Proof. This will follow from [Reference Akcoglu, Bellow, Jones, Losert, Reinhold-Larsson and WierdlAkc96, Theorem 2.2] and Proposition 3.2 below.

Proof. Let $P>P_0$ be an arbitrary integer. Choose $\eta>0$ small. It will be sufficient to show that for any $\tilde {E}\in \beta $ we have

(3.6)

To prove (3.6), we need the following lemma.

Lemma 3.3. Under the hypothesis of Proposition 3.2, there exists $J_0=J_0(\eta )\in {\mathbb {N}}$ such that for every Lebesgue measurable set $E\subset {\mathbb {R}}$ the following holds:

(3.7)

Proof. Let us assume that the conclusion is not true. Then for every integer $J_k$ , there exists a Lebesgue measurable set $E_k\subset {\mathbb {R}}$ such that

(3.8)

Fix such a $J_k$ and $E_k.$ Observe that the set $E_k':= E_k\cap [0,J_k+s_P]$ also satisfies (3.8). Then for any $J>J_k+s_P$ , we have

(3.9)

Letting $J\to \infty $ , we get

(3.10)

But this contradicts our hypothesis (3.4). This finishes the proof of our lemma.

We now establish (3.6). Let $\tilde {E}$ be an arbitrary element of ${\beta }$ . For a fixed element $x\in \tilde {X}$ , define by

(3.11)

Applying (3.7) on the set $E_x$ , we get

(3.12)

By definition of the sets $E_x$ and $\tilde {E}$ , we get

Substituting the above equality in (3.12), we get

(3.13)

Introducing the set

(3.14)

we can rewrite equation (3.13) as

which implies

Integrating the above equation with respect to the x variable, using (3.11) in the right-hand side and applying Fubini’s theorem, we get

(3.15)

By using the fact that $U^t$ is measure-preserving we get

(3.16) $$ \begin{align} {\mathfrak{m}} (B) J_0 \leq (C+\eta) {\mathfrak{m}}(\tilde{E}) J_0. \end{align} $$

This gives us that

(3.17) $$ \begin{align} {\mathfrak{m}}(B)\leq (C+\eta) {\mathfrak{m}} (\tilde{E}), \end{align} $$

finishing the proof of the desired maximal inequality (3.6). This completes the proof.

Proof. We want to show that the maximal inequality (3.20) transfers to ${\mathbb {R}}$ . If we can show that, then Proposition 3.4 will finish the proof.

Let $\phi $ be a measurable function. After a standard reduction, we can assume that $\gamma =1$ and $\phi $ is positive. It will be sufficient to show that for any fixed P we have

(3.21) $$ \begin{align} \overline{d}\Big(t|\max_{N\leq P} \displaystyle\mathbb{A}_{n\in [N]}\phi(s_n+t)\geq 1\Big)\leq C\overline{d} (\phi). \end{align} $$

Define

(3.22) $$ \begin{align} G:=\Big\{t: \max_{N\leq P}\mathbb{A}_{n\in [N]}\phi({s_{n}}+t)>1\Big\}. \end{align} $$

Let $(L_n)$ be a sequence such that

(3.23) $$ \begin{align} \overline{d}(G)=\lim_{n\to \infty} \frac{1}{L_n-s_P}\unicode{x3bb}([0,L_n-s_P]\cap G). \end{align} $$

Fix $L_n$ large enough so that it is bigger than $s_P$ . Then by applying the ${\mathbb {R}}$ -action version of the Rohlin tower [Reference MirekLin75, Theorem 1], we can find a set $F\subset X$ , such that $T^{[0,L_n]}(F)$ is disjoint, measurable and $\mu (T^{[0,L_n]}F)>1-\eta $ , where $\eta>0$ is small. It also has the additional property that if we define the map $\nu $ by $\nu (A):=\mu (T^A(F))$ for any Lebesgue measurable subset A of $[0,L_n]$ , then $\nu $ becomes a constant multiple of the Lebesgue measure.

Define $f :X\to {\mathbb {R}}$ as follows:

$$ \begin{align*} f(x)= \begin{cases} \phi(t)&\text{when } x\in T^t(F),\\ 0&\text{when } x\not\in T^{[0,L_n]}(F). \end{cases} \end{align*} $$

Now we observe that for each $x\in T^{t}(F)$ with $t\in [0,L_n-s_P]$ and $N\leq P$ we have

$$ \begin{align*} \max_{N\leq P}\frac{1}{N}\displaystyle\sum_{n\leq N}\phi({s_{n}}+t)=\max_{N\leq P}\frac{1}{N}\displaystyle\sum_{n\leq N}f(T^{s_{n}}x) \end{align*} $$

Substituting this into (3.20), we get

(3.24) $$ \begin{align} \mu(x|x\in T^{t}(F) \text{ for some } t\in [0,L_n-s_P]\cap G)\leq C\int_X f\,d\mu. \end{align} $$

Then (3.24) can be rewritten as

(3.25) $$ \begin{align} \mu(T^{[0,L_n-s_P]\cap G} (F))\leq C\int_X f\,d\mu. \end{align} $$

