Borel sets which are null or non-σ-finite for every translation invariant measure

Márton Elekes^{∗}

Rényi Alfréd Institute Budapest, POB 127, H-1364 Hungary emarci@renyi.hu and

Tamás Keleti^{†}

Eötvös Loránd University Department of Analysis Budapest, H-1117 Hungary elek@cs.elte.hu

December 21, 2004

Abstract

We show that the set of Liouville numbers is either null or non-σ-finite with respect to every translation invariant Borel measure on ℝ, in particular, with respect to every Hausdorff measure H^{g} with gauge function g. This answers a question of D. Mauldin. We also show that some other simply defined Borel sets like non-normal or some Besicovitch-Eggleston numbers, as well as all Borel subgroups of ℝ that are not F_{σ} possess the above property. We prove that, apart from some trivial cases, the Borel class, Hausdorff or packing dimension of a Borel set with no such measure on it can be arbitrary.

∗Corresponding author. Partially supported by Hungarian Scientific Foundation grant no. 37758 and F 43620.

†Partially supported by Hungarian Scientific Foundation grant no. F 43620.

MSC codes: Primary 28C10; Secondary 28A78, 43A05 Key Words: translation invariant Borel measure, Liouville numbers, σ-finite

Introduction

In many branches of mathematics a standard tool is that ‘nicely defined’ sets admit natural probability measures. For example, limit sets in the theory of Iterated Function Systems or Conformal Dynamics as well as self-similar sets in Geometric Measure Theory are usually naturally equipped with an invariant Borel measure, very often with a Hausdorff or packing measure. In many situations the sets in consideration are unbounded, for example periodic, so we cannot hope for an invariant probability measure. Similarly, the trajectories of the Brownian motion are of positive σ-finite H^{g}-measure with probability 1, where the gauge function g is g(t) = t^{2} log log \frac{1}{t} in case of planar Brownian motion and g(t) = t^{2} log \frac{1}{t} log log log \frac{1}{t} in dimension 3 and higher. Therefore the natural notion to work with is that of an invariant positive and σ-finite measure.

It is natural to ask if there is some sort of unified theory behind the existence of these measures, for example, one is tempted to ask if every Borel subset of ℝ^{n} of some ‘regular structure’ is positive and σ-finite for some Hausdorff measure, or at least admits a positive and σ-finite invariant measure. In particular, R. D. Mauldin ([13], [4] and see [3] and [2] for partial and related results) formulated this question about a specific well-known set of very nice structure; the set of Liouville numbers, denoted by L:

Definition 0.1

L = \{x ∈ ℝ \ ℚ : ∀n ∈ ℕ ∃p, q ∈ ℕ (q ≥ 2) such that |x − \frac{p}{q}| < \frac{1}{q^{n}} \} .

Question 0.2 (Mauldin) Is there a translation invariant Borel measure on ℝ such that the set of Liouville numbers is of positive and σ-finite measure?

Note that we of course do not require that the measure be σ-finite on ℝ. Not only because Hausdorff measures are non-σ-finite on ℝ, but also because it is well-known that every σ-finite translation invariant Borel measure on the real line is a constant multiple of Lebesgue measure.

As we will answer this question in the negative, we introduce a definition.

Definition 0.3 A nonempty Borel set B ⊂ ℝ is said to be immeasurable if it is either null or non-σ-finite for every translation invariant Borel measure on ℝ.

The main result of this paper will be Theorem 1.1 stating that the set of Liouville numbers is immeasurable. Then in the second part of the paper we show using various methods that there are other well-known ‘nice’ immeasurable sets. Specifically, we show that the set of non-normal numbers, the complement of the set of Besicovitch-Eggleston numbers, BE(1, 0) (one of the Besicovitch-Eggleston classes itself) are all immeasurable. One of the main tool is Theorem 2.10 stating that every Borel but not F_{σ} additive subgroup of ℝ is immeasurable. Using this we also show that there are immeasurable sets of arbitrary Borel class (except of course open, as sets of positive Lebesgue measure are obviously not immeasurable). Similarly, we provide examples of immeasurable sets of arbitrary Hausdorff or packing dimension.

We note here that it is not only the regular structure of the sets considered here that makes it difficult to prove immeasurability. Even it is highly nontrivial to construct some immeasurable set. The two papers [10] and [5] containing the two known examples are entirely devoted to the constructions of the two immeasurable sets.

We would like to point out that in this paper a Borel measure is a measure defined on a σ-algebra containing the Borel sets. A group is always an additive subgroup of ℝ. Otherwise the notation we follow throughout the paper can be found for example in [11] or [8] and [9]. λ denotes Lebesgue measure, int(A) is the interior of the set A, A+B = {a+b : a ∈ A, b ∈ B}, A+t = {a+t : a ∈ A}, card (X) is the cardinality of the set X, a G_{δ} set is the countable intersection of open sets, a G_{δσ} set is the countable union of G_{δ} sets, etc. By Cantor set we mean a set homeomorphic to the classical ‘middle-thirds’ Cantor set.

1 Liouville numbers

In this section we answer the question of Mauldin.

Theorem 1.1 The set of Liouville numbers is immeasurable; that is, either null or non-σ-finite for every translation invariant Borel measure on ℝ.

It is well-known and easy to check that L is of Lebesgue measure zero, dense, G_{δ} and periodic mod ℚ (that is, L + q = L for every q ∈ ℚ). Hence the above theorem is a corollary to the following.

