On the convergence of Σ_{n=1}^{∞}f(nx) for measurable functions

by

Zoltán Buczolich^{1} Department of Analysis Eötvös Loránd University Budapest, Hungary email: buczo@ludens.elte.hu

and

R. Daniel Mauldin^{2} Department of Mathematics University of North Texas Denton, Texas 76203

May 26, 1999

ABSTRACT. We answer questions of Haight and of Weizsäcker by proving the following theorem: Theorem 1. There exists a measurable function f : (0, +∞) → {0, 1} and two nonempty intervals I_{F} , I_{∞} ⊂ [\frac{1}{2}, 1) such that for every x ∈ I_{∞} we have ∑_{n=1}^{∞} f(nx) = +∞ and for almost every x ∈ I_{F} we have ∑_{n=1}^{∞} f(nx) < +∞. The function f is the characteristic function of an open set E.

1 Research supported by the Hungarian National Foundation for Scientific Research Grant No. T 019476 and FKFP 0189/1997

2 Research supported by NSF Grant DMS-9502952. AMS(MOS) subject classifications(1980). Primary 28A35; Secondary 28A80 Key words and phrases. Borel-Cantelli lemma, simultaneous approximation.

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Buczolich and Mauldin Page 1

INTRODUCTION

Recently one of us was reminded of a problem of Weizsäcker [W]: Given a Lebesgue measurable function f : R^{+} → R^{+}, is it true that either for Lebesgue measure almost every x > 0, the series Σ_{n=1}^{∞}f(nx) converges or, else for Lebesgue measure almost every x > 0, the series Σ_{n=1}^{∞}f(nx) diverges? Weizsäcker investigated this problem in his Diplomarbeit and in particular showed that if the function f is in L_{1}, then the series converges almost surely. At about the same time, Haight [H1] showed that there is a Lebesgue measurable subset E of the positive real line with infinite measure such that if t and s are two distinct numbers in E then t/s ∉ N and for each positive x, there are only finitely many positive integers n such that nx ∈ E. Thus, letting f be the characteristic function of E, we have a measurable function which is not integrable and yet the series Σ_{n=1}^{∞}f(nx) converges for all x > 0. Haight generalized his construction in [H2] and reiterated his question: If E is a Lebesgue measurable subset of the positive real line with infinite measure and N(x, E) = card{n ∈ N | nx ∈ E}, is it true that either N(x, E) = ∞ for almost all x or else N(x, E) < ∞ for almost all x? In this note we shall construct an open set E which shows that the answer to both questions is no.

DEFINITIONS AND NOTATION

For x ∈ [1, ∞) we set Φ(x) = [\frac{1}{2}, 1) ∩ \{\frac{x}{n} : n ∈ N\}.

The intervals used in the Theorem will be defined as I_{F} = (\frac{8}{9}, 1) and I_{∞} = [\frac{16}{25}, \frac{16}{24}].

For β ≠ 0 we set ||x||_{β} = min{|x − nβ| : n ∈ Z}. If β = 1, we simply write ||x|| instead of ||x||_{1}. Observe that ||−x||_{β} = ||x||_{β}, ||x+y||_{β} ≤ ||x||_{β} +||y||_{β} and ||x|| = q||\frac{x}{q}||_{1/q} when q > 0.

The Lebesgue measure of the set A is denoted by |A|. We denote by χ_{A}(x) the characteristic function of A, that is, χ_{A}(x) = 1 for x ∈ A, and χ_{A}(x) = 0 otherwise.

In this paper we denote by log x the logarithm in base 2.

PRELIMINARY RESULT

We will use Kronecker’s Theorem on simultaneous inhomogenous approximation [C, p. 53]. Here we state a special case of it which will be used later.

Kronecker’s Theorem. Assume θ_{1}, ..., θ_{L} ∈ R and (α_{1}, ..., α_{L}) is a real vector. The following two statements are equivalent:

A) For every ϵ > 0, there exists p ∈ Z such that

||θ_{j}p − α_{j}|| < ϵ, for 1 ≤ j ≤ L.

B) If (u_{1}, ..., u_{L}) is a vector consisting of integers and

u_{1}θ_{1} + ... + u_{L}θ_{L} ∈ Z,

then u_{1}α_{1} + ... + u_{L}α_{L} ∈ Z.

MAIN RESULT

Theorem 1 easily follows from the following Lemma.

Lemma. There exists K_{0} ∈ N such that for every k ≥ K_{0}, there exists N_{k} with the property that for each integer ν ≥ N_{k}, there is an open set H_{k} ⊂ (2^{ν−1}, 2^{ν}) for which I_{∞} ⊂ Φ(H_{k}) and |I_{F} ∩ Φ(H_{k})| < 5 · 2^{−k}.