Applying the change-of-variable formula, we can rewrite (3.25) as

$$ \begin{align*} \nu([0,L_n-s_P]\cap G)&\leq C\int_{[0,L_n]} \phi(t)\,d\nu(t). \end{align*} $$

Since $\nu $ is a constant multiple of the Lebesgue measure, we get

$$ \begin{align*} \unicode{x3bb}([0,L_n-s_P]\cap G)&\leq C\int_{[0,L_n]} \phi(t)\, dt. \end{align*} $$

Dividing both sides by $(L_n-s_P)$ and taking limits inferior and superior, we have

$$ \begin{align*} \displaystyle\liminf_{n\to \infty} \frac{1}{L_n-s_P}\unicode{x3bb}([0,L_n-s_P]\cap G)&\leq C,\quad \displaystyle\limsup_{n\to \infty} \frac{L_n}{L_n-s_P}\frac{1}{L_n}\int_{[0,L_n]} \phi(t)\,dt. \end{align*} $$

This implies, by (3.23), that

$$ \begin{align*} \overline{d}(G)&\leq C\overline{d}(\phi). \end{align*} $$

This is the desired maximal inequality on ${\mathbb {R}}.$ We now invoke Proposition 3.4 to finish the proof.

Proof. The above lemma is a consequence of the well-known Kronecker diophantine theorem.

Proof. Suppose, if possible, that there exist $\unicode{x3bb} _i\in {\mathbb {Q}}, \text { for } \ i\in [N]$ , such that $\sum _{i\leq N}\unicode{x3bb} _i s_i=0$ is a non-trivial relation in S. We can express this relation as $P(p_1^{{1}/{m}},p_2^{{1}/{m}},\ldots ,p_r^{{1}/{m}})~=~0$ , where P is a polynomial, and for all $i\in [r]$ , $p_i$ is a prime divisor of ${s_j}^m$ for some $j\in [N].$ Since S is a good set, we can reduce this relation to a relation $P'(p_1^{{1}/{m}},p_2^{{1}/{m}},\ldots ,p_r^{{1}/{m}})=0$ such that each coefficient of $P'$ is a non-zero constant multiple of the corresponding coefficient of P, and the degree of $P'$ with respect to $p_i^{{1}/{m}}$ is $<m$ for all $i\in [r].$ But this implies, by [Reference BesicovitchBes40, Corollary 1], that all coefficients of $P'$ vanish. Hence, all the coefficients of P will also vanish. But this contradicts the non-triviality of the relation that we started with. This completes the proof.

Proof of Proposition 3.8

We will apply Theorem 3.1 to prove this result. So, let $\epsilon \in (0,1), P_0$ and $C>0$ be arbitrary. Choose $\rho>1$ such that $\delta :={\rho }{(1-\epsilon )}<1$ . By using condition (a), we can choose K large enough so that $\overline {d}(R_K)> \delta .$

This implies that we can find a subsequence $(s_{n_j})$ of $(s_n)$ and $N_0\in {\mathbb {N}}$ such that

(3.28) $$ \begin{align} \frac{\#\{s_n\in R_K:n\leq n_j\}}{n_j}>\delta\quad \text{for all } j\geq N_0. \end{align} $$

Define $C_K:=S\setminus R_K=\bigcup _{j>K}S_j$ .

Let P be a large number which will be determined later. For an interval $J\subset [P]$ of indices, we consider the average $\mathbb {A}_{n\in J}f(U^{s_n}x)$ , where $U^t$ is a measure-preserving flow on a space which will be a torus with high enough dimensions for our purposes.

We let $\tilde {S}$ be the truncation of the given sequence S up to the Pth term. Define $\tilde {S}_k$ and $\tilde {C}_K$ as the corresponding sections of $S_k$ and $C_K$ in $\tilde {S}$ respectively, that is,

$$ \begin{align*} \tilde{S} &:= (s_n)_{n\in [P]}, \ \ \ \tilde{S}_k:= \tilde{S}\cap {S_k}\quad \text{for }k\in[K], \\ \tilde{C}_K &:= \tilde{S}\cap C_K. \end{align*} $$

The partition $\tilde {S}_1, \tilde {S}_2,\ldots ,\tilde {S}_K, \tilde {C}_K$ of $\tilde {S}$ naturally induces a partition of the index set $[P]$ into $K+1$ sets $\mathcal {N}_k, k\leq K+1$ , where $\tilde {S}_k$ corresponds to $\mathcal {N}_k$ for $k\in [K]$ and $\tilde {C}_K$ corresponds to $\mathcal {N}_{K+1}.$ The averages $\mathbb {A}_{n\in J}f(U^{s_n}x)$ can be written as

(3.29) $$ \begin{align} \displaystyle\mathbb{A}_{n\in J}f(U^{s_n}x)= \frac{1}{\# J}\sum_{k\leq {K+1}}\sum_{n\in J\cap \mathcal{N}_{k}}f(U^{s_n} {x}). \end{align} $$