Theorem 1.2 Let B ⊂ ℝ be a nonempty G_{δ} set of Lebesgue measure zero. Assume that {t ∈ ℝ : B + t ⊂ B} is dense in ℝ. Then B is immeasurable.

The proof of Theorem 1.2 will be based on two lemmas. The first one is reminiscent of [16], where similar results are proved for finite measures using more complicated methods.

Lemma 1.3 Let B be a Borel set of Lebesgue measure zero and μ a Borel measure on ℝ for which B is positive and σ-finite. Then

(i) μ(B ∩ (B + t)) = 0 for λ-a.e. t,

(ii) there exists a Borel set B′ ⊂ B with μ(B′) > 0 and int(B′ − B′) = ∅,

(iii) there exists a compact set C ⊂ B with μ(C) > 0 and int(C − C) = ∅

Lemma 1.4 Let B be a dense G_{δ} set such that {t ∈ ℝ : B + t ⊂ B} is dense in ℝ, and C ⊂ B be a compact set with int(C −C) = ∅. Then there are uncountably many (in fact, continuum many) disjoint translates of C inside B.

It is easy to see that applying Lemma 1.4 to C of Lemma 1.3 (iii) yields Theorem 1.2

In the rest of the section we prove the two lemmas. Proof. (Lemma 1.3) (i) Let μ : S → [0, ∞] be the given measure, where S is a σ-algebra of subsets of ℝ containing all Borel sets. Define a new measure μ_{B} by

μ_{B}(S) = μ(B ∩ S) for every S ∈ S.

μ_{B} is clearly a σ-finite Borel measure on ℝ. Define

B̃ = {(x, y) ∈ ℝ^{2} : x + y ∈ B}.

This is clearly a Borel set, hence μ_{B} × λ-measurable. As both measures are σ-finite, we can apply the Fubini theorem to B̃. Vertical sections of B̃ are of the form

{y ∈ ℝ : x + y ∈ B} = {y ∈ ℝ : y ∈ B − x} = B − x,

therefore are all of Lebesgue measure zero. By Fubini, [μ_{B} × λ](B̃) = 0, and so λ-a.e. horizontal section of B̃ is of μ_{B}-measure zero. A horizontal section is {x ∈ ℝ : x + y ∈ B} = B − y, therefore for λ-a.e. y we obtain 0 = μ_{B}(B − y) = μ(B ∩ (B − y)). Replacing y by −t yields the result.

(ii) By (i) we can choose a countable dense set D ⊂ ℝ such that μ(B ∩ (B + d)) = 0 for every d ∈ D. Define

B′ = B \ ⋃_{d∈D}(B + d).

It is easy to check that μ(B′) = μ(B) > 0 and D ∩ (B′ − B′) = ∅, so the proof of (ii) is complete.

(iii) It is sufficient to find a compact set C ⊂ B′ of positive μ-measure. Since B′ ⊂ B, B′ is σ-finite for μ. Let B′ = ∪_{n=0}^{∞}S_{n}, where S_{n} ∈ S and μ(S_{n}) < ∞. Since μ(B′) > 0, there exists S = S_{n} ⊂ B′ such that 0 < μ(S) < ∞. Define

μ_{S}(A) = μ(S ∩ A) for every Borel set A ⊂ ℝ.

Note that in contrast with the above μ_{B}, this time the measure is defined only for Borel sets.

μ_{S} is clearly a finite measure on the Borel sets, hence inner regular w.r.t. compact sets [9, Thm 17.11]. Apply this to B′, a Borel set with μ_{S}(B′) = μ(S∩B′) = μ(S) > 0, and obtain a compact set C ⊂ B′ with μ_{S}(C) = μ(S ∩ C) > 0. So μ(C) > 0 follows, and the proof of Lemma 1.3 is complete. □

Proof. (Lemma 1.4) Let

T = {t ∈ ℝ : C + t ⊂ B}.

B is G_{δ}, so there are open sets U_{n} such that B = ∩_{n=0}^{∞}U_{n}. Clearly C + t ⊂ ∩_{n=0}^{∞}U_{n} iff C + t ⊂ U_{n} holds for every n ∈ ℕ. Therefore T = ∩_{n=0}^{∞}G_{n}, where G_{n} = {t ∈ ℝ : C + t ⊂ U_{n}}. As C is compact and U_{n} is open, G_{n} is open. Note that G_{n} is also dense by our assumption.

It is clearly sufficient to construct a Cantor set P ⊂ T with the property that (C+p_{0})∩(C+p_{1}) = ∅ holds for every pair of distinct points p_{0}, p_{1} ∈ P. We define P via a usual Cantor scheme as follows. Let 2^{n} stand for the set of 0–1 sequences of length n. We define nondegenerate compact intervals I_{s} for every n ∈ ℕ and s ∈ 2^{n} by induction on n (we also make sure that at level n all intervals are of length at most \frac{1}{n}). Fix I_{∅} such that I_{∅} ⊂ G_{0}. Once I_{s} is already defined for some s ∈ 2^{n}, we pick x ∈ I_{s}∩G_{n+1}. As G_{n+1} is open dense and C−C is nowhere dense we can find y ∈ [I_{s} ∩ G_{n+1}] \ [(C − C) + x]. This ensures (C + x) ∩ (C + y) = ∅, as otherwise c_{0} + x = c_{1} + y for some c_{0}, c_{1} ∈ C, so y = (c_{0} − c_{1}) + x, a contradiction. By compactness we can find disjoint I_{s^{⌢}0}, I_{s^{⌢}1} ⊂ I_{s} ∩G_{n+1} such that x ∈ I_{s^{⌢}0}, y ∈ I_{s^{⌢}1} and (C + I_{s^{⌢}0}) ∩ (C + I_{s^{⌢}1}) = ∅.

