes the following: Theorem.There exists a purely (H 1 , 1) unrectifiable Borel measurable subset S ⊆ 2 with H 1 (S) = 1, such that the set of directions from which S is visible, We now briefly indicate why this is the case, and Section 2 consists of a detailed version of the argument.It is relevant to observe on Figure 1 that there actually exist "exceptional" lines on which C projects to a nonnegligible set -it projects for instance to a nondegenerate interval I on the orange line L. As a matter of fact the restriction of the projection P L C is nearly injective, so that C is nearly a graph.Specifically, if we remove the corners of the countably many solid squares used in the inductive construction of C, we obtain a set C which is the graph of f : I \ D → L ⊥ , where D is countable and By general descriptive set theory f is Borel measurable.In fact f is continuous, as the happy reader will verify.However f cannot possibly be Lipschitz, for otherwise C would be 1 rectifiable -nor even approximately differentiable on a set of positive measure, for otherwise C would not be purely (H1 , 1) unrectifiable.Figure 1 suggests to consider a sequence f j j of step functions approximating f , in the obvious way when f 1 takes four distinct values, f 2 takes sixteen distinct values, etc.Details are provided in Subsection 2.1.We use these to define The fact that f is approximately differentiable almost nowhere should somehow imply the same about γ.This in turn should say that S = im γ is purely (H 1 , 1) unrectifiable.In fact the domain I of f j is divided into 4 j intervals I j,k on which f j is constant, and if we define t ∈ I, then the reader will easily apply the proof of Subsection 2.5 to showing that if inf j δ j (t) = 0 then γ is not approximately differentiable at t, as should be intuitively sound.Furthermore the Lebesgue density Theorem clearly implies that is conegligible in I. Therefore S = im γ is purely (H 1 , 1) unrectifiable, because γ has the Luzin (N) property (adapt Subsections 2.3 and 2.6 1 ).Yet the projection of S = im γ onto a subspace generated by finitely many coordinates, consists of finitely many line segments, a "very" rectifiable set whose projections should have positive measure in many directions, if only we had made sense of such statement in ∞ .
We now seek for a modification of γ taking their values in 2 rather than in ∞ .With each sequence β j j ∈ 2 we can of course associate The problem is that by doing so we have likely destroyed the almost everywhere non approximate differentiability property that we used to imply the image of γ is purely unrectifiable.This is because β j → 0. Hope suggests to investigate the case when this convergence is slow.The point here is that the coordinates f j j are stochastically independent, thus the Borel-Cantelli Lemma yields an improved version of the use of the Lebesgue density Theorem above, see Subsection 2.7.In Subsection 2.8 we establish the existence of a proper choice of β j j ∈ 2 such that our new γ is approximately differentiable almost nowhere.
It may be worth pointing out that this can be interpreted as a stronger version of the nondifferentiability property of the original function f arising in R 2 .In other words, the four corners Cantor set of Figure 1 is very much purely unrectifiable, in some yet unspecified sense.Whether or not this is generic behavior of purely unrectifiable subsets of the Euclidean plane remains unsettled.
We finally need to evoke why S = im γ projects to a set of positive measure onto "many" lines L = span{u}, u ∈ S 2 .From the definition of γ we infer that the measure of this projection equals the measure of the image of Even though we arranged everything so far in order that t → j 1 β j f j (t) be very much nonLipschitz, we can now play with the coefficients u j j , hopefully allowed to converge fast enough to 0, to compensate for this wild behavior, in fact to guarantee t → j 1 β j u j f j (t) is Lipschitz with small Lipschitz constant, when restricted to an appropriate nonnegligible subset of its domain.This associated with the choice of a first coordinate u 0 t with u 0 close to 1 yields a projection of positive measure, for Aronszajn nonnegligeably many u's, see Subsection 2.10.