By hypothesis (b) and Lemma 3.6, it follows that each $\tilde {S}_k$ has the property that if it is partitioned into Q sets, so that $\tilde {S}_k= \bigcup _{q\leq Q} \tilde {S}_{k,q}$ , then there is an integer r such that

(3.30) $$ \begin{align} rs\in I_q \text{ mod } 1\quad \text{ for } s\in \tilde{S}_{k,q}\text{ and } q\leq Q. \end{align} $$

where

(3.31) $$ \begin{align} I_q:= \bigg(\frac{q-1}{Q},\frac{q}{Q}\bigg). \end{align} $$

The space of action is K-dimensional torus ${\mathbb {T}}^K$ , subdivided into little K-dimensional cubes C of the form

(3.32) $$ \begin{align} C= I_{q(1)}\times I_{q(2)}\times\cdots \times I_{q(k)} \quad\text{for some } q(k)\leq Q \text{ for } k\leq K. \end{align} $$

At this point, it is useful to introduce the following vectorial notation to describe these cubes C. For a vector $\textsf {q}=(q(1),q(2),\ldots ,q(K))$ , with $q(k)\leq Q$ , define

(3.33) $$ \begin{align} I_{\textsf{q}}:= I_{q(1)}\times I_{q(2)}\times \cdots \times I_{q(K)}. \end{align} $$

Since each component $q(k)$ can take the values $1,2,\ldots ,Q$ , we divide ${\mathbb {T}}^K$ into $Q^K$ cubes.

We also consider the ‘bad’ set E defined by

(3.34) $$ \begin{align} E:= \displaystyle\bigcup_{k\leq K} (0,1)\times(0,1)\times\cdots \times \underbrace{(I_1\cup I_2)}_{k\text{th coordinate}}\times\cdots \times (0,1). \end{align} $$

Defining the set $E_k$ by

(3.35) $$ \begin{align} E_k:= (0,1)\times (0,1) \times \cdots \times \underbrace{(I_1\cup I_2)}_{k\text{th coordinate}}\times \cdots \times (0,1), \end{align} $$

we have

(3.36) $$ \begin{align} E=\displaystyle\bigcup_{k\leq K}E_k\quad \text{and}\quad \unicode{x3bb}^{(K)}(E_k)\leq \frac{2}{Q}\quad\text{for every } k\leq K \end{align} $$

where $\unicode{x3bb} ^{(K)}$ is the Haar-Lebesgue measure on ${\mathbb {T}}^K\!.$

By (3.36), we have

(3.37) $$ \begin{align} \unicode{x3bb}^K(E)\leq \frac{2K}{Q}. \end{align} $$

Observe that by taking Q very large, one can make sure that the measure of the bad set is

(3.38) $$ \begin{align} \unicode{x3bb}^{(K)}(E)<\frac{1}{C}. \end{align} $$

Now the idea is to have averages that move each of these little cubes into the supports of the set E. The two-dimensional version of the process is illustrated in Figure 1.

Figure 1

Illustration of the two-dimensional case. Here the ‘bad set’ E is the orange-colored region. Let $(x_1,x_2)$ be an arbitrary point (which belongs to $B_{6,3}$ in this case). We need to look at an average where $r_1s_n\in ({4}/{10},{5}/{10})$ for all $n\in \tilde {S}_1$ and $r_2s_n\in ({7}/{10},{8}/{10})$ for all $n\in \tilde {S}_2.$ Then it would give us $(x_1,x_2)+(r_1s_n,r_2s_n)\in E$ for all $n\in \tilde {S}_1\cup \tilde {S}_2$ . The picture suggests the name ‘grid method’.

Since we have $Q^K$ little cubes, we need to have $Q^K$ averages $\mathbb {A}_J$ . This means we need to have $Q^K$ disjoint intervals $J_i$ of indices. The length of these intervals $J_i$ needs to be ‘significant’. More precisely, we will choose $J_i$ so that it satisfies the following two conditions.

1. (i)

   (3.39) $$ \begin{align} \ J_0=[1,P_0], \frac{\#J_i}{\sum\limits_{0\leq j\leq i}\#J_{j}}>\frac{1}{\rho}\quad \text{for all } i\in [Q^K]. \end{align} $$

2. (ii) If $J_i=[N_{i},N_{i+1}],$ then $(N_{i+1}-N_{i})$ is an element of the subsequence $(n_j)$ and

   (3.40) $$ \begin{align} (N_{i+1}-N_{i})>N_0\quad \text{for all } i\in [Q^K]. \end{align} $$

So we have $Q^K$ little cubes $C_i$ , $i\leq Q^K$ , and $Q^K$ intervals $J_i$ , $i\leq Q^K$ . We match $C_i$ with $J_i$ . The large number P can be taken to be the end-point of the last intervals $J_{Q^K}.$ We know that each $C_i$ is of the form

(3.41) $$ \begin{align} C_i= I_{\textsf{q}_i}, \end{align} $$

for some K-dimensional vector $\textsf {q}_i= (q_i(1),q_i(2),\ldots q_i(K)) $ with $q_i(k)\leq Q$ for every $k\leq K.$ The interval $J_i$ is partitioned as