Now define

P = ⋂_{n=0}^{∞} ⋃_{s∈2^{n}} I_{s}.

Then clearly P is a Cantor set, P ⊂ T = ∩_{n=0}^{∞}G_{n} as I_{s} ⊂ G_{n} for every n and s ∈ 2^{n}. Moreover, if p_{0}, p_{1} ∈ P, p_{0} ≠ p_{1} then there are n and s ∈ 2^{n} such that p_{0} ∈ I_{s^{⌢}0} and p_{1} ∈ I_{s^{⌢}1} (or the other way around), but then (C+p_{0})∩(C+p_{1}) = ∅ holds since (C + I_{s^{⌢}0}) ∩ (C + I_{s^{⌢}1}) = ∅. This completes the proof of Lemma 1.4. □

Remark 1.5 Theorem 1.2 holds (with essentially the same proof) in ℝ^{n}, more generally in locally compact abelian Polish groups as well. Of course, Lebesgue measure is replaced by Haar measure.

No assumptions of Theorem 1.2 can be omitted. The empty set is not immeasurable by definition. The example of ℚ and the counting measure shows that the G_{δ} assumption is essential. Sets of positive Lebesgue measure are clearly not immeasurable. The density of {t ∈ ℝ : B + t ⊂ B} is also important, as the example B = L ∪ C_{1/3} shows, where C_{1/3} is the classical middle-thirds Cantor set. Indeed, log 2/ log 3-dimensional Hausdorff measure is positive and finite on B, since it is easy to see that dim_{H} L = 0; that is, Hausdorff dimension of L is zero (see [11] for the definition).

Theorem 1.2 provides a lot of immeasurable sets in the following sense. Let A be a (nonempty) set of Lebesgue measure zero, and let B be a G_{δ} set of measure zero containing A + ℚ. Then ∩_{q∈ℚ}(B + q) fulfills all conditions of Theorem 1.2, hence it is immeasurable. Clearly A ⊂ ∩_{q∈ℚ}(B + q), therefore every Lebesgue nullset can be covered by an immeasurable set. Combining this with Lemma 2.2 (i) we obtain that the σ-ideal generated by the immeasurable sets is the Lebesgue null ideal.

2 Non-normal numbers, classes of Besicovitch-Eggleston numbers and Borel groups

In this section we provide some more examples of well-known sets that are immeasurable.

The following definition is classical.

Definition 2.1 A real number x ∈ [0, 1) is called normal if its decimal expansion x = 0.d_{1}d_{2} … satisfies

lim_{n→∞} \frac{∑_{i=1}^{n} d_{i}}{n} = 4.5.

A real number x ∈ ℝ is called normal if its fractional part, {x} is normal. Note that for negative numbers this is not the usual decimal expansion. The set of normal numbers is denoted by N.

It is well-known and easy to check using the Strong Law of Large Numbers that λ-a.e. real number is normal.

Our next goal is to prove that the set of non-normal numbers, ℝ \ N is immeasurable. Note that the Hausdorff dimension of this set is 1 (see [1]), while we already mentioned that dim_{H} L = 0. Moreover, ℝ \ N is not G_{δ}, since its exact Borel class is G_{δσ} (that is, ℝ \ N ∈ G_{δσ} \ F_{σδ}, see [9, Ex 23.7]).

Before the proof we need one more easy lemma.

Lemma 2.2 (i) The class of immeasurable sets is closed under countable unions.

(ii) If the symmetric difference A△B is countable then A is immeasurable iff B is immeasurable.

Proof. (i) This clearly follows from the definition.

(ii) If A and B are nonempty countable then counting measure works for both. Otherwise all measures we consider are continuous; that is, singletons are of measure zero. □

Theorem 2.3 The set ℝ \ N of non-normal numbers is immeasurable; that is, either null or non-σ-finite for every translation invariant Borel measure on ℝ.

Proof. By Lemma 2.2 (ii) it is enough to show that (ℝ\N)\ℚ is immeasurable. Write this set as

(ℝ \ N) \ ℚ = ⋃_{k=1}^{∞} A_{k} ∪ ⋃_{k=1}^{∞} B_{k},

where A_{k} = \{x ∈ ℝ \ ℚ : lim sup_{n→∞} ^{\frac{∑_{i=1}^{n} d_{i}}{n}} ≥ 4.5 + \frac{1}{k} \} and similarly B_{k} =

\{x ∈ ℝ \ ℚ : lim inf_{n→∞} ^{\frac{∑_{i=1}^{n} d_{i}}{n}} ≤ 4.5 − \frac{1}{k} \}. By Lemma 2.2 (i) it is enough to prove that the sets A_{k} and B_{k} are immeasurable for every k. The other case being similar we check this only for A_{k}.

We show that Theorem 1.2 applies. As a.e. number is normal, A_{k} is of Lebesgue measure zero. Moreover, it is easy to see that it is dense and periodic modulo a dense set, namely the set of real numbers with finitely many nonzero digits. Therefore what remains to be checked is that A_{k} is G_{δ}.