Preliminaries
The ambient space of this paper is the infinite dimensional separable Hilbert space 2 (N), sometimes abbreviated 2 .We let e 0 , e 1 , e 2 , . . .be its canonical orthonormal basis.The norm in 2 is denoted • or • 2 , and the inner product •, • .Occasionally we consider the finite dimensional n 2 , i.e.R n equipped with its usual inner product.Whether X = 2 or X = n 2 we let S X denote the unit sphere of X.
1.1.Hausdorff measure.Given S ⊆ 2 (N) and δ > 0 we recall that and diam S j δ for every j , and everywhere.This is well-known and will be used in Subsection 2.6.As it also happens to be easily established, we include a sketch of proof.
For each x ∈ 2 (N) we define x and we notice Lip f x Lip f .Note also that whenever both f x and f y are differentiable at t. Choose a dense sequence x j j in 2 (N) and let N j be an , and we denote the corresponding limit by g(t).We ought to show that the convergence to g(t) is in fact strong for L 1 almost every t.Recall that the weak convergence is promoted to strong convergence when norms converge as well, according to the parallelogram law.Since readily Note the function t → g(t) is L 1 measurable.Observe next that for every t, t + h ∈ [a, b] and every j one has x j t+h t g(s) dL 1 (s).
Extracting from x j j a sequence that converges to f (t + h) − f (t) we infer from the above that It is now clear that (2) holds whenever t is a Lebesgue point of g .

1.
3. Pure unrectifiability.We say R ⊆ 2 (N) is 1 rectifiable if there exists a bounded set A ⊆ R and a Lipschitz map Using a Whitney decomposition of R \ clos A one shows that each f as above admits a Lipschitz extension f : R → 2 (N), see e.g.[15].This follows alternatively from Kirszbraun's Theorem [11, 2.10 ). Therefore Lip h 1 and in turn h (s) 1 for L 1 almost every s.The Area Theorem from [17] then implies that h (s) = 1 for L 1 almost every s.This will be called an arclength parametrization.
1.4.Aronszajn null sets.Let X be a separable Banach space.A Borel subset B ⊆ X is Aronszajn null if the following holds.For every sequence v j j whose span is dense in X, there exists a decomposition B = ∪ j B j into Borel sets B j subject to the following requirement: For each j, the intersection of B j with each line parallel to v j is negligible, i.e.H 1 (B j ∩ (y + span{v j })) = 0 for every y ∈ X.We say that an arbitrary subset of X is Aronszajn null if it is contained in an Aronszajn null Borel subset of X.
It is important in this definition that the sets B j be Borel measurable.Indeed W. Sierpiński established, under the continuum hypothesis, the existence of a partition ) is null on every horizontal (respectively vertical) line, see e.g.[16,Chapter 4].It follows from Fubini's Theorem applied to the Lebesgue measure L 2 that one of E 1 and E 2 -and therefore both -must be L 2 nonmeasurable.
It also follows from Fubini's Theorem applied to Lebesgue's measure L n , and from the Borel regularity of L n , that in case X = R n is finite dimensional the Aronszajn null sets coincide with the Lebesgue null sets.
We now let X be a separable Hilbert space, i.e.X = n 2 for some n ∈ N or X = 2 .By G(X, 1) we denote the collection of 1 dimensional linear subspaces of X.We consider the hereditary σ ideal N X consisting of those Aronszajn null sets in X.We use the map ψ : X \ {0} → G(X, 1) : x → span{x} to define a hereditary σ ideal N G(X,1) on G(X, 1) in the following way: In case X = n 2 the Coarea Theorem [9, 3.4.
Proof: This is because t → g(4 i t) itself is constant on each I j,k whenever 0 i < j.
(B) For every j ∈ N \ {0} and every k ∈ {0, 1, . . ., 4 j − 2} one has Proof: The initial case j = 1 follows readily from the explicit definition of f 1 = g above.In fact, for the sake of this proof, it is useful to notice that if s, t ∈ R + \ {0} belong to two distinct members of the partition {(m/4, (m + 1)/4] : m ∈ N} and belong to the same interval (k, k + 1] for some k ∈ N, then |g(s) − g(t)| 1  2 .Also, |g(s) − g(t)| 1 regardless of the relative position of s and t.