(3.42) $$ \begin{align} J_i= \displaystyle\bigcup_{k\leq K+1}(J_i\cap \mathcal{N}_k). \end{align} $$

For a given $k\leq K$ , let us define the set of indices $\mathcal {N}_{k,q}$ , for $q\leq Q$ , by

(3.43) $$ \begin{align} \mathcal{N}_{k,q} := \bigcup\limits_{i\leq Q^K, q_i(k)=q} (J_i\cap \mathcal{N}_k). \end{align} $$

Since the sets $\tilde {S}_{k,q} := \{s_n | n\in \mathcal {N}_{k,q}\}$ form a partition of $\tilde {S}_k$ , by the argument of (3.30), we can find $r_k$ so that

(3.44) $$ \begin{align} r_ks_n\in I_{Q-q} \text{ (mod }1)\quad \text{for } n\in \mathcal{N}_{k,q}\quad\text{and}\quad q\leq Q. \end{align} $$

Define the flow $U^t$ on the K-dimensional torus ${\mathbb {T}}^K$ by

(3.45) $$ \begin{align} U^t(x_1,x_2,\ldots, x_K):= (x_1+r_1t,x_2+r_2t,\ldots , x_K+r_Kt). \end{align} $$

We claim that

(3.46)

This will imply that

(3.47)

To see this implication, let $\textbf {x}\in {\mathbb {T}}^K$ and .

Assume that $J_i=[N_i,N_{i+1}]$ . Then

where the second line holds since , and the third holds by (3.39). This proves (3.47).

Now let us prove our claim (3.46).

Let $\textbf {x}\in C_i$ and consider the average . We have

by (3.45). If we can prove that for each $k\in [K]$ ,

(3.48) $$ \begin{align} (x_1+r_1s_n,x_2+r_2s_n,\ldots, x_k+r_ks_n,\ldots, x_K+r_Ks_n)\in E_k\quad\text{if }n\in J_i\cap \mathcal{N}_k, \end{align} $$

then we will have

Using the notation $J_i=[N_i,N_{i+1}]$ ,

using (3.28) and (3.40) for the last inequality. We have proved (3.46).

We are yet to prove (3.48).

Since $\textbf {x}\in C_i= I_{\textsf {q}}$ we have $x_k\in I_{q_i(k)}$ for every k. By the definition of $r_k$ in (3.44) we have $r_ks_n\in I_{Q-q_i(k)}$ if $n\in J_i\cap \mathcal {N}_k$ . It follows that

(3.49) $$ \begin{align} x_k+r_ks_n\in I_{q_{i(k)}}+I_{Q-q_i(k)}\quad \text{if } n\in J_i\cap\mathcal{N}_k. \end{align} $$

Since $I_{q_{i(k)}}+I_{Q-q_{i(k)}}\subset I_1\cup I_2$ , we get

(3.50) $$ \begin{align} x_k+r_ks_n\in I_1\cup I_2\quad \text{if } n\in J_i\cap \mathcal{N}_k. \end{align} $$

By the definition of $E_k$ in (3.35), this implies that

(3.51) $$ \begin{align} (x_1+r_1s_n,x_2+r_2s_n,\ldots x_k+r_ks_n,\ldots, x_K+r_Ks_n)\in E_k \end{align} $$

as claimed.

Thus, what we have so far is the following: for every $C>0,\epsilon \in (0,1)$ and $P_0\in {\mathbb {N}}$ , there exist an integer $P>P_0$ and a set E in the dynamical system $({\mathbb {T}}^K,\Sigma ^{(K)},\unicode{x3bb} ^{(K)},U^t)$ such that

The right-hand side of this inequality is true by (3.38).

Now Theorem 3.1 finishes the proof.

Proof of Theorem 1.1

If we can prove that the sequence $(n^\alpha )$ is strong sweeping out, then part (b) of remark (at the end of §2) will give us the desired result.

To prove that the sequence $(n^\alpha )$ is strong sweeping out, we want to apply Proposition 3.8 with $S=(n^{\alpha })$ .

Observe that the set $S_0:=\{s\in S: s \text { is an integer}\}$ has relative density $0$ in the set S. Hence, we can safely delete $S_0$ from S and rename the modified set as S. Let $\alpha ={a}/{b} \text {with gcd}(a,b)=1 \text { and } b\geq 2$ .

By applying the fundamental theorem of arithmetic, one can easily prove that every positive integer n can be uniquely written as $n={j^b}\tilde {n}$ , where $j,\tilde {n}\in {\mathbb {N}}$ , $\tilde {n}$ is b-free. This observation suggests us that we can partition the set S as follows.

