2 Preliminaries and main result

We say that a real-valued function $\mu ^{\star }$ defined on the power set $\mathcal P(\mathbb N)$ of $\mathbb N$ is an arithmetic upper density (on $\mathbb N$ ) if, for all $X, Y \subseteq \mathbb N$ , the following conditions are satisfied:

1. (F1) $\mu ^{\star }(X) \le \mu ^{\star }(\mathbb N) = 1$ ;

2. (F2) $\mu ^{\star }$ is monotone, that is, if $X \subseteq Y$ , then $\mu ^{\star }(X) \le \mu ^{\star }(Y)$ ;

3. (F3) $\mu ^{\star }$ is subadditive, that is, $\mu ^{\star }(X \cup Y) \le \mu ^{\star }(X) + \mu ^{\star }(Y)$ ;

4. (F4) $\mu ^{\star }(k \cdot \, X + h) = \mu ^{\star }(X)/k$ for every $k \in \mathbb N^+$ and $h \in \mathbb {N}$ .

Moreover, we call $\mu ^{\star }$ an arithmetic upper quasidensity (on $\mathbb N$ ) if it satisfies (F1), (F3) and (F4).

We let the conjugate of an arithmetic upper quasidensity $\mu ^{\star }$ be the function $ \mu _{\star } \colon \mathcal {P}(\mathbb N) \to \mathbb R \colon X\mapsto 1-\mu ^{\star }(\mathbb N\setminus X)$ , and we refer to the restriction $\mu $ of $\mu ^{\star }$ to the set

$$ \begin{align*} \mathcal D := \{X \subseteq \mathbb N \colon \mu^{\star}(X) = \mu_{\star}(X)\} \end{align*} $$

as the arithmetic quasidensity induced by $\mu ^{\star }$ , or simply as an arithmetic quasidensity (on $\mathbb N$ ) if explicit reference to $\mu ^{\star }$ is unnecessary. Accordingly, we call $\mathcal D$ the domain of $\mu $ and denote it by $\text {dom}(\mu )$ .

Arithmetic upper (quasi)densities and arithmetic (quasi)densities were introduced in [Reference Leonetti and Tringali9] and further studied in [Reference Leonetti and Tringali8, Reference Leonetti and Tringali10, Reference Leonetti and Tringali11], though we are adding here the adjective ‘arithmetic’ to emphasise that they assign precise values to APs (see Proposition 3.1(iv) below).

Notable examples of arithmetic upper densities include the upper asymptotic, upper Banach, upper analytic, upper logarithmic, upper Pólya and upper Buck densities (see [Reference Leonetti and Tringali9, Section 6 and Examples 4, 5, 6 and 8]). In particular, we recall that the upper Buck density (on $\mathbb N$ ) is the function

$$ \begin{align*} \mathfrak{b}^{\star} \colon \mathcal{P}(\mathbb N) \to \mathbb R \colon \quad X \mapsto \inf_{A \in \mathscr{A},\, X\subseteq A}\textsf{d}^{\star}(A), \end{align*} $$

where $\textsf {d}^{\star }$ is the upper asymptotic density (on $\mathbb N$ ), that is, the function

$$ \begin{align*} \mathcal P(\mathbb N) \to \mathbb R \colon \quad X \mapsto \limsup_{n \to \infty} \frac{| X \cap [1, n] |}{n}. \end{align*} $$

We will write $\mathfrak b_{\star }$ and $\mathfrak b$ , respectively, for the conjugate of and the density induced by $\mathfrak b^{\star }$ .

We are ready to state the main theorem of the paper, whose proof we postpone to Section 3.

Note that the analogous statement to Theorem 2.3 may not hold for a set B with a nonzero upper Buck density and a real number $\alpha \in [\mathfrak {b}^\star (B),1]$ . For example, if $B \in \mathscr {A}$ , then the upper Buck density of $A+B$ takes only a finite number of values as A varies over the subsets of $\mathbb N$ .

The sets $B\subseteq \mathbb {N}$ such that $\mathfrak {b}(B)=0$ have been studied in [Reference Leonetti and Tringali10], where they are called small sets. Since $\mathfrak {b}$ is monotone and subadditive, it is clear that the family of small sets is closed under finite unions and subsets. Examples of small sets include the finite sets, the factorials, the perfect powers and the primes. One may be tempted to conjecture that a set $B\subseteq \mathbb {N}$ is small if (and only if) it is, in some sense, ‘sufficiently sparse’. However, the property of being small depends on the distribution of B through the APs of $\mathbb {N}$ (see Proposition 3.1(v)), so much so that the set $\{n!+n \colon n \in \mathbb {N}\}$ is not small (its upper Buck density is $1$ ), in spite of being sparse, by any standards.

