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The "classical Beurling density" of a subset of the natural numbers is $d(A)=lim_{n\rightarrow\infty}\frac{|A\cap[1,n]|}{n}$, when it exists. It defines a finitely additive probability measure on the natural numbers which is invariant with respect to the sum. Here is my question: does there exist a "nice formula" to describe a finitely additive probability measure on $\mathbb N$ which is invariant with respect to the multiplication?

A couple of remarks: I don't know if it is trivial that such a measure exists, but anyway it follows from the application of a general result of Vern Paulsen (on arxiv "Syndetic sets and amenability"). Another problem would be that of finding the measure of particular sets. What about the measure of {$1!,2!,3!,4!...$}? Sets with measure differente from 0 and 1? For example the set of numbers whose first digit through 4 to 9 seems to have measure $=\log_{10}4$... any other?

Thanks in advance, Valerio

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  • $\begingroup$ when you say invariant with respect to the sum do you mean that $\lbrace k+2 \mid k \in A \rbrace$ has the same measure as $A$? $\endgroup$
    – Aaron Meyerowitz
    Commented Mar 21, 2011 at 21:52
  • $\begingroup$ Any finitely additive probability measure (also called a content) is defined on a field. But the collection of subsets $A$ of the natural numbers having a density $d(A)$ is not a field. $\endgroup$
    – Did
    Commented Mar 21, 2011 at 22:37
  • $\begingroup$ That is, one can write down two sets that have density, but whose intersection does not have density. $\endgroup$
    – Gerald Edgar
    Commented Mar 21, 2011 at 23:48
  • $\begingroup$ One example would be a measure assigning $\mu(A)=1$ if $A$ contains $0$, and $\mu(A)=0$ otherwise. $\endgroup$
    – Daniel Litt
    Commented Mar 22, 2011 at 0:59
  • $\begingroup$ @Daniel, for many of us, $0$ is not a natural number. $\endgroup$
    – Gerry Myerson
    Commented Mar 22, 2011 at 4:31

3 Answers 3

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The natural thing to do here is to replace the intervals $[1,n]$ (and $n = |[1,n]|$) in the definition of $d(A)$ with a sequence $F_n$ of subsets of $\mathbb{N}$ which is multiplicatively asymptotically invariant (or, in other words, a Folner sequence for the semigroup $(\mathbb{N},\cdot)$). For an exploration of this idea, as well as applications, see for instance this article by Vitaly Bergelson:

Multiplicatively large sets and ergodic Ramsey theory, Israel Journal of Mathematics 148 (2005), 23-40.

EDIT: One particular example (mentioned in the article) is to take $F_n$ to be the set of all positive integers which can be written as a product of powers of the first $n$ primes, where the powers are allowed to be any non-negative integer which is less than or equal to $n$. The motivation for choosing $F_n$ this way is that just as $1$ generates the additive semigroup $(\mathbb{N},+)$, the primes generate the multiplicative semigroup. Think of balls in the corresponding Cayley graphs with radius getting larger and larger. The article contains many other examples of such $F_n$, and each of them gives a notion of "multiplicative density" by setting $d(A) = \lim_{n \to \infty} \frac{|A \cap F_n|}{|F_n|}$ (if the limit exists. Otherwise one usually considers the limsup and liminf).