Lemma 3.10. $S{\ =\ \ }\cdot \hspace {-8.4pt}\bigcup _{k\in {\mathbb {N}}}S_{k}$ , where we define

(3.52) $$ \begin{align} S _1:= \{n^{{a}/{b}}: n \text{ is }b\text{-}\mathrm{free}\}\quad\text{and}\quad S_k:=k^aS_1=\{k^a n^{{a}/{b}}:n\text{ is } b\text{-}\mathrm{free}\}. \end{align} $$

Proof. It can be easily seen that $S=\bigcup _{k\in {\mathbb {N}}}S_{k}.$

To check the disjointness let, if possible, there exist $s\in S_{k_1}\cap S_{k_2}, k_1\neq k_2$ . Then there exist b-free integers $n_1,n_2$ such that

(3.53) $$ \begin{align} s=k_{i}^a{n_i^{{a}/{b}}},\quad i=1,2. \end{align} $$

Without loss of any generality, we can assume gcd $(k_1,k_2)=1$ .

Equation (3.53) implies that

(3.54) $$ \begin{align} k_1^{b}n_1=k_2^{b}n_2. \end{align} $$

Observe that gcd $(k_1,k_2)=1\!\implies \!\text {gcd}(k_1^b,k_2^b)=1$ . Hence, (3.54) can hold only if $\mathrel {k_1^b \mid n_2}$ and $\mathrel {k_2^b\mid n_1}$ . Since $n_2$ is b-free, $k_1^b$ can divide $n_2$ only when $k_1=1.$ Similarly, $k_2^b$ can divide $n_1$ only when $k_2=1.$ But this contradicts our hypothesis $k_1 \neq k_2$ .

We will be done if we can show that each $S_{k}$ satisfies the hypothesis of Proposition 3.8. We need to show that the upper density of $R_K := \bigcup _{k\in [K]}S_k$ goes to 1 as $K\to \infty $ .

In fact, we will show that the lower density of $R_K := \bigcup _{k\in [K]}S_k$ goes to 1 as $K\to \infty $ . Let us look at the complement $C_{K} := S\setminus R_{K}.$ It will be enough to show that the upper density of $C_{K}$ goes to $0$ as $K\to \infty .$

An element $n^\alpha $ of S belongs to $C_{K}$ only when $n^{\alpha }\in k^a S$ for some $k>K$ . This means that there is $k>K$ so that $k^b \mid n$ . Hence, we have

(3.55) $$ \begin{align} C_K\subset \bigcup\limits_{k>K}k^a S=\bigcup\limits_{k>K}k^a S_1=\bigcup\limits_{k>K} S_k. \end{align} $$

To see that the first equality of (3.55) is true, let $ x\in \bigcup _{k>K}k^a S $ . Then we must have that $x=k_1^a n^\alpha $ for some $k_1>K.$ Write $n=j^b \tilde {n}$ where $\tilde {n}$ is b-free. Then $x=(jk_1)^a\tilde {n}^\alpha $ . Since $jk_1>K$ , $x\in \bigcup _{k>K}k^a S_1.$ The reverse inclusion is obvious. We want to show that the upper density of $A_K:=\bigcup _{k>K}S_k$ goes to $0$ as $K\to \infty .$ This would imply that the upper density of $C_K$ also goes to $0$ , implying that the lower density of its complement $R_K$ goes to 1. We will see that we even have a rate of convergence of these lower and upper densities in terms of K.

Let us see why the upper density of $A_K$ goes to $0$ as $K\to \infty $ .

We have

$$ \begin{align*} \#S(N)&:=\{n^\alpha\leq N:n\in {\mathbb{N}}\}\\ &=\lfloor{N^{a/b}}\rfloor,\\ \#S_k(N)&:=\{(k^bn)^\alpha\leq N:n\in {\mathbb{N}}\}\\ &=\bigg\lfloor{\frac{N^{a/b}}{k^b}}\bigg\rfloor. \end{align*} $$

Hence, ${\#S_k(N)}/{\#S(N)} \leq {1}/{k^b}$ .

We now have

$$ \begin{align*} \frac{\#A_K(N)}{\#S(N)}&=\frac{\#\left(\bigcup_{k>K} S_k(N)\right)}{\#S(N)}\\ &\leq \frac{\sum_{k>K}\#S_k(N)}{\#S_N}\\ &\leq \sum_{k>K}\frac{1}{k^b} \end{align*} $$

Since $\sum _{k\geq K} {1}/{k^b} \to 0$ as $K\to \infty $ , we must have that the upper density of $A_K$ goes to $0$ as $K\to \infty $ . This proves that $S_k$ satisfies the density condition.

To satisfy condition (b), first observe that every element $s_j$ of $S_1$ can be thought of as a product of the form $p_1^{{c_1 a}/{b}}p_2^{{c_2a}/{b}}\ldots p_k^{{c_ka}/{b}}$ where the $c_i$ depend on $p_i$ and $s_j$ for $i\in [k]$ . We will apply Lemma 3.7 on the set $S_1$ with $m=b.$

To check (i), note that since $S_1$ consists of b-free elements, we have $0\leq c_i<b$ . Assume $c_i>0$ . By assumption we have gcd $(a,b)=1.$ This implies that $ac_i\neq 0$ (mod b) for $i\in [k]$ .