Proof of Theorem 1 based on the Lemma. Using the Lemma, choose a sequence of integers ν_{K_{0}} < ν_{K_{0}+1} < ... such that for each ν_{k}, (k ≥ K_{0}), there exists an H_{k} satisfying the conclusions of the Lemma for ν = ν_{k}.

Let f(x) = ∑_{k=K_{0}}^{∞} χ_{H_{k}}(x). It is clear that for every x ∈ I_{∞} and for each k = K_{0}, K_{0} + 1, ... there exists n_{k} such that n_{k}x ∈ H_{k}. Since the sets H_{k} are pairwise disjoint, n_{k} ≠ n_{k′}, if k ≠ k′ and therefore ∑_{n=1}^{∞} f(nx) = ∞ on I_{∞}. On the other hand, by the Borel-Cantelli Lemma for almost every x ∈ I_{F}, there exists K_{x} such that nx ∉ H_{k} for all k ≥ K_{x} and n ∈ N. Hence, ∑_{n=1}^{∞} f(nx) is finite almost everywhere on I_{F} . This completes the proof of Theorem 1.

Proof of the Lemma. Fix k. It is clear from the Prime Number Theorem that there is a positive integer N_{k} ≥ 3 such that if ν ≥ N_{k}, then there are 2^{k} primes p_{1}, ..., p_{2^{k}} with

\frac{23}{16}2^{ν} < p_{1} < ... < p_{2^{k}} < \frac{24}{16}2^{ν}.

For each ν ≥ N_{k}, set L = 2^{k} + 2^{ν−2} + 1 and define α_{j} and θ_{j} as follows:

α_{j} = \{ \begin{matrix}\frac{j}{2^{k}} & j = 1, ..., 2^{k}, \\ 0 & 2^{k} < j ≤ L; \end{matrix}

and

θ_{j} = \{ \begin{matrix}log p_{j} & j = 1, ..., 2^{k}, \\ log n_{j} & 2^{k} < j ≤ L, \end{matrix}

where n_{j} = 7/8 · 2^{ν} + j − 2^{k} − 1, for 2^{k} < j ≤ L. We note that n_{j} runs through all the 2^{ν−2} + 1 integers beginning with 7/8 · 2^{ν} and ending with 9/8 · 2^{ν}. We show that condition B of Kronecker’s theorem holds. Indeed, if a vector (u_{1}, ..., u_{L}) consists of integers and u_{1}θ_{1} + ... + u_{L}θ_{L} = t ∈ Z, then p_{1}^{u_{1}} ⋯ p_{2^{k}}^{u_{2^{k}}} · ∏_{2^{k}<j≤L} n_{j}^{u_{j}} = 2^{t}. Note that if 1 ≤ j ≤ 2^{k} < j′ ≤ L, then p_{j} > \frac{23}{16}2^{ν} > \frac{9}{8}2^{ν} ≥ n_{j′}. It follows from the Fundamental Theorem of Arithmetic that u_{j} = 0 for all j, 1 ≤ j ≤ 2^{k}. Since α_{j} = 0 for 2^{k} < j ≤ L, we have

u_{1}α_{1} + ... + u_{L}α_{L} = 0 ∈ Z.

This shows that Condition B of Kronecker’s Theorem holds and hence Condition A is also true. Thus, for ϵ = \frac{1}{4·2^{k}}, we can choose q ∈ Z such that

||θ_{j}q − α_{j}|| < ϵ holds for all j ≤ L.

The choice of ϵ and α_{j} = \frac{j}{2^{k}} for j ≤ 2^{k} implies that q ≠ 0. If q > 0, set q′ = q and α′_{j} = α_{j}. If q < 0, set q′ = −q and α′_{j} = 1 − α_{j}. Then in both cases ||θ_{j}q′ − α′_{j}|| < ϵ holds for j ∈ I. Observe that in both cases the set {α′_{j} : j ≤ 2^{k}} equals (modulo 1) the set \{\frac{j}{2^{k}} : j = 1, ..., 2^{k}\}, and for 2^{k} < j ≤ L, 0 = α_{j} equals (modulo 1) α′_{j} = 1. Since these are the only properties we use, we can assume without limiting generality, that q > 0 and in the sequel we use q and α_{j} instead of q′ and α′_{j}. Dividing by q we find that

||θ_{j} − \frac{α_{j}}{q} ||_{\frac{1}{q}} < \frac{ϵ}{q} holds for j ≤ L.

This means that if j ≤ 2^{k}, we have

|| log p_{j} − \frac{j}{q · 2^{k}} ||_{\frac{1}{q}} < \frac{1}{q · 4 · 2^{k}} ,

while for 2^{k} < j ≤ L, we have

|| log n_{j}||_{\frac{1}{q}} < \frac{1}{q · 4 · 2^{k}} .