To date, it is not known whether nonmonotone arithmetic quasidensities exist (see Remark 2.1). However, arithmetic quasidensities satisfy a weak form of monotonicity (implicit in the proof of Proposition 3.1) that will be enough for our goals.

3 Proofs

To start with, we collect some basic properties of (upper and lower) quasidensities that will be used, possibly without further comment, in the proof of Theorem 2.3.

Proof. See [Reference Leonetti and Tringali11, Proposition 2.1] for items (i)–(iv) and [Reference Leonetti and Tringali10, Proposition 2.6] for item (v). As for item (vi), let $X \in \mathscr {A}$ and $Y \subseteq \mathbb N$ . There then exist $k \in \mathbb {N}^+$ and $H\subseteq [\![ 0,k-1]\!]$ such that $X=k\cdot \, \mathbb {N}+H$ , and hence

$$ \begin{align*} X+Y=\bigcup_{y \in Y}(X+y)=\bigcup_{h \in H+Y}(k\cdot\, \mathbb{N}+h)=k\cdot\, \mathbb{N}+H', \end{align*} $$

where $H'$ is the finite set $\bigcup _{i}\{\min ((H+Y)\cap (k\cdot \, \mathbb {N}+i)\}$ and the union is extended over all $i \in [\![ 0,k-1]\!]$ such that $(H+Y)\cap (k\cdot \, \mathbb {N}+i) \ne \varnothing $ (with the understanding that an empty union is the empty set).

We are ready for the proof of our main result. Note that the special case of a nonempty finite $B \subseteq \mathbb N$ was settled in [Reference Leonetti and Tringali11, Theorem 1.2] by a different argument.

Proof of Theorem 2.3.

If $\alpha = 1$ , then the conclusion is obvious (by taking $A = \mathbb N$ ). So, we assume from now on that $0 \le \alpha < 1$ . We divide the remainder of the proof into a series of three claims.

Claim 1. There exist a sequence $(H_n)_{n \ge 1}$ of (nonempty) subsets of $\mathbb N$ with $H_n \subseteq [\![ 0, n! - 1 ]\!]$ and a sequence $(h_n)_{n \ge 1}$ with $h_n \in H_n$ such that, for all $n \ge 1$ , the following hold:

1. (i) $\mathfrak b^{\star }(n! \cdot \, \mathbb N + H_n' + B) \le \alpha < \mathfrak b^{\star }(n! \cdot \, \mathbb N + H_n + B)$ , where $H_n' := H_n \setminus \{h_n\}$ ;

2. (ii) $n! \cdot \, \mathbb N + H_n' \subseteq H_{n+1} + (n+1)! \cdot \, \mathbb N \subseteq n! \cdot \, \mathbb N + H_n$ .

To prove Claim 1(i), we proceed by induction. The base case is clear by taking $H_1:=\{0\}$ and $h_1:=0$ .

As for the inductive step, fix $m \ge 1$ and suppose we have already found a set $H_m \subseteq [\![ 0, m! - 1 ]\!]$ and an integer $h_m \in H_m$ such that the claimed inequality holds for $n = m$ . Since

$$ \begin{align*} m! \cdot\, \mathbb N + H_m = (m! \cdot\, \mathbb N + H_m') \cup ((m+1)! \cdot\, \mathbb N + h_m + m! \cdot\, [\![ 0, m ]\!]) \end{align*} $$

and $\mathfrak b^{\star }(m! \cdot \, \mathbb N + H_m' + B) \le \alpha < \mathfrak b^{\star }(m! \cdot \, \mathbb N + H_m + B)$ , there is a minimal $k_{m+1} \in [\![ 0, m ]\!]$ such that

(3.1) $$ \begin{align} \mathfrak b^{\star}((m! \cdot\, \mathbb N + H_m') \cup ((m+1)! \cdot\, \mathbb N + h_m + m! \cdot\, [\![ 0, k_{m+1} ]\!]) + B)> \alpha. \end{align} $$

Consequently, we define

$$ \begin{align*} H_{m+1}:=(H_m'+ m! \cdot\, [\![ 0,m]\!]) \cup (h_m + m! \cdot\, [\![ 0, k_{m+1} ]\!]) \quad\text{and}\quad h_{m+1} := h_m + k_{m+1} \cdot\, m!\!. \end{align*} $$

It is thus clear from (3.1) and the minimality of $k_{m+1}$ that the claimed inequality is also true for $n = m+1$ . By induction, this is enough to complete the proof.