For (ii), let, if possible, $s_1=p_1^{{c_1a}/{b}}p_2^{{c_2a}/{b}}\ldots p_k^{{c_ka}/{b}}$ and $s_2=p_1^{{d_1a}/{b}}p_2^{{d_2a}/{b}}\ldots p_k^{{d_ka}/{b}}$ be two distinct elements of $S_1$ satisfying $ac_i=ad_i$ (mod b) for all $i\in [k].$ This implies $a(c_i-d_i)=0$ (mod b) for all $i\in [k]\!\implies \!(c_i-d_i)=0$ (mod b) for all $i\in [k].$ But this is not possible, since elements of $S_1$ are b-free, and we must have $c_i,d_i<b$ for all $i\in [k].$

Hence, we conclude that $S_1$ is a good set. Since $S_k$ is an integer multiple of $S_1$ , it follows that $S_k$ is also linearly independent over ${\mathbb {Q}}$ . Thus condition (b) is also satisfied. This finishes the proof.

Proof of Theorem 1.4

Define

$$ \begin{align*} S_1:=\{n_1^{\alpha_1}n_2^{\alpha_2}\ldots n_l^{\alpha_l} :n_i\in{\mathbb{N}} \text{ such that } n_i \text{ is }b_i\text{ -free for all } i\in[l]\}. \end{align*} $$

First, let us see that the elements of $S_1$ can be written uniquely. Suppose not. Then there are two different representation of an element:

(3.56) $$ \begin{align} n_1^{\alpha_1}n_2^{\alpha_2}\ldots n_l^{\alpha_l}=m_1^{\alpha_1}m_2^{\alpha_2}\ldots m_l^{\alpha_l}. \end{align} $$

Let us introduce the notation

(3.57) $$ \begin{align} \overline{b_i}=b_1b_2\ldots b_{i-1}b_{i+1}b_l. \end{align} $$

Equation (3.56) implies that

(3.58) $$ \begin{align} n_1^{a_1\overline{b_1}}n_2^{a_2\overline{b_2}}\ldots n_l^{a_l\overline{b_l}}=m_1^{a_1\overline{b_1}}m_2^{b_2\overline{b_2}}\ldots m_l^{a_l\overline{b_l}}. \end{align} $$

After some cancellation if needed, we can assume that $\gcd (n_i,m_i)=1.$ There exists i for which $n_i>1$ , otherwise there is nothing to prove. Without loss of generality, let us assume that $n_1>1$ and p is a prime factor of $n_1$ . Let $p^{c_i}$ be the highest power of p that divides $m_i$ or $n_i$ for $i\in [l].$ Let us focus on the power of p in equation (3.58). We have

$$ \begin{align*} c_1 a_1 \overline{b_1}&=\pm c_2 a_2 \overline{b_2}\pm c_3 a_3 \overline{b_3}\pm \cdots \pm c_l a_l \overline{b_l}. \end{align*} $$

This implies that

$$ \begin{align*} c_1a_1\overline{b_1}&=0\quad (\text{mod}\ b_1). \end{align*} $$

Since $(a_1\overline {b_1},b_1)=1$ , and $c_1<b_1$ , we have $c_1=0.$ This is a contradiction! This proves that $n_i=1=m_i$ for all $i\in [l].$ This finishes the proof that every element of $S_1$ has a unique representation. Now let $(e_k)$ be the sequence obtained by arranging the elements of the set $\{\prod _{i\leq l} n_i^{b_i}, n_i\in {\mathbb {N}} \}$ in an increasing order. Define $S_k:=e_kS_1=\{e_kn_1^{\alpha _1}n_2^{\alpha _2}\ldots n_l^{\alpha _l} :n_i\in ~{\mathbb {N}} \text { and } n_i \text { is }b_i\text { -free} \text { for all } i\in [l]\}.$ Next we will prove that $S = \cdot \hspace {-8.4pt}\bigcup _{k\in {\mathbb {N}}} S_k$ . One can easily check that $S=\bigcup _{k\in {\mathbb {N}}} S_k.$ To check the disjointness, we will use an argument similar to above. Suppose to the contrary that there exists $k_1\neq k_2$ such that $S_{k_1}=S_{k_2}.$ So we have

(3.59) $$ \begin{align} e_{k_1} n_1^{\alpha_1}n_2^{\alpha_2}\ldots n_l^{\alpha_l}=e_{k_2} m_1^{\alpha_1}m_2^{\alpha_2}\ldots m_l^{\alpha_l}. \end{align} $$

This implies, using the same notation as before, that

(3.60) $$ \begin{align} e_{k_1}^{b_1b_2\ldots b_l}n_1^{a_1\overline{b_1}}n_2^{a_2\overline{b_2}}\ldots n_l^{a_l\overline{b_l}}=e_{k_2}^{b_1b_2\ldots b_l}m_1^{a_1\overline{b_1}}m_2^{b_2\overline{b_2}}\ldots m_l^{a_l\overline{b_l}}. \end{align} $$