A_{k} = ⋂_{l=1}^{∞} ⋂_{m=1}^{∞} ⋃_{n=m}^{∞} \{x ∈ ℝ \ ℚ : \frac{∑_{i=1}^{n} d_{i}}{n} > 4.5 + \frac{1}{k} − \frac{1}{l}\} ,

and \{x ∈ ℝ \ ℚ : ^{\frac{∑_{i=1}^{n} d_{i}}{n}} > 4.5 + \frac{1}{k} − \frac{1}{l}\} is clearly an open subset relative to ℝ \ ℚ, hence A_{k} is G_{δ} relative to ℝ \ ℚ, hence A_{k} is G_{δ} in ℝ as well. □

Our next example concerns the so called Besicovitch-Eggleston numbers. These are the numbers for which ‘the asymptotic frequency of every digit exists’: Let {x} = 0.d_{1}d_{2} … be the expansion of the fractional part of the real number x in base b ≥ 2. If there are two such expansions, we choose the one with finitely many nonzero digits.

Definition 2.4 Let α_{0}, …, α_{b−1} be real numbers such that α_{d} ≥ 0 for every d = 0, …, b − 1 and ∑_{d=0}^{b−1} α_{d} = 1.

BE(α_{0}, …, α_{b−1}) =

\{x ∈ ℝ : lim_{n→∞} \frac{card (\{i : 1 ≤ i ≤ n, d_{i} = d\})}{n} = α_{d} for every d = 0, …, b − 1\} .

The union of all these classes (for a fixed positive integer b) is denoted by BE_{b}.

Theorem 2.5 For every integer b ≥ 2 the set ℝ \ BE_{b} is immeasurable.

Proof. The proof is very similar to that of Theorem 2.3. We can again restrict ourselves to the set of irrational numbers. Write

(ℝ \ BE_{b}) \ ℚ = ⋃_{d=0}^{b−1} ⋃_{\begin{matrix}p<q \\ p,q∈ℚ∩[0,1] \end{matrix}} (C_{q}^{d} ∩ D_{p}^{d}) ,

where

C_{q}^{d} = \{x ∈ ℝ \ ℚ : lim sup_{n→∞} \frac{card (\{i : 1 ≤ i ≤ n, d_{i} = d\})}{n} ≥ q\}

and

D_{p}^{d} = \{x ∈ ℝ \ ℚ : lim inf_{n→∞} \frac{card (\{i : 1 ≤ i ≤ n, d_{i} = d\})}{n} ≤ p\} .

It is again easy to see using the Strong Law of Large Numbers that Lebesgue a.e. number is in BE_{b}, namely in BE(\frac{1}{b}, …, \frac{1}{b}), hence every C_{q}^{d} and D_{p}^{d} is of Lebesgue measure zero. Moreover, these sets are invariant under all translations from the dense set of numbers with finitely many nonzero digits. So it is enough to show that C_{q}^{d} and D_{p}^{d} are G_{δ} in ℝ \ ℚ. But this is clear, as

C_{q}^{d} = ⋂_{l=1}^{∞} ⋂_{m=1}^{∞} ⋃_{n=m}^{∞} \{x ∈ ℝ \ ℚ : \frac{card (\{i : 1 ≤ i ≤ n, d_{i} = d\})}{n} > q − \frac{1}{l}\} ,

and similarly for D_{p}^{d}. □

So far all immeasurable sets in this paper were comeager. Our next example is easily seen to be meager. Not surprisingly we need new methods to prove immeasurability here.

Definition 2.6 Let A be a set of natural numbers. The expression

lim_{n→∞} \frac{card (A ∩ [1, n])}{n}

(if exists) is called the density of A.

It is easy to see that sets of zero density form an ideal on the natural numbers. Dually, sets of density 1 form a filter.

Theorem 2.7 BE(1, 0) is immeasurable; that is, the set of real numbers with the property that in the dyadic expansion the 1’s form a set of zero density is either null or non-σ-finite for every translation invariant Borel measure on ℝ.

This result will follow from the following more general statement.

Theorem 2.8 Let B be a Borel set such that B + B ⊂ B. If B − B, the group generated by B, is not F_{σ} then B is immeasurable.

Proof. Assume towards a contradiction that μ is a translation invariant measure for which B is positive and σ-finite. As in the proof of Lemma 1.3 (iii) we obtain that there exists a compact set C ⊂ B of positive μ-measure. We get a contradiction by showing that C has uncountably many disjoint translates in B. We define the transfinite sequence of numbers {t_{α} : α < ω_{1}} by transfinite induction, so that the sets {C + t_{α} : α < ω_{1}} are pairwise disjoint. Clearly (C + t_{α}) ∩ (C + t_{β}) = ∅ iff t_{α} ∉ C − C + t_{β}. Throughout the induction we choose the numbers from B, which implies C + t_{α} ⊂ B by our assumption, so what remains to show is that the induction cannot get stuck at some countable ordinal. At step α our task is to find t_{α} ∈ B so that t_{α} ∉ C − C + t_{β} for every β < α. These latter sets are in B − B because this is a group, and we claim that they cannot cover B. Indeed, if B ⊂ ∪_{β<α}(C − C + t_{β}) then B − B = ∪_{β,γ<α}[(C − C + t_{β}) − (C − C + t_{γ})], so B − B is F_{σ}, which is a contradiction. □

Before we finish the proof of Theorem 2.7 we need one more lemma, which provides an example of an immeasurable group.

