Real Analysis Exchange Vol. 27(1), 2001/2002, pp. 941–944

Krzysztof Ciesielski^{∗}, Department of Mathematics, West Virginia University, Morgantown, WV 26506-6310, USA. e-mail: K Cies@math.wvu.edu, internet:

http://www.math.wvu.edu/~kcies

MEASURE ZERO SETS WHOSE ALGEBRAIC SUM IS NON-MEASURABLE

Abstract In this note we will show that for every natural number n > 0 there exists an S ⊂ [0, 1] such that its n-th algebraic sum nS = S + ⋯ + S is a nowhere dense measure zero set, but its n + 1-st algebraic sum nS + S is neither measurable nor it has the Baire property. In addition, the set S will be also a Hamel base, that is, a linear base of ℝ over ℚ.

We use the standard notation as in [2]. Thus symbols ℝ, ℚ, ℤ, and 𝔠 stand for the set of real numbers, the set of rational numbers, the set of integers, and the cardinality of ℝ, respectively. The set of natural numbers {0, 1, 2, … } will be denoted by either ℕ or ω, and |X| will stand for the cardinality of a set X. For A, B ⊆ ℝ we put A + B = {a + b: a ∈ A & b ∈ B} and LIN_{ℚ}(A) will stand for the linear subspace of ℝ over ℚ spanned by A. In addition for 0 < n < ω symbol [X]^{n} will stand for the family of all n-element subsets of X and nA for the n-th algebraic sum of A, that is,

nA = {∑_{i=1}^{n} a_{i}: a_{i} ∈ A for all i = 1, 2, …, n} .

Thus 2S = S + S. For a Polish space X we say that B ⊂ X is a Bernstein set (in X) provided B and X \ B intersect every perfect set P ⊂ X. Clearly every Bernstein subset of an interval in ℝ is neither measurable nor it has the Baire property.

For 0 < p < q < ω let Z_{q}^{p} stand for the set of all numbers x ∈ [0, 1] which in base q have representations formed with digits 0, …, p, that is,

Z_{q}^{p} = \{∑_{i=1}^{∞} \frac{a(i)}{q^{i}}: a(i) ∈ \{0, …, p\} for every i = 1, 2, … \} .

Key Words: sets, algebraic sum, measurability.

Mathematical Reviews subject classification: 28A05

Received by the editors December 27, 2000

∗Papers authored or co-authored by a Contributing Editor are managed by a Managing Editor or one of the other Contributing Editors.

Clearly if p < q −1 then the set Z_{q}^{p} is closed, nowhere dense, and has Lebesgue measure zero. Also, for every k ∈ ℕ

if 0 < kp < q then kZ_{q}^{p} = Z_{q}^{kp}.  (1)

In what follows we will also need the following lemma.

Lemma 1. Let n > 0 be a natural number, q = 2n+2, and A ⊂ ℝ with |A| < 𝔠. Then for every x ∈ [0, 1] \ LIN_{ℚ}(A) there exists {x_{0}, …, x_{n}} ∈ [Z_{q}^{2}]^{n+1} \ A such that x = x_{0}+⋯+x_{n} and {x_{0}, …, x_{n}}∪A is linearly independent over ℚ.

Proof. In the proof any sequence ⟨z(i) ∈ {0, …, q − 1}: i = 1, 2, … ⟩ will be treated as a base q representation of a number z ∈ [0, 1], that is, z = ∑_{i=1}^{∞} \frac{z(i)}{q^{i}}.

Let x ∈ [0, 1]\LIN_{ℚ}(A). Then x > 0 and it can be represented as sequence ⟨x(i) ∈ {0, …, q − 1}: i = 1, 2, … ⟩. We can also assume that the set T = {i: x(i) > 0} is infinite, since any number with almost all x(i)’s being zero has also another representation with almost all x(i)’s equal to q − 1.

Let {T_{0}, …, T_{n−1}} be a partition of T onto infinite sets. For each i ∉ T and j ≤ n define x_{j}^{∗}(i) = 0. For i ∈ T_{k} we choose x_{j}^{∗}(i) ∈ {0, 1, 2} such that

x_{k}^{∗}(i) ∈ {0, 1}, x_{n}^{∗}(i) ∈ {1, 2}, and x_{0}^{∗}(i) + ⋯ + x_{n}^{∗}(i) = x(i).  (2)

Such a choice can be made since 0 < x(i) ≤ 2n + 1 for any such i. Next, by induction on k < n, we will choose the sequences ⟨s_{k}(i) ∈ {0, 1}: i < ω⟩ such that

s_{k}(i) = 0 for all i ∉ T_{k}.  (3)

We aim for x_{n} = x_{n}^{∗} − ∑_{j=0}^{n−1} s_{j} and x_{k} = x_{k}^{∗} + s_{k} for every k < n. Notice that, by (2) and (3), such a definition ensures that x_{j}’s belong to Z_{q}^{2} and that their sum is equal to x. The freedom of choice of s_{i}’s will allow us to enforce required linear independence.

In our inductive construction we will use the following notation for Z ⊂ ℝ and T ⊂ {1, 2, 3, … }:

Z ↾ T = \{z ↾ T: z maps \{1, 2, 3, … \} into \{0, …, q − 1\} & ∑_{i=1}^{∞} \frac{z(i)}{q^{i}} ∈ Z\} .