Assume without loss of generality that $e_{k_1}>1$ , $\gcd (e_{k_1},e_{k_2})=1$ and $\gcd (n_i,n_j)=1$ for $i\neq j.$ Let p be a prime which divides $e_{k_1}$ . Assume $p^{c_i}$ is the highest power of p that divides $m_i$ or $n_i$ , for $i=1,2,\ldots l$ , and $p^r$ is the highest power of p that divides $e_{k_1}$ . We now focus on the power of p in (3.60). We have

$$ \begin{align*} rb_1b_2\ldots b_l&=\pm c_1 a_1 \overline{b_1}\pm c_2 a_2 \overline{b_2}\pm c_3 a_3 \overline{b_3}\pm \cdots \pm c_l a_l \overline{b_l}. \end{align*} $$

This implies that for all $i\in [l]$ we have

$$ \begin{align*} c_ia_i\overline{b_i}&=0\ (\text{mod } b_i). \end{align*} $$

Since $(a_i\overline {b_i},b_i)=1$ , and $c_i<b_i$ , we get $c_i=0\text { for all }i\in [l].$ This contradicts the assumption that $p \mid e_{k_1}$ . Hence, we conclude that $e_{k_1}=1.$ Similarly, $e_{k_2}=1$ . This finishes the proof of disjointness of the $S_k$ .

We now check the density condition. For any sequence A, denote by $A(N)$ the set $\{a\in A:a\leq N\}$ . As before, we will show that the lower density of $B_K := \bigcup _{k\in [K]}S_k$ goes to 1 as $K\to \infty $ . Let us look at the complement $C_K:=S \setminus B_K$ . We want to show that the upper density of $C_K$ goes to $0$ as $K \to \infty $ . Since the sets $C_K$ are decreasing, it will be sufficient to show that the upper density of the subsequence $C_M$ converges to $0$ as $M\to \infty $ , where $M={e_K}^{a_1+a_2+\cdots +a_l}$ . But this is equivalent to saying that upper density of $C_M$ converges to $0$ as $K\to \infty .$ This will follow from the following two lemmas.

Lemma 3.11. $C_M(N)\subset \bigcup _{i\in [l]}\bigcup _{L\geq e_K}D_i^L(N)$ where

$$ \begin{align*} D_i^L(N):=\{n_1^{\alpha_1}n_2^{\alpha_2}\ldots n_{i-1}^{\alpha_{i-1}} (L^{b_i}n_i)^{\alpha_i} n_{i+1}^{\alpha_{i+1}} \ldots n_l^{\alpha_l}\leq N: n_j \in {\mathbb{N}} \text{ for }j\in[l] \}. \end{align*} $$

Lemma 3.12. For every fixed L and i,

(3.61) $$ \begin{align} \frac{\#D_i^L(N)}{\#S(N)}\leq \frac{1}{L^{b_i}}. \end{align} $$

Proof. Fix $n_j=r_j$ for $j=1,2,\ldots ,i-1,i+1,\ldots ,l$ . Then

$$ \begin{align*} \#\{r_1^{\alpha_1}r_2^{\alpha_2}\ldots r_{i-1}^{\alpha_{i-1}} (L^{b_i}n_i)^{\alpha_i}r_{i+1}^{\alpha_{i+1}} \ldots r_l^{\alpha_l}\leq N: n_i \in {\mathbb{N}}\} =\bigg\lfloor{\frac{N^{{1}/{\alpha_i}}}{\kappa L^{b_i}}}\bigg\rfloor, \end{align*} $$

where $\kappa =(r_1^{\alpha _1}r_2^{\alpha _2}\ldots r_{i-1}^{\alpha _{i-1}}r_{i+1}^{\alpha _{i+1}} \ldots r_l^{\alpha _l})^{{1}/{\alpha _i}}$ , and

$$\begin{align*}\#\{r_1^{\alpha_1}r_2^{\alpha_2}\ldots r_{i-1}^{\alpha_{i-1}} (n_i)^{\alpha_i} r_{i+1}^{\alpha_{i+1}}\ldots r_l^{\alpha_l}\leq N: n_i \in {\mathbb{N}}\}=\bigg\lfloor{\frac{N^{{1}/{\alpha_i}}}{\kappa}}\bigg\rfloor. \end{align*}$$

The proof follows by observing the following two equalities:

$$ \begin{align*} \#S{(N)}&=\#\{n_1^{\alpha_1}n_2^{\alpha_2}\ldots n_l^{\alpha_l}\leq N: n_j \in {\mathbb{N}} \text{ for }j\in[l] \}\\ &= \sum\#\{r_1^{\alpha_1}r_2^{\alpha_2}\ldots r_{i-1}^{\alpha_{i-1}} (n_i)^{\alpha_i} r_{i+1}^{\alpha_{i+1}}\ldots r_l^{\alpha_l}\leq N: n_i \in {\mathbb{N}}\}\end{align*} $$