Assuming the claim holds for j, we proceed to prove it for j + 1.We notice that (3) f j+1 (t) = f j (t) + 4 −j g(4 j t).
Letting s ∈ I j+1,k and t ∈ I j+1,k+1 we infer from (3) that The second case occurs when I j+1,k ⊆ I j,k and I j+1,k+1 ⊆ I j,k +1 for some k ∈ {0, 1, . . ., 4 j − 2}.Choosing s ∈ I j+1,k and t ∈ I j+1,k+1 we infer from (3) and the induction hypothesis that 2.2.A Borel measurable "curve" in 2 .Here we assume that (H1) and with it we associate where e 0 , e 1 , e 2 , . . . is the canonical orthonormal basis of 2 (N).We also abbreviate S = im γ.In Subsection 2.3 we show S has finite H 1 measure.In Subsection 2.5 we consider further restrictions regarding the parameters β j j in order that γ be approximately differentiable almost nowhere.This in turn implies S is purely unrectifiable, Subsection 2.6.In Subsections 2.7 and 2.8 we show how to calibrate the parameters β j j so that all these conditions are met.Finally, in Subsection 2.10 we exhibit "many" lines in 2 (N) on which S projects to a nonnegligible set.
For now we observe that (D) γ is Borel measurable, and its image S is Borel measurable as well.
Proof: Letting P n : 2 (N) → 2 (N) be the orthogonal projection onto span{e 0 , e 1 , . . ., e n } we notice that each P n • γ is Borel measurable, and that P n • γ n converges (uniformly) to γ.The Borel measurability of γ easily follows, and in turn that of S becomes a consequence of the injectivity of γ, see e.g.[22, Theorem 4.5.4].
2.3.The Luzin (N) property of γ and the Hausdorff measure of its image.
Proof: Letting as above P 0 denote the orthogonal projection on span{e 0 } we notice that P 0 (S) = I 0,0 and therefore H 1 (S) 1.It thus remains only to show that H 1 (S) 1. We shall first establish the inequality in case E = I j0,k0 for some j 0 ∈ N and k 0 ∈ {0, 1, . . ., 4 j0 − 1}.Given n j 0 we define We also let so that readily I j0,k0 = ∪ k∈Kn I n,k .With each k ∈ K n we associate the finite sequence of integers k j ∈ {0, 1, . . ., 4 j − 1}, j = 1, . . ., n, characterized by the relations where the c j,k 's are defined in Subsection 2.1(A).Letting P n : 2 (N) → 2 (N) still denote the orthogonal projection onto span{e 0 , e 1 , . . ., e n }, and abbreviating P ⊥ n = id 2 −P n , we next observe that and that according to Subsection 2.1(C).In other words γ(I n,k ) ⊆ S n,k where , it follows from the definition of Hausdorff measure that Letting n → ∞ we conclude that H 1 (γ (I j0,k0 )) L 1 (I j0,k0 ) .Now if E ⊆ I 0,0 is arbitrary and ε > 0 we choose an open set U ⊆ R containing E and such that L 1 (U ) < ε + L 1 (E).We further extract a disjointed sequence J i i from the family {I j,k : j ∈ N and k = 0, 1, . . ., 4 j − 1} such that U ∩ I 0,0 = ∪ i J i .We note that Since ε is arbitrary the proof is complete.