Now, since A_{0} = (LIN_{ℚ}(A ∪ {x})) ↾ T_{0} has cardinality less than 𝔠 there is an s_{0} as in (3) for which ⟨x_{0}^{∗}(i) + s_{0}(i): i ∈ T_{0}⟩ ∉ A_{0}. This clearly implies that x_{0} = x_{0}^{∗} + s_{0} ∉ LIN_{ℚ}(A ∪ {x}). In general, if for some 0 < k < n the sequences s_{i}’s (so also x_{i}’s) are already defined for all i < k we put Measure Zero Sets Whose Algebraic Sum Is Non-Measurable 943

A_{k} = (LIN_{ℚ}(A ∪ {x} ∪ {x_{i}: i < k})) ↾ T_{k} and choose s_{k} as in (3) for which ⟨x_{k}^{∗}(i) + s_{k}(i): i ∈ T_{k}⟩ ∉ A_{k}. This ensures that

x_{k} = x_{k}^{∗} + s_{k} ∉ LIN_{ℚ}(A ∪ {x} ∪ {x_{i}: i < k}).  (4)

This finishes the inductive construction.

Finally, by (4), x_{n} = x − (x_{0} + ⋯ + x_{n−1}) ∉ LIN_{ℚ}(A ∪ {x_{i}: i < n}) since x ∉ LIN_{ℚ}(A).

Theorem 2. For every natural number n > 0 there exists a Hamel basis H ⊂ Z_{2n+2}^{2} such that for every natural number m > n the set [0, 1] ∩ mH is Bernstein in [0, 1].

Before we prove the theorem, we like to list some of its corollaries. First note that if H is as in the theorem then, by (1), for every natural number k ≤ n we have kH ⊂ Z_{2n+2}^{2k}. So kH is nowhere dense and it has measure 0. In particular,

Corollary 3. For every natural number n > 0 there exists an S ⊂ [0, 1] such that for every k ∈ ℕ, kS is Lebesgue measurable if and only if k ≤ n.

Corollary 3 used with n = 1 implies that there exists a measure 0 subset S of [0, 1] such that S + S is non-measurable. This fact has been known for quite a while and was used by several authors. (See e.g. [5, 3, 1].) However, we have not been able to locate its proof in the literature. This prompted the author to write this note. At the same time we should note here that it is very easy to find a meager measure zero set E ⊂ ℝ such that for some natural number n the n-th algebraic sum nE of E is neither measurable nor it has the Baire property. For this take a meager measure zero Hamel basis H ⊂ ℝ and note that E = {qh: q ∈ ℚ & h ∈ H} has these properties, since ℝ = ⋃_{n<ω} nE. Sets with the properties similar to these of our set E has been also investigated in [4].

Notice also, that if we put S = ℤ + H then we get the following

Corollary 4. For every natural number n > 0 there exists an S ⊂ ℝ such that for every k ∈ ℕ

• if k ≤ n then kS is meager and it has Lebesgue measure 0;

• if k > n then kS is a Bernstein set in ℝ.

It is also not difficult to complicate a bit shifts in the definition of S from Corollary 4 to ensure that S is still a Hamel basis.

Corollary 5. There exists a Bernstein set B ⊂ ℝ such that kB is also Bernstein for every natural k > 0.

Proof. Just put B = S +S where S is from Corollary 4 used with n = 1. Proof of Theorem 2. Fix an n > 0 and let {P_{ξ}: ξ < 𝔠} be an enumeration of all perfect subsets of [0, 1]. By transfinite induction on ξ < 𝔠 construct a sequence ⟨H_{ξ} ∈ [Z_{2n+2}^{2}]^{n+1} : ξ < 𝔠⟩ such that for every ξ < 𝔠

• ∑_{y∈H_{ξ}} y ∈ P_{ξ} and H_{ξ} is linearly independent over LIN_{ℚ} (⋃_{ζ<ξ} H_{ζ}).

To make such a choice, just put A = ⋃_{ζ<ξ} H_{ζ}, pick an x ∈ P_{ξ} \ LIN_{ℚ}(A), and apply Lemma 1 to find H_{ξ}.

The set H̄ = ⋃_{ξ<𝔠} H_{ξ} ⊂ Z_{2n+2}^{2} is linearly independent over ℚ. Since (n + 1)Z_{2n+2}^{2} ⊃ [0, 1] we can find an H ⊂ Z_{2n+2}^{2} containing H̄ which is a Hamel basis. We will show that H is as desired.

For this notice that

(∗) for every m > n the set mH intersects every P ⊂ [0, 1].

This is clear for m = n + 1, since for ξ < 𝔠 such that P = P_{ξ} the number ∑_{y∈H_{ξ}} y ∈ P_{ξ} belongs to P and (n + 1)H_{ξ} ⊂ (n + 1)H.

So, take an m > n + 1 and let k = m − (n + 1). Taking a subset of P, if necessary, we can assume that there exists an ε > 0 such that P ⊂ (ε, 1]. By the part with m = n+1 the set H contains arbitrary small elements. So, there exists a z ∈ (0, ε) ∩ kH. Then P − z ⊂ [0, 1] and, again by case m = n + 1, there exists an x ∈ (P − z) ∩ (n + 1)H. But then x + z ∈ P ∩ mH. So (∗) is proved.

To finish the theorem it is enough to note that for every m > n and perfect set P ⊂ [0, 1] by (∗) we have P ∩ (m + 1)H ≠ ∅, while (m + 1)H ⊂ ℝ \ mH (since H is a Hamel basis).

References

[1] J. Marshall Ash, Stefan Catoiu, and Ricardo Ríos-Collantes-de-Terán, On the n-th quantum derivative, preprint.

[2] K. Ciesielski, Set Theory for the Working Mathematician, London Math. Soc. Stud. Texts 39, Cambridge Univ. Press 1997.

[3] H. Fejzić, C. E. Weil, Repairing the proof of a classical differentiation result, Real Anal. Exchange 19 (1993-94), 639–643.

[4] F. B. Jones, Measure and other properties of a Hamel basis, Bull. Amer. Math. Soc. 48 (1942), 472–481.

[5] B. Thomson, Symmetric Properties of Real Functions, Marcel Dekker,