# Additivity of asymptotic density of periodic sets

It is well known that the asymptotic density of an infinite union of disjoint sets of integers may not be the sum of their individual asymptotic densities.

Can this failure of countable additivity occur even if each of the sets being unioned is periodic (calling a set periodic if there is some nonzero integer such that it is closed under addition and subtraction of this integer)?

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For each $r\in\mathbb{Z}$, let $a_{r}$ be a positive real number so that $\sum_{r\in\mathbb{Z}}a_{r}<1$. Let $A_{r}\subseteq\mathbb{Z}$ be a periodic set with $\mu(A_{r})<a_{r}$ and $r\in A_{r}$. Then $\mathbb{Z}=\bigcup_{r\in\mathbb{Z}}A_{r}$. Now reenumerate the set $\{A_{r}|r\in\mathbb{R}\}=\{B_{n}|n\in\mathbb{N}\}$. Then $\sum_{n}\mu(B_{n})<1$. Let $C_{n}=B_{n}\setminus(B_{0}\cup...\cup B_{n-1})$. Then $\mathbb{Z}=\bigcup_{n}C_{n}$, but $\sum_{n}\mu(C_{n})\leq\sum_{n}\mu(B_{n})<1$, so this measure is not countably additive.
Ah, nice. Here's a followup question I'm also interested in, which you will perhaps resolve with just as much ease: Suppose now we consider an increasing series of subsets $A_0 \subseteq A_1 \subseteq A_2 \subseteq ...$ of the integers, where each of these is not merely periodic but in fact of the form "Any multiple of a member of F", for some finite set F. Can the density of their union still fail to match the supremum of their individual densities? – Sridhar Ramesh Jun 22 '14 at 4:09