Lemma 2.9 Denote

G = {x ∈ ℝ : d_{n} = d_{n+1} for all n but a set of density zero} ,

where {x} = 0.d_{1}d_{2} … is the dyadic expansion of the fractional part of x. Then

(i) BE(1, 0) − BE(1, 0) = G,

(ii) G is a Borel but not F_{σ} group.

Proof. (i) Put B = BE(1, 0) and define S(x) = {n ∈ ℕ : d_{n} ≠ d_{n+1}}, where {x} = 0.d_{1}d_{2} … is the dyadic expansion of the fractional part of x. Then G = {x ∈ ℝ : S(x) is of density zero}. We have to show that B − B = G. Note that all dyadic rationals (numbers with only finitely many nonzero digits) are in B as we chose the expansion that is eventually zero.

First we check B − B ⊂ G. Fix b_{1}, b_{2} ∈ B, and write b_{1} − b_{2} = b_{1} + (−b_{2}). If b_{2} is a dyadic rational then so is −b_{2}, hence all but finitely many digits of b_{1} + (−b_{2}) and b_{1} coincide, hence b_{1} − b_{2} ∈ B. So we can assume that b_{2} is not a dyadic rational, and then we obtain the expansion of (the fractional part of) −b_{2} by replacing 0’s with 1’s and vice versa. Let d_{i}, e_{i} and f_{i} be the digits of the expansions of b_{1}, −b_{2} and b_{1} + (−b_{2}), respectively. Then the set A = {i : d_{i} = d_{i+1} = 0, e_{i} = e_{i+1} = 1} is of density 1. If i ∈ A then either f_{i} = f_{i+1} = 0 or f_{i} = f_{i+1} = 1 (depending on whether there is a carried digit from the i + 2^{nd} place), hence i ∉ S(b_{1} + (−b_{2})). Therefore S(b_{1} + (−b_{2})) is of density zero and so b_{1} + (−b_{2}) ∈ G.

Now we show that G ⊂ B−B. Given x ∈ G with expansion {x} = 0.d_{1}d_{2} … we construct b ∈ B with expansion {b} = 0.e_{1}e_{2} … such that x + b ∈ B. We can assume that x is not a dyadic rational, otherwise b = 0 works. Define b as follows: let e_{i} = 1 iff d_{i} = 1 and d_{i+1} = 0. Since x ∈ G we have that S(x) is of density zero, therefore b ∈ B. Let {x + b} = 0.f_{1}f_{2} . . .. It suffices to check that f_{i} = 1 iff d_{i} = 0 and d_{i+1} = 1, hence then {i : f_{i} = 1} is a subset of S(x), therefore is of density zero, thus x + b ∈ B. Suppose first that d_{i+1} = 0. Then e_{i+1} = 0, so there is no carried digit from this place. Moreover e_{i} = d_{i}, so f_{i} = 0. Suppose now that d_{i+1} = 1. Let k be the minimal integer such that i+2 ≤ k and d_{k} = 0. Then e_{j} = 0 for i ≤ j ≤ k −2 and for j = k, and e_{k−1} = 1. Consequently, there is no carried digit from the k^{th} place, so f_{k−1} = 0 and there is a carried digit from the k − 1^{st} place, and so on there is a carried digit for every i + 1 ≤ j ≤ k − 1, and at the end f_{i} = 1 iff d_{i} = 0.

(ii) B = BE(1, 0) and G are clearly Borel sets. As B is easily seen to be closed under addition, G = B − B is a group. So we have to check that G is not F_{σ}.

Suppose towards a contradiction that it is. First we construct a continuous map f : [0, 1) \ 𝔻 → [0, 1) \ 𝔻 (where 𝔻 is the set of dyadic rationals), with the property that f^{−1}(G ∩ ([0, 1) \ 𝔻)) = B ∩ ([0, 1) \ 𝔻). For x = 0.d_{1}d_{2} … construct the digits of f(x) = 0.e_{1}e_{2} … as follows. Let d_{1} = e_{1}, and then define e_{n} recursively such that e_{n+1} = e_{n} iff d_{n} = 0. It is easy to check that f satisfies the requirements. By assumption G is F_{σ}, hence G∩([0, 1)\𝔻) is F_{σ} in [0, 1)\𝔻, so B ∩ ([0, 1) \ 𝔻) is also F_{σ} in [0, 1) \ 𝔻. Then B ∩ [0, 1) is G_{δσ}, which is a contradiction, as by [9, Ex 23.7] it is not G_{δσ}. □

Proof. (Theorem 2.7) As we already mentioned, BE(1, 0) is clearly Borel and closed under addition, hence Lemma 2.9 shows that Theorem 2.8 applies. □

We also formulate the following immediate corollary to Theorem 2.8, which is interesting in its own right.

Theorem 2.10 If G ⊂ ℝ is a Borel but not F_{σ} additive subgroup then G is immeasurable. □

Remark 2.11 In fact this theorem holds in every Polish group, and F_{σ} can be weakened to K_{σ} (that is, countable union of compact sets). In case the group is not abelian we have to use left-translations everywhere (including the definition of immeasurable.)

3 Immeasurable sets with arbitrary Borel class, Hausdorff or packing dimension

A nonempty open set is of positive Lebesgue measure, hence cannot be immeasurable. Our next theorem shows that we can find immeasurable sets in all other Borel classes.

