# Additivity of upper densities with respect to arithmetic progressions of integers

Let $\mathsf{d}^\star$ be the asymptotic upper density, defined on the power set of positive integers $\mathbf{N}^+$, so that $$\mathsf{d}^\star\colon \mathcal{P}(\mathbf{N}^+) \to\mathbf{R}\colon X\mapsto \limsup_{n\to \infty} \frac{|X\cap [1,n]|}{n}.$$ It is easy to verify that if $k\cdot \mathbf{N}^++h:=\{kx+h\colon x \in \mathbf{N}^+\}$ is an arithmetic progression of $\mathbf{N}^+$, and $X$ is a set of positive integers having no elements in common with $k\cdot \mathbf{N}^++h$, then $$\mathsf{d}^\star(X\cup (k\cdot \mathbf{N}^++h))=\mathsf{d}^\star(X)+\frac{1}{k}.$$ The same reasoning can be extended to the upper analytic, upper logarithmic, upper Banach and upper Buck densities, at least. That's why one may ask if this property holds in general:

We say that a set function $\mu^\star\colon \mathcal{P}(\mathbf{N}^+)\to \mathbf{R}$ is an "upper density" whenever it satisfies the following axioms:

(F1) $\mu^\star(\mathbf{N}^+)=1$

(F2) $\mu^\star(X) \le \mu^\star(Y)$ if $X\subseteq Y$

(F3) $\mu^\star(X\cup Y) \le \mu^\star(X)+\mu^\star(Y)$

(F4) $\mu^\star(k\cdot X+h)=\mu^\star(X)/k$

for all positive integers $k,h$ and sets $X,Y \subseteq \mathbf{N}^+$. Then, is it true that if $X \cap (k\cdot \mathbf{N}^++h)=\emptyset$ then $$\mu^\star(X \cup (k\cdot \mathbf{N}^++h))=\mu^\star(X)+\frac{1}{k}\,\,?$$

• Do you have any counterexample to the stronger (and more natural?) statement that $\mu^\ast(X \cup Y) = \mu^\ast(X) + \mu^\ast(Y)$ whenever $X,Y \subseteq \mathbf N^+$, $X \cap Y = \emptyset$, and $\mu^\ast(Y) + \mu^\ast(Y^c) = 1$, i.e. $Y$ belongs to the domain of the density induced by $\mu^\ast$? Aug 18, 2015 at 12:01
• I agree that your question would be natural (and obviously stronger); the reason why I asked in this form is that, on the one hand, I guess this is not simpler from the general case, and, on the other hand, this would be enough to prove that also its associated lower density $\mu_\star \colon \mathcal{P}(\mathbf{N}^+)\to \mathbf{R}\colon X\mapsto 1-\mu^\star(X^c)$ satisfies (F4) too.. Aug 20, 2015 at 8:38
• I didn't ask you to agree with me (on something); I asked if you have a counterexample (to something). Anyway, here is an idea: Let $\mathcal A^\sharp$ be the collection of all subsets of $\mathbf N^+$ that can be expressed as a finite union of sets of arithmetic progressions of ${\bf N}^+$, or which differ from these by finitely many integers. Given $X\subseteq{\bf N}^+$, can you show that there exists a nonincreasing (resp., nondecreasing) sequence $(U_n)_{n \ge 1}$ of $\mathcal A^\sharp$ such that $X\subseteq U_n$ (resp., $U_n\subseteq X$) for each $n$ and $\lim_n\mu^\ast(U_n)=\mu^\ast(X)$? Aug 21, 2015 at 9:43
• I am almost sure that what you are asking holds if and only if $X$ belongs to the domain of the Buck's density, which holds if and only if $X$ belongs to the topology generated by the set of arithmetic progressions $\mathscr{A}$, or differ by these by a finite set of elements. Anyway, for a negative answer to your question, it sufficient to provide a counterexample (which holds for all $\mu^\star)$. Set $X=\cup_{n\ge 1}\{(2n)!,\ldots,(2n+1)!\}$, then it contains no AP, and the smallest AP containing $X$ is $\mathbf{N}^+$ itself. Aug 21, 2015 at 9:56
• I've just realized that the answer to my previous question is negative: Looking at the case of the upper asymptotic density, ${\sf d}^\ast$, on $\mathbf N^+$, fix $\alpha \in {]0,1[}$ and consider as $X$ the set $\bigcup_{n \ge 1} [\![\alpha(2n-1)! + (1-\alpha)(2n)!+1, (2n)!]\!]$; then recall from mathoverflow.net/questions/207522 that ${\sf d}^\ast(X) = \alpha$. Edit: I see we added a comment almost at the same time. Aug 21, 2015 at 9:56

Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function $$h := (\alpha f^q + (1-\alpha) g^q)^{\frac{1}{q}}$$ is an upper density too (in particular, condition (F3) follows from Minkowski's inequality, which is why we need $q \ge 1$).
Next, fix a set $X \subseteq 2\cdot\mathbf N^+$, let $x := 2f(X)$, $y := 2g(X)$ and $Y := 2 \cdot \mathbf N^+ + 1$, and suppose to a contradiction that $h$ is weakly additive'' (that is, $h(A \cup B) = h(A) + h(B)$ for all disjoint $A, B \subseteq \mathbf N^+$ such that $B$ is an (infinite) arithmetic progression), regardless of the actual values of the parameters $\alpha$ and $q$. Then, also $f$ and $g$ are weakly additive, and using that $f(Y) = g(Y)=\frac{1}{2}$, we obtain $$\begin{split} 2h(X \cup Y) & = 2(\alpha (f(X \cup Y))^q + (1-\alpha) (g(X \cup Y))^q)^{\frac{1}{q}} \\ & = 2(\alpha (f(X) + f(Y))^q + (1-\alpha) (g(X) + g(Y))^q)^{\frac{1}{q}} \\ & = (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} \end{split}$$ and $$\begin{split} h(X) + h(Y) & = (\alpha (f(X))^q + (1-\alpha) (g(X))^q)^{\frac{1}{q}} + \frac{1}{2} \\ & = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}}+ \frac{1}{2} \end{split}$$ which, together with $h(X \cup Y) = h(X) + h(Y)$, yields $$(\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}} + 1.$$ On the other hand, an appropriate choice of $f$, $g$ and $X$ makes it possible to have $x$ equal to zero while $y$ takes any prescribed value in the interval $[0,1]$: This can be achieved, for instance, by letting $f$ be the upper asymptotic density (on $\mathbf N^+$), $g$ the upper Banach density, and $X$ a suitable subset of the intersection, $S$, of $\bigcup_{n \ge 1} [\![2^n, 2^n + n]\!]$ and $2 \cdot\mathbf N^+$, and by considering that (i) the upper asymptotic density of $S$ is $0$, (ii) the upper Banach density of $S$ is $\frac{1}{2}$, (iii) the upper asymptotic and upper Banach densities are upper densities, and (iv) upper densities have the strong, and hence the weak, Darboux property (by the main theorem here).
Accordingly, we should have $$(\alpha + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (1-\alpha)^{\frac{1}{q}}y + 1$$ for all $\alpha, y \in [0,1]$ and $q \in [1,\infty[$, which, however, is blatantly false. []
Added later. If you assume $\alpha = \frac{1}{2}$ and $q = 2$ in the last displayed equation, you don't even need to know that the upper Banach density has the weak Darboux property, since then you end up with the equation $$\sqrt{1 + (y+1)^2} = y + \sqrt{2},$$ which has a unique solution for $y \in \bf R$ (namely, $y = 0$).