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  • $\begingroup$ Perhaps I am misunderstanding you, but I don't think this works; this is what Gerald Edgar attempts in his answer. It seems to me that $\mathbb{N}$ still has density $1$, whereas $2\mathbb{N}$ has density $1/2$. $\endgroup$
    – Daniel Litt
    Commented Mar 22, 2011 at 19:48
  • $\begingroup$ No, $A_k = [1,k!]$ is not a Folner sequence for $(\mathbb{N},\cdot)$, since $2[1,k!]=[2,2 \cdot k!]$ and $[1,k!] \triangle [2,2 \cdot k!]$ has cardinality $1+ k!$, which is not $o(|A_k| = k!)$. $\endgroup$
    – Mark
    Commented Mar 22, 2011 at 21:01
  • $\begingroup$ Ah, I see. (Btw, $2[1, k!]$ is only the evens in $[2, 2k!]$, but your computations still works). What is an example of a Folner sequence for $\mathbb{N}$? $\endgroup$
    – Daniel Litt
    Commented Mar 22, 2011 at 22:09
  • $\begingroup$ Right. The notation confused me with intervals in the real numbers. As for your question, I've edited my answer to include such an example. $\endgroup$
    – Mark
    Commented Mar 22, 2011 at 23:09
  • $\begingroup$ Thanks a lot, Mark. Even if I can't open the article, from Bergelson's home page directly either.. don't understand why! $\endgroup$
    – Valerio Capraro
    Commented Mar 25, 2011 at 8:10
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Here is an example. Let $d(A)$ be as you define it in the question, namely $$d(A)=\lim_{n\to \infty} \frac{|A\cap [1,n]|}{n}.$$ Let $U_n=\{k\in\mathbb{N} \text{ such that } k \text{ is a mutiple of } n!\}$. Then define $$\mu(A)=\lim_{n\to \infty} n!\cdot d(A\cap U_n).$$ Then I claim that $\mu$ is finitely additive and multiplicatively invariant. Finite additivity is obvious. For multiplicative invariance, note that for $s>k$, we have $(s+k)!\cdot d(kA\cap U_{s+k})=s!\cdot d(A\cap U_{s})$, unless I've screwed something up.

EDIT: Note by the way that one can replace the limit in the definition with $d$ with the Cesaro mean, for example, giving a much broader class of sets with defined measure. For example, with this addition, the set of natural numbers with a fixed leading digit in a fixed prime base $p$ has density $1/(p-1)$.

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  • $\begingroup$ What would be an example of a set with density strictly between zero and one? $\endgroup$
    – Gerry Myerson
    Commented Mar 22, 2011 at 6:08
  • $\begingroup$ It seems to me that numbers whose first digit is $1$ in base $3$ works, for example. $\endgroup$
    – Daniel Litt
    Commented Mar 22, 2011 at 6:14
  • $\begingroup$ (That is, leading, not trailing.) $\endgroup$
    – Daniel Litt
    Commented Mar 22, 2011 at 6:15
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    $\begingroup$ Another, more interesting example, seems to be--natural numbers $n$ such that the exponent of the greatest power of $2$ dividing $n$ equals twice that of the greatest power of $3$ dividing $n$, say $\pm 2$. $\endgroup$
    – Daniel Litt
    Commented Mar 22, 2011 at 8:03
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How about: $$ \lim_{k\to\infty} \frac{|A \cap [1,k!]|}{k!} $$ when it exists?

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  • $\begingroup$ Then all the positive integers have density 1, but the even integers still have density 1/2. Right? $\endgroup$
    – JBL
    Commented Mar 22, 2011 at 0:13
  • $\begingroup$ (In fact, your proposed definition must agree with the usual one whenever the usual one exists.) $\endgroup$
    – JBL
    Commented Mar 22, 2011 at 0:14
  • $\begingroup$ I would argue that the two measures disagree. Take the set of integers $A:=\cup_{n\in\mathbb{N}}[n!,\mathrm{E}(\sqrt{n}n!)]$. If I'm not mistaken, you should have that the sequence $\frac{A\cap[1,k]}{k}$ has both 0 and 1 as adherent values. $\endgroup$
    – Olivier Bégassat
    Commented Mar 22, 2011 at 0:36
  • $\begingroup$ while $\frac{|A\cap [1,k!]|}{k!}$ tends to $0$. $\endgroup$
    – Olivier Bégassat
    Commented Mar 22, 2011 at 0:38
  • $\begingroup$ Which is consistent with what I wrote: Gerald Edgar's density is the limit of a subsequence of the sequence whose limit is the usual density. $\endgroup$
    – JBL
    Commented Mar 22, 2011 at 0:48

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