2.4.The random variables k j and δ j .We define a countable set D = I 0,0 ∩ {k4 −j : j ∈ N and k = 1, . . ., 4 j }, and for each t ∈ I 0,0 \D we let k j (t) j be the unique sequence of integers such that t ∈ I j,kj (t) .We further define the relative distance of t to the pair of endpoints of I j,kj (t) by δ j (t) = dist(t, bdry I j,kj (t) ) 4 −j , and we notice that 0 < δ j (t) 1 2 .The k j and δ j are clearly Borel measurable.Occasionally we will regard these as random variables on the probability space (I 0,0 , B(I 0,0 ), L 1 ), where L 1 is the Lebesgue measure on the σ-algebra B(I 0,0 ) of Borel subsets of I 0,0 .We observe that the random variables k j mod 4, j ∈ N, are mutually independent.

Whether γ is not approximately differentiable.
If t ∈ I 0,0 \ D is so that Proof: Given Λ > 0 we define By definition of ap lim sup we need to prove that Θ * 1 (L 1 B Λ , t) > 0. Hypothesis (H2) guarantees that there exists a subsequence β j(n) n of β j j such that β j(n) δ j(n) (t) Λ for every n.We consider one j = j(n) of these indices, which we assume sufficiently large for 0 = k j (t) = 4 j − 1.For either k = k j (t) − 1 or k = k j (t) + 1 we have Since I j,kj (t) and I j,k are consecutive it follows from Subsection 2.1(B) that .
Accordingly, B(t, 2δ j (t)4 −j ) ∩ I j,k ⊆ B Λ and therefore Recalling this holds for each j = j(n) and letting n → ∞ we obtain , and the proof is complete.
Proof: Assume if possible that there exists a Lipschitz λ : R ⊇ A → 2 (N) such that H 1 (S ∩ im λ) > 0. In view of Subsection 1.3 there is no restriction to assume that A = [a, b] and that λ is injective and an arclength parametrization.According to Subsection 1.2 λ is differentiable at each s ∈ (a, b) \ N where N ⊆ (a, b) is L 1 negligible.We define and we notice H 1 (S 1 ) = H 1 (S) > 0. For each x ∈ S 1 we let s x ∈ (a, b) be such that x = λ(s x ) and we define a unit vector v x = λ (s x ).Routine verifications show that v x is tangent to S 1 in the following sense: For every ε > 0 there exists r(x, ε) > 0 such that where D is defined in Subsection 2.4 and Z ⊆ S 1 consists of the points that are isolated in S 1 \ γ(D).We notice that H 1 (S 2 ) = H 1 (S 1 ) > 0 because both γ(D) and Z are countable.We claim that for every x ∈ S 2 one has (5) v x = ±e 0 .
In order to prove this we choose a sequence y i i in (S 1 \ γ(D)) \ {x} converging to x and we define t, t+h i ∈ I 0,0 \D uniquely by the relations γ(t) = x and γ(t + h i ) = y i .Given j 1 we observe that As |h i | y i − x → 0 and t ∈ D we infer from Subsection 2.1(A) that y i − x, e j = 0 if i is large enough.Therefore v x , e j = 0. Since j 1 is arbitrary (5) is established.

2.7.
We should now proceed to showing that there actually exists a choice of parameters β j j in 2 (N) that verifies hypothesis (H3) of Subsection 2.6.This will be done in two steps.We start with the following observation.

Assume that
(1) λ n n is a sequence in N \ {0} and j(n) n is the sequence defined by j(1) = 1 and j(n + 1) = j(n) and therefore carries a Borel probability measure µ = ⊗ ∞ j=0 µ j where each µ j is a normalized Lebesgue measure supported on the j th compact interval factor.If K η,θ,ε were Aronszajn null, it would decompose into a countable union of Borel sets B j , j ∈ N, such that H 1 (B j ∩ (y + span{e j })) = 0 for every y ∈ span{e j } ⊥ , and therefore also µ j (B j ∩ (y + span{e j })) = 0. Fubini's Theorem and the Borel measurability of B j then imply µ(B j ) = 0. Since j is arbitrary we conclude µ(K η,θ,ε ) = 0, a contradiction.

Figure 1 .
Figure 1.Four corners Cantor set in Euclidean plane.