Set

G = \{x ∈ (log \frac{8}{9} + ν, ν) : ||x||_{\frac{1}{q}} < \frac{1}{q · 2^{k}} \}.

Clearly, G is open as is H_{k} ⊂ (2^{ν−1}, 2^{ν}) which is defined by H_{k} = {2^{x} : x ∈ G}.

We next show I_{∞} ⊂ Φ(H_{k}). Let y ∈ I_{∞} and let x = log y. Since {α_{j} : j ≤ 2^{k}} equals (modulo 1) the set \{\frac{j}{2^{k}} : j = 1, ..., 2^{k}\}, we can choose j_{x} ≤ 2^{k} such that

||x + \frac{j_{x}}{q · 2^{k}} ||_{\frac{1}{q}} ≤ \frac{1}{q · 2 · 2^{k}} .

Then

||x + log p_{j_{x}}||_{\frac{1}{q}} ≤ ||x + \frac{j_{x}}{q · 2^{k}} ||_{\frac{1}{q}} + || log p_{j_{x}} − \frac{j_{x}}{q · 2^{k}} ||_{\frac{1}{q}} ≤ \frac{1}{q · 2 · 2^{k}} + \frac{1}{q · 4 · 2^{k}} < \frac{1}{q · 2^{k}} .

Since log \frac{16}{25} ≤ x ≤ log \frac{16}{24} and log(\frac{23}{16}2^{ν}) < log p_{j_{x}} < log(\frac{24}{16}2^{ν}), we obtain

ν + log \frac{8}{9} < ν + log \frac{23}{25} ≤ x + log p_{j_{x}} < ν.

Thus, x + log p_{j_{x}} ∈ G which means p_{j_{x}}y ∈ H_{k}. Hence, I_{∞} ⊂ Φ(H_{k}).

Finally, we show |I_{F} ∩ Φ(H_{k})| < 5 · 2^{−k}, if k is sufficiently large. Towards this end, set G′ = (log \frac{8}{9}, 0) ∩ \{x : ||x||_{\frac{1}{q}} < \frac{2}{q·2^{k}} \}. We have the estimate

|G′| < card\{n : q log \frac{8}{9} − \frac{2}{2^{k}} < n < \frac{2}{2^{k}} \} · \frac{4}{q · 2^{k}} ≤ (\frac{4}{2^{k}} + q log \frac{9}{8}) · \frac{4}{q · 2^{k}} .

Thus, for large values of k (that is, k ≥ K_{0}), we have |G′| < \frac{5}{2^{k}} and letting H′ = \{2^{x} : x ∈ G′\}, we have |H′| < \frac{5}{2^{k}} as well. We claim I_{F} ∩ Φ(H_{k}) ⊂ H′. To see this, suppose y ∈ I_{F} ∩Φ(H_{k}), that is, there exists n such that ny ∈ H_{k}. Set x = log y. Since log \frac{8}{9} < x < 0 and x + log n ∈ G, we have

||x + log n||_{\frac{1}{q}} < \frac{1}{q · 2^{k}} ,

and

ν + log \frac{8}{9} ≤ x + log n < ν.

Hence,

ν − x + log \frac{8}{9} ≤ log n < ν − x,

and using 0 < −x < log \frac{9}{8}, we obtain

ν + log \frac{8}{9} < log n < ν + log \frac{9}{8}.

Thus,

\frac{7}{8}2^{ν} < \frac{8}{9}2^{ν} < n < \frac{9}{8}2^{ν}.

This implies n = n_{j}, for some j, 2^{k} < j ≤ L and therefore,

|| − log n||_{\frac{1}{q}} = || log n||_{\frac{1}{q}} < \frac{1}{q · 4 · 2^{k}};

so

||x||_{\frac{1}{q}} ≤ ||x + log n||_{\frac{1}{q}} + || − log n||_{\frac{1}{q}} < \frac{1}{q · 2^{k}} + \frac{1}{4 · q · 2^{k}} < \frac{2}{q · 2^{k}} .

We infer that x ∈ G′ and y ∈ H′. This completes the proof of the Lemma.

REFERENCES

[C] J.W.S. Cassels, An Introduction to Diophantine Approximation, Cambridge University Press, 1957.

[H1] J.A. Haight, A linear set of infinite measure with no two points having integral ratio, Mathematika 17(1970), 133-138.

[H2] J.A. Haight, A set of infinite measure whose ratio set does not contain a given sequence, Mathematika 22(1975), 195-201.

[W] H. v. Weizsäcker, Zum Convergenzverhalten der Reihe Σ_{n=1}^{∞}f(nt) für λmeasurable Functionen f : R^{+} → R^{+}, Diplomarbeit, Universität München, 1970.