For 1 ≤ α < ω_{1} denote Σ_{α}^{0}, Π_{α}^{0} and Δ_{α}^{0} the additive, multiplicative and ambiguous Borel classes, respectively (see e.g. [9]). We say that a Borel set B is of exact class Σ_{α}^{0} (Π_{α}^{0}) if B ∈ Σ_{α}^{0} \ Π_{α}^{0} (B ∈ Π_{α}^{0} \ Σ_{α}^{0}). We say that B is of exact class Δ_{α}^{0} if B ∈ Δ_{α}^{0} but B ∉ Σ_{β}^{0} ∪ Π_{β}^{0} for β < α.

Theorem 3.1 For every 1 ≤ α < ω_{1} there exist immeasurable sets of exact Borel class Π_{α}^{0}. There exist immeasurable sets of exact Borel class Σ_{α}^{0} and Δ_{α}^{0} iff 2 ≤ α < ω_{1}.

Proof. Δ_{1}^{0}: Impossible, as open sets are not immeasurable.

Σ_{1}^{0}: Impossible, for the same reason.

Π_{1}^{0}: Davies [5] constructed an immeasurable Cantor set D.

Δ_{2}^{0}: D minus a point is immeasurable by Lemma 2.2 (ii).

Σ_{2}^{0}: D + ℚ is immeasurable by Lemma 2.2 (i). To see that its exact class is F_{σ}, first note that as D is immeasurable, it is of Lebesgue measure zero. Moreover it is closed, hence nowhere dense, therefore D + ℚ is a dense meager F_{σ} set. Thus it cannot be G_{δ}, as dense G_{δ} sets are comeager.

Π_{2}^{0}: The set of Liouville numbers is immeasurable, see Theorem 1.2. Larman’s example [10] is also G_{δ}. Both examples are easily seen to be non-F_{σ} using the fact that nonmeager F_{σ} sets have interior points.

All other classes: These examples can be chosen to be groups: in [12] Borel groups of exact class Π_{α}^{0}, Σ_{α}^{0} and Δ_{α}^{0} are constructed (for 3 ≤ α < ω_{1}), and these are all immeasurable by Theorem 2.10. □

Our next theorem shows that there are lots of immeasurable sets from the viewpoint of Hausdorff dimension as well.

Theorem 3.2 For every 0 ≤ α ≤ 1 there exists an immeasurable subset of ℝ of Hausdorff dimension α. In particular, these sets can be chosen to be additive subgroups of ℝ.

Proof. First we consider the case α ≠ 1.

In [7] groups G_{α} ⊂ ℝ of Hausdorff dimension α are constructed for every 0 < α < 1. In fact, the proof also yields a nontrivial group G_{0}, but for our purposes it is sufficient to put G_{0} = {0}. We will also use the fact that G_{α} ⊂ G_{β} for α < β.

Unfortunately G_{α} is F_{σ}. Our goal is to find a group H ‘sufficiently orthogonal to G_{α}’ such that G′_{α} = G_{α} + H is a non-F_{σ} group of dimension α.

The following lemma is probably well-known, however, we were unable to find suitable references for its second half, so we include a proof here.

Lemma 3.3 Let M be a meager subset of ℝ \ {0}. Then the typical compact subset C of ℝ (that is, comeager many elements of the space of compact subset of ℝ endowed with the Hausdorff metric) possesses the following properties.

(i) C is a Cantor set which is linearly independent over the rationals.

(ii) The group generated by C is disjoint from M.

Proof. (i) This is essentially [9, Ex 19.2] which follows from [9, Thm 19.1].

(ii) The complement of M contains a set that can be written as the intersection of countably many dense open sets. Hence it is enough to prove that for a fixed dense open set U, the group generated by C is in U.

For (n_{1}, …, n_{k}, n′_{1}, …, n′_{l}) ∈ (ℕ \ {0})^{k+l} define

C(n_{1}, …, n_{k}; n′_{1}, …, n′_{l}) =

= {n_{1}c_{1} + … + n_{k}c_{k} − n′_{1}c′_{1} − … − n_{l}c′_{l} : c_{i}, c′_{j} ∈ C are all distinct}.

Set C(∅) = {0}. The union of all these sets as (n_{1}, …, n_{k}, n′_{1}, …, n′_{l}) ranges over the countable sets (ℕ\{0})^{k+l} is the group generated by C, hence it suffices to prove that

C(n_{1}, …, n_{k}; n′_{1}, …, n′_{l}) ⊂ U

for the typical C. Define the map f : ℝ^{k+l} → ℝ by

f(x_{1}, …, x_{k}, x′_{1}, …, x′_{l}) = n_{1}x_{1} + … + n_{k}x_{k} − n′_{1}x′_{1} − … − n_{l}x′_{l}.

This map is clearly continuous and open, therefore f^{−1}(U) is dense and open in ℝ^{k+l}. Then [9, Thm 19.1] states precisely that for the typical C the set

(C)^{k+l} = {(c_{1}, …, c_{k}; c′_{1}, …, c′_{l}) : c_{i}, c′_{j} ∈ C are all distinct}

is contained in f^{−1}(U). Therefore C(n_{1}, …, n_{k}; n′_{1}, …, n′_{l}) ⊂ U, which finishes the proof of the lemma. □

The last easy statement we need before we define our group H is that every Cantor subset K of ℝ contains a Cantor set K_{0} of packing dimension zero. We need this additional step as the typical compact set has packing dimension 1. See [11] for the definition. Packing dimension (which is the same as upper packing dimension) of a set B will be denoted by dim_{p} B, upper Minkowski dimension of B is denoted by \overline{dim}_{M}B. We will use the well-known inequality dim_{p} B ≤ \overline{dim}_{M}B.

