Measure on the positive integers respecting independence of prime divisors

I want to know if there exists a measure $\mu$ on the positive integers (equipped with the $\sigma-$algebra of all subsets) satisfying:

1) For all $n > 1$, $\mu(A_n) = 1/n$, where $A_n = \{n, 2n, \ldots \}$ is the multiples of $n$, and

2) The events $A_m, A_n$ are independent when $m$ and $n$ are relatively prime.

If such a measure existed, it would mimic the notion of drawing a "random integer" from all of $\mathbb{N}$, at least with respect to these divisibility properties.

1) and 2) hold approximately for the uniform distribution on $\mathbb{N} \cap [1,X]$ for large $X$; a similar weakening of the question is to use asymptotic density, i.e. define $\mu(A) = \lim_{n \to \infty} \frac{|A \cap \{1, 2, 3, \ldots, n\}|}{n}$, which satisfies 1) and 2) but is not countably additive.

The zeta distribution, given by (for any $s > 1$) the density $\nu_s(n) = \frac{1}{\zeta(s)} n^{-s}$, comes very close: it is an easy check that 2) holds, but $\nu_s(A_n) = n^{-s}$ for all $n$. (Interestingly, as $s \to 1^+$, for a fixed subset $A \subset \mathbb{N}, \nu_s(A)$ converges to the asymptotic density of $A$, if it exists.)

I suspect the answer is that no such measure exists: the fact that the zeta distribution fails makes me think 1 and 2 are somehow at odds.

I believe any $\mu$ satisfying 1) and 2) should have $\mu(\mathbb{N}) = \infty$, but I don't have a proof.

Thanks!

Edit: As people have pointed out, 1) clearly implies 2). (Whoops!) And @Zhoraster has given a nice proof that no measure can exist in this case. Here's what (I think) will be a harder question:

Can we find a measure $\mu$ satisfying

1a) $\mu(A_p) = 1/p$ for primes $p$, and

2a) The events $A_p$ and $A_q$ are independent when $p$ and $q$ are prime.

Now it's not obvious that $1$ and $2$ hold for products of many primes: we do get $\mu(A_{pq}) = 1/pq$ for distinct primes $p$ and $q$, but that's it. I don't think the inclusion-exclusion proof will work as stated, but perhaps a similar idea can be used...

• View $\mu$ as a bounded linear map $l^\infty \to \mathbb{R}$ where $A_n$ is the sequence indicating the multiples of $n$. Then $\mu$ is the natural density. It exists by the Hahn-Banach theorem applied to $l^\infty$, but it may be very strange for weird subsets of $\mathbb{N}$. Also, you can restrict to the closed subspace $E = \{ (a_k) \in l^\infty, \lim_{N \to \infty} \frac{1}{N} \sum_{k=1}^N a_k \text{ exists } \}$ of $l^\infty$, in that case $\mu$ is well-defined. – reuns Feb 14 '17 at 10:54
• As I said in the post, the 'natural density' or 'asymptotic density' is not a measure, since it isn't countably additive. – J Richey Feb 14 '17 at 22:00
• As a linear operator, it can be extended to any bounded sequence by Hahn-Banach. Not sure there is much more to say. And try to be more clear next time – reuns Feb 14 '17 at 22:44
• The linear operator you're describing is perfectly valid, but it doesn't answer the question, since it doesn't represent a measure. For instance, the natural density of any finite subset of $\mathbb{N}$ (your operator $\mu$ applied to an element of $l^\infty$ with only finitely many non-zero terms) is zero, but the natural density of the infinite union of all even integers is 1/2. As you say, the 'measure' is finitely additive, (or 'linear' as an operator on $l^\infty$), but it isn't countably additive. What is exactly is unclear? I'm happy to clarify further. – J Richey Feb 15 '17 at 0:54
• @Richey By the way, isn't $(2)$ a consequence of $(1)$ ? – Charles Madeline Feb 18 '17 at 11:09

Such measure doesn't exist. The idea is to show that the only measure satisfying 1 (which, by the way, obviously implies 2) is $$\mu(\{n\}) = \frac{1}{n\zeta(1)} = 0, n\ge 1,$$ which is a contradiction.