and

$$ \begin{align*} \#D_i^L(N)&=\#\{n_1^{\alpha_1}n_2^{\alpha_2}\ldots n_{i-1}^{\alpha_{i-1}} (L^{b_i}n_i)^{\alpha_i} n_{i+1}^{\alpha_{i+1}} \ldots n_l^{\alpha_l}\leq N: n_j \in {\mathbb{N}} \text{ for }j\in[l] \}\\ &= \sum\#\{r_1^{\alpha_1}r_2^{\alpha_2}\ldots r_{i-1}^{\alpha_{i-1}} (L^{b_i}n_i)^{\alpha_i} r_{i+1}^{\alpha_{i+1}}\ldots r_l^{\alpha_l}\leq N: n_i \in {\mathbb{N}}\}, \end{align*} $$

where the summations are taken over all the $(l-1)$ -tuple $(r_1,r_2,\ldots ,r_{i-1},r_{i+1},\ldots ,r_l)$ such that their products $r_1^{\alpha _1}r_2^{\alpha _2}\ldots r_{i-1}^{\alpha _{i-1}} r_{i+1}^{\alpha _{i+1}}$ are distinct.

Hence, the relative upper density of the set $C_{M}$ in S is less than or equal to $\sum _{i\in [l]}\sum _{L\geq e_K} {1}/{L^{b_i}}$ , which goes to $0$ as $K\to \infty .$

We now show that each $S_k$ is linearly independent. It will be sufficient to show that $S_1$ is linearly independent. After realizing every element $s_j$ of $S_1$ as $\prod _{p\in \mathcal {P}} p^{{v_{j,p}}/{b_1b_2\ldots b_l}}$ , we apply Lemma 3.7 on $S_1$ with $m=b_1b_2\ldots b_l.$

(i) First let us show that for any given $j_0\in J$ , and prime $p_0\in \mathcal {P}$ , if $v_{j_0,p_0}\neq 0$ , then $v_{j_0,p_0}\not \equiv 0$ (mod m). By definition of the set $S_1$ , $v_{j_0,p_0}$ can be written as $v_{j_0,p_0}=c_1a_1\overline {b_2}+c_2a_2\overline {b_2}+\cdots +c_la_l\overline {b_l}$ , where $0\leq c_i<b_i$ , and $c_i$ depends on $s_{j_0,p_0}$ for $i\in [l]$ . Here $\overline {b_i}$ is same as in (3.57).

To the contrary, suppose $v_{j_0,p_0}\equiv 0$ (mod m). Hence, we have

$$ \begin{align*} c_1a_1\overline{b_1}+c_2a_2\overline{b_2}+\cdots+c_la_l\overline{b_l}&\equiv 0 \text{ (mod }m).\end{align*} $$

This means that

$$ \begin{align*} c_1a_1\overline{b_1}+c_2a_2\overline{b_2}+\cdots+c_la_l\overline{b_l} &\equiv 0 \text{ (mod }b_i)\quad \text{for all }i\in [l]. \end{align*} $$

But this implies that

$$ \begin{align*} c_ia_i\overline{b_i}&\equiv 0 \text{ (mod }b_i)\quad \text{for all }i\in [l],\end{align*} $$

implying that

$$ \begin{align*} c_i&=0\quad \text{for all } i \in [l]. \end{align*} $$

So, we get

$$ \begin{align*} v_{j_0,p_0}&=0. \end{align*} $$

This is a contradiction! Hence, we conclude that if $v_{j_0,p_0}\neq 0$ , then $v_{j_0,p_0}\not \equiv 0$ (mod m).

(ii) Now let us assume that $v_{j,p}\equiv v_{k,p}$ (mod $m)$ , for all $p~\in ~\mathcal {P}$ .

Write $v_{j,p}=c_1a_1\overline {b_2}+c_2a_2\overline {b_2}+\cdots +c_la_l\overline {b_l}$ and $v_{k,p}=d_1a_1\overline {b_2}+d_2a_2\overline {b_2}+\cdots +d_la_l\overline {b_l}$ , where $0\leq c_i,d_i<b$ , and $c_i$ depends on $p,s_j$ and $d_i$ depends on $p,s_k$ for all $i\in [l]$ . By assumption, we have

$$ \begin{align*} c_1a_1\overline{b_2}+c_2a_2\overline{b_2}+\cdots+c_la_l\overline{b_l}&\equiv d_1a_1\overline{b_2}+d_2a_2\overline{b_2}+\cdots+d_la_l\overline{b_l} \text{ (mod }m). \end{align*} $$

This implies that

$$ \begin{align*} (c_i-d_i)a_i\overline{b_i}&\equiv 0 \text{ (mod }b_i)\quad \text{for all } i\in [l].\end{align*} $$

Hence, we get

$$ \begin{align*} (c_i-d_i)&=0\quad \text{for all } i\in [l]. \end{align*} $$

So, we conclude that

$$ \begin{align*} v_{j,p}=v_{k,p}\quad \text{for all } p\in \mathcal{P}. \end{align*} $$

But we know that every element of $S_1$ has a unique representation. Hence, we get $j=k$ .

Thus we conclude that S is a good set and hence linearly independent over the rational. This completes the proof.