In order to prove this statement, we perform a Cantor-type construction as follows. We can clearly assume K ⊂ [0, 1]. For every integer m ≥ 1 divide [0, 1] into 2^{m} subintervals of length \frac{1}{2^{m}}. One can check that it is possible to choose recursively for every m a nonempty subfamily I_{m} of these intervals such that

• ∪I_{m} ⊃ ∪I_{n} for m < n,

• card(I_{m}) ≤ m

• ∀I ∈ I_{m} ∃n ∈ ℕ and J, J′ ∈ I_{n}, J ≠ J′, such that J, J′ ⊂ I,

• ∀I ∈ I_{m} int(I) ∩ K ≠ ∅.

Define K_{0} = ⋂_{m=1}^{∞} ∪I_{m}. This set is easily seen to be a Cantor set, and using the “box-counting” version of the upper Minkowski dimension (see [11, page 78]) yield \overline{dim}_{M}K_{0} = 0. Hence dim_{p} K_{0} = 0.

Now we complete the proof of Theorem 3.2. The only nonmeager Borel subgroup of ℝ is ℝ itself (since by [14, page 93] if B is a nonmeager Borel set than B − B contains an interval). Therefore the sets G_{1− \frac{1}{n}} \ \{0\} are meager. Apply Lemma 3.3 to M = ∪_{n∈ℕ}(G_{1− \frac{1}{n}} \\{0\}) and obtain a compact set C. By the previous argument we can choose a Cantor set C_{0} ⊂ C such that dim_{p} C_{0} = 0. Fix a Borel but not F_{σ} subset B ⊂ C_{0}, and define H as the group generated by B. We have to check that G′_{α} = G_{α} + H is a Borel but not F_{σ} group of Hausdorff dimension α for every 0 ≤ α < 1.

As G_{α} ⊂ G_{β} for α < β, we obtain G_{α} ∩ H = {0} for every 0 ≤ α < 1.

First we show that H is Borel. As H is the group generated by B, it is the union of all the sets B(n_{1}, …, n_{k}; n′_{1}, …, n′_{l}) (the notation was defined in the proof of Lemma 3.3). Therefore it is sufficient to show that these latter sets are all Borel. But since B is linearly independent, such a set is the continuous one-to-one image of the Borel set (B)^{k+l} (as above, using the notation of the proof of Lemma 3.3), hence Borel.

Similarly, the natural map from the Borel set G_{α}×H ⊂ ℝ^{2} onto G′_{α} = G_{α}+H is continuous. Moreover, it is one-to-one, since g_{1} + h_{1} = g_{2} + h_{2} implies g_{1} − g_{2} = h_{2} − h_{1} and G_{α} and H are groups satisfying G_{α} ∩ H = {0}. Hence G′_{α} is Borel.

In order to show that G′_{α} is not F_{σ} will prove G′_{α} ∩ C_{0} = B, which is clearly sufficient as C_{0} is compact and B is not F_{σ}. Suppose g′ = g + h ∈ C_{0}, where g ∈ G_{α}, h ∈ H. We want to show that g′ ∈ B. Clearly, g = g′ − h ∈ C_{0} − H.

But C_{0} −H is a subset of the group generated by C, and G_{α} ⊂ M, so by Lemma 3.3 (ii) we obtain g = 0. Hence h ∈ C_{0}, and h is in the group generated by B ⊂ C_{0}. But C_{0} is linearly independent, therefore h ∈ B and we are done, as h = g′.

What remains to be shown is that dim_{H} G′_{α} = α. Obviously α = dim_{H} G_{α} ≤ dim_{H} G′_{α}, so it suffices to prove dim_{H} G′_{α} ≤ α. It is well-known [11, Thm 8.10] that dim_{p}(X × Y) ≤ dim_{p} X + dim_{p} Y, which implies dim_{p}(X_{1} × … × X_{n}) ≤ dim_{p} X_{1} + … + dim_{p} X_{n}. First we prove dim_{p} H = 0. As H is the countable union of the sets B(n_{1}, …, n_{k}; n′_{1}, …, n′_{l}), it is sufficient to show that dim_{p} B(n_{1}, …, n_{k}; n′_{1}, …, n′_{l}) = 0. This set can be covered by a Lipschitz image of the set B^{k+l}, and Lipschitz images do not increase dimension, so it is enough to check dim_{p} B^{k+l} = 0. But this clearly holds as B ⊂ C_{0}, so dim_{p} B ≤ dim_{p} C_{0} = 0, hence dim_{p}(B ×…×B) ≤ dim_{p} B +…+dim_{p} B = 0+…+0 = 0. So dim_{p} H = 0.

Another part of [11, Thm 8.10] states that dim_{H}(X×Y) ≤ dim_{H} X+dim_{p} Y. Choosing X = G_{α} and Y = H yields dim_{H}(G_{α} × H) ≤ α + 0 = α. But G′_{α} is a Lipschitz image of G_{α} × H, hence dim_{H} G′_{α} ≤ α. This completes the proof for α < 1.