(ii) This is now a straightforward consequence of the recursive construction of the sequences $(H_n)_{n \ge 1}$ and $(h_n)_{n \ge 1}$ , as given in the inductive step of the proof of item (i). This proves our claim.

Claim 2. Set $A := \bigcap _{n \ge 1} A_n$ , where $A_n := n! \cdot \, \mathbb N + H_n$ . Then, $A \in \mathrm {dom}(\mathfrak b)$ .

To prove our claim, pick $n \in \mathbb N^+$ . It follows from Claim 1(ii) that $A_n\setminus (n!\cdot \, \mathbb {N}+h_n) \subseteq A \subseteq A_n$ . Considering that $A_n$ and $A_n\setminus (n!\cdot \, \mathbb {N}+h_n)$ are both in $\mathscr {A}$ , and $\mathscr A$ is contained in $\mathrm {dom}(\mathfrak b)$ , we obtain

(3.2) $$ \begin{align} \mathfrak{b}(A_n)-\frac{1}{n!}\le \mathfrak{b}_{\star}(A)\le \mathfrak{b}^{\star}(A) \le \mathfrak{b}(A_n). \end{align} $$

However, Claim 1(ii) gives $A_{n+1} \subseteq A_n$ . Since $\mathfrak b$ is monotone, it follows that $\mathfrak b(A_n)$ tends to a limit as $n \to \infty $ . So, (3.2) implies that $\mathfrak {b}_{\star }(A) = \mathfrak {b}^{\star }(A) = \lim _n \mathfrak b(A_n)$ and hence, $A \in \mathrm {dom}(\mathfrak b)$ . This proves our claim.

Claim 3. $A+B \in \mathrm {dom}(\mathfrak b)$ and $\mathfrak b(A+B) = \alpha $ .

To prove our claim, fix $n \in \mathbb N^+$ . We gather from Claim 1(ii) that

$$ \begin{align*} A_n\setminus (n!\cdot\, \mathbb{N}+h_n) +B\subseteq A+B \subseteq A_n+B. \end{align*} $$

Considering that, by Proposition 3.1(vi), $X \in \mathscr {A}$ yields $X+B \in \mathscr {A}$ and $\mathscr {A}\subseteq \mathrm {dom}(\mathfrak {b})$ , it follows that

(3.3) $$ \begin{align} \mathfrak{b}(A_n\setminus (n!\cdot\, \mathbb{N}+h_n) +B) \le \mathfrak{b}_{\star}(A+B) \le \mathfrak{b}^{\star}(A+B) \le \mathfrak{b}(A_n +B). \end{align} $$

Now, fix $\varepsilon> 0$ . Since $\mathfrak {b}(B)=0$ , there exists $n_\varepsilon \in \mathbb N^+$ such that B covers at most $\varepsilon \cdot \, n!$ residue classes modulo $n!$ for all $n\ge n_\varepsilon $ . Consequently, we obtain from the subadditivity of $\mathfrak b^{\star }$ and Claim 1(i) that

$$ \begin{align*} \mathfrak{b}(A_{n_\varepsilon} + B)\le \mathfrak{b}(A_{n_\varepsilon} \setminus (n_\varepsilon! \cdot\, \mathbb{N} + h_{n_\varepsilon}) +B)+\mathfrak{b}(n_\varepsilon! \cdot\, \mathbb{N} + h_{n_\varepsilon} + B) \le \alpha + \varepsilon. \end{align*} $$

However, the first inequality in the last display and Claim 1(i) imply

(3.4) $$ \begin{align} \mathfrak{b}(A_{n_\varepsilon} \setminus ({n_\varepsilon}! \cdot\, \mathbb{N} + h_{n_\varepsilon}) + B) \ge \mathfrak{b}(A_{n_\varepsilon} + B) - \mathfrak{b}({n_\varepsilon}! \cdot\, \mathbb{N} + h_{n_\varepsilon} + B)> \alpha - \varepsilon. \end{align} $$

Therefore, (2) and (3) imply that $\alpha - \varepsilon < \mathfrak b_{\star }(A + B) \le \mathfrak b^{\star }(A + B) \le \alpha + \varepsilon $ for every $\varepsilon> 0$ , which suffices to conclude that $A+B \in \mathrm {dom}(\mathfrak b)$ and $\mathfrak b(A + B) = \alpha $ . This proves our claim.

By Proposition 3.1(iii) and Claims 2 and 3, this is enough to finish the proof of the theorem.

As a final remark, we point out that, mutatis mutandis, all the results of this paper carry over to arithmetic (upper) quasidensities on $\mathbb Z$ , in the same spirit of [Reference Leonetti and Tringali8–Reference Leonetti and Tringali11].