Finally, we have to deal with the case α = 1. As G′_{1− \frac{1}{n}} is an increas-

ing sequence of immeasurable groups of Hausdorff dimension 1 − \frac{1}{n}, the union ∪_{n=1}^{∞}G_{1− \frac{1}{n}} is a group of Hausdorff dimension 1 which is immeasurable by Lemma 2.2 (i). □

Remark 3.4 In [7] only dim_{H} G_{α} = α is proved, but dim_{p} G_{α} = α also holds (see sketch of proof below). Therefore one can easily check that a minor modification of the above proof yields dim_{p} G′_{α} = α, hence Theorem 3.2 is also valid for packing instead of Hausdorff dimension.

Now we sketch the proof of dim_{p} G_{α} = α. Since α = dim_{H} G_{α} ≤ dim_{p} G_{α}, it is enough to prove dim_{p} G_{α} ≤ α. Write G_{α} = ∪_{k_{0}∈ℕ} ∪_{κ∈ℕ} ∪_{n∈ℤ}(G_{α}(k_{0}, κ) + n), where G_{α}(k_{0}, κ) is the set of points in [0, 1] satisfying the requirement in the definition of G_{α} for fixed k_{0} and κ (see the construction in [7]). It is sufficient to show dim_{p} G_{α}(k_{0}, κ) ≤ α for fixed k_{0} and κ. This set is a Cantor set, and it is not hard to see that the ‘natural measure’ on it (we always split the measure into equal parts in the Cantor-type construction) satisfies all conditions of [11, Theorem 6.11 (2)] with A = G_{α}(k_{0}, κ), λ = 1 and s = α + ε, where ε > 0 is arbitrary. Hence dim_{p} G_{α}(k_{0}, κ) ≤ α.

4 Concluding remarks

There are numerous questions arising naturally concerning immeasurability. Is it really weaker to require some translation invariant Borel measure than a Hausdorff or packing measure? The answer to the second version turns out to be positive, as Peres [15] showed that some of the so called Bedford-McMullen carpets are of zero or non-σ-finite packing measure for every gauge function, while by recent results of the authors [6] none of these carpets are immeasurable. We do not know the answer to the other version: Does the existence of a translation invariant Borel measure that is positive and σ-finite on a Borel set B imply the existence of a Hausdorff measure with a suitable gauge function that is positive and σ-finite on B?

We may be overlooking something simple, but the following questions also seem to be open. We say that a Borel set B is measured if it is not immeasurable; that is, there is a translation invariant Borel measure for which B is positive σ-finite. Is the union of two measured sets also measured? (It can be shown using the construction in [5] that this is false for countably many sets.) Is the set of Liouville numbers a finite/countable union of measured sets? Is every Borel set a finite/countable union of measured sets?

Acknowledgement Originally, the results of this paper were proven for measures defined on the Borel sets. The authors are indebted to D. H. Fremlin for pointing out how to generalize the results for measures defined on σ-algebras containing the Borel sets. We are also grateful to R. D. Mauldin for helpful discussions.

References

[1] L. Barreira, B. Saussol and J. Schmeling, Distribution of frequencies of digits via multifractal analysis, J. Number Theory 97 (2002), no. 2, 410– 438.

[2] C. E. Bluhm, Liouville numbers, Rajchman measures, and small Cantor sets, Proc. Amer. Math. Soc. 128 (2000), no. 9, 2637–2640.

[3] Y. Bugeaud, M. M. Dodson and S. Kristensen, Zero-infinity laws in Diophantine approximation, arXiv:math.NT/0310433 v2.

[4] M. Csörnyei, Open Problems. Compiled and edited by M. Csörnyei, Proceedings from the conference ‘Dimensions and Dynamics’, Miskolc, Hungary, July 20-24, 1998. Periodica Math. Hung. 37 (1998), 227–237.

[5] R. O. Davies, Sets which are null or non-sigma-finite for every translation-invariant measure. Mathematika 18 (1971), 161–162.

[6] M. Elekes and T. Keleti, Translation invariant measures for some selfaffine sets, manuscript.

[7] P. Erdős and B. Volkmann, Additive Gruppen mit vorgegebener Hausdorffscher Dimension, J. Reine Angew. Math. 221 (1966), 203–208.

[8] K. J. Falconer, The geometry of fractal sets. Cambridge Tracts in Mathematics No. 85, Cambridge University Press, 1986.

[9] A. S. Kechris: Classical Descriptive Set Theory. Graduate Texts in Mathematics No. 156, Springer-Verlag, 1995.

[10] D. G. Larman, The approximation of G_{δ}-sets, in measure, by F_{σ}-sets, Proc. Cambridge Philos. Soc. 61 (1965), 105–107.

[11] P. Mattila: Geometry of Sets and Measures in Euclidean Spaces. Cambridge Studies in Advanced Mathematics No. 44, Cambridge University Press, 1995.

[12] R. D. Mauldin, On the Borel subspaces of algebraic structures, Indiana Univ. Math. J. 29 (1980), no. 2, 261–265.

[13] R. D. Mauldin, Problem Session at the Millennium Symposium, Denton, Texas, May, 2000.

[14] J. C. Oxtoby: Measure and Category. A survey of the analogies between topological and measure spaces. Second edition. Graduate Texts in Mathematics No. 2, Springer-Verlag, 1980.

[15] Y. Peres, The packing measure of self-affine carpets, Math. Proc. Cambridge Philos. Soc. 115 (1994), no. 3, 437–450.

[16] V. Prokaj, A characterization of singular measures, Real Anal. Exchange 29 (2003/04), no. 2, 805–812.