So let $\mu$ satisfy 1. Denote by $\mathbb{P}$ the set of primes. Then, for any $n\in \mathbb N$, we have by the inclusion-exclusion formula (all series below are divergent, but one can proceed by finite approximations) $$\mu(\{n\}) = \mu(A_n) - \mu\Big(\bigcup_{p\in \mathbb{P}} A_{np}\Big) \\ = \frac{1}{n} - \sum_{p\in \mathbb P}\mu(A_{np}) + \sum_{\substack{p_1,p_2\in \mathbb P\\p_1\neq p_2 }}\mu(A_{np_1}\cap A_{np_2}) - \sum_{\text{distinct } p_1,p_2,p_3\in \mathbb P}\mu(A_{np_1}\cap A_{np_2}\cap A_{np_3}) + \dots \\ = \frac1n + \sum_{k=1}^\infty (-1)^k \sum_{\text{distinct } p_1,p_2,\dots,p_k\in \mathbb P}\mu\Big(\bigcap_{j=1}^k A_{np_j}\Big)\\ = \frac1n + \sum_{k=1}^\infty (-1)^k \sum_{\text{distinct } p_1,p_2,\dots,p_k\in \mathbb P}\mu\Big( A_{np_1p_2\cdots p_k}\Big) \\ = \frac1n + \sum_{k=1}^\infty (-1)^k \sum_{\text{distinct } p_1,p_2,\dots,p_k\in \mathbb P}\frac1{np_1p_2\cdots p_k} = \frac1n\prod_{p\in \mathbb P}\Big(1-\frac{1}{p}\Big) = 0,$$ as claimed.

Similarly, it can be shown that the unique measure satisfying $\mu_s(A_n) = n^{-s}$ with $s>1$ is $$\mu_s(\{n\})=\frac1n\prod_{p\in \mathbb P}\Big(1-\frac{1}{p^s}\Big) = \frac{1}{n^s \zeta(s)}, n\ge 1.$$

• Great answer which is obviously more appropriate than mine ; minor detail : on your seventh line you wrote twice $\mu (A_n)-\mu \Big( \bigcup \limits_{p \in \mathbb{P}} A_{np} \Big)$ – Charles Madeline Feb 18 '17 at 14:25
• @charMD, thanks, fixed. – zhoraster Feb 18 '17 at 14:42
• Nice solution! I had thought to use this inclusion-exclusion formula, but didn't realize it's (very simple) connection with the zeta function. I added an edit weakening the assumptions a little bit... do you think a similar proof will work in the case where we only assume 1) and 2) hold for distinct primes $p$ and $q$? – J Richey Feb 18 '17 at 21:39
• @JRichey If you only assume 1 and 2 for primes then, I think, such measure may be easily constructed. – zhoraster Feb 19 '17 at 15:16
• @Zhoraster What do you have in mind? I don't have any ideas on how to construct one. – J Richey Feb 19 '17 at 18:33

Such a measure can not be a probability (i.e. we can't have $\mu(\mathbb{N})=1$).

Ad absurdum, suppose we have a probability $P$ satisfying $(1)$, $(2)$.

Let $(p_n)_{n \ge 1}$ be an ordered enumeration of prime numbers. There is no positive integer which has an infinity of prime divisors, and thus $$\bigcap \limits_{n \in \mathbb{N}} \bigcup \limits_{k \ge n} A_{p_k} = \varnothing.$$

Hence $P \Big(\bigcap \limits_{n \in \mathbb{N}} \bigcup \limits_{k \ge n} A_{p_k} \Big) = 0$, so $$P \Big( \bigcup \limits_{n \in \mathbb{N}} \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) > 0.$$

However, $P$ is continuous below so \begin{align*} P \Big( \bigcup \limits_{n \in \mathbb{N}} \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) & = \lim \limits_{n \to \infty} P \Big( \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) \\ & = \lim \limits_{n \to \infty} \prod \limits_{k \ge n} P (\overline{A_{p_k}}) \end{align*}

because $P$ is a probability, the $A_{p_k}$ are mutually independant, and thus so are the $\overline{A_{p_k}}$. Hence \begin{align*} P \Big( \bigcup \limits_{n \in \mathbb{N}} \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) & = \lim \limits_{n \to \infty} \prod \limits_{k \ge n} \Big( 1 - P (A_{p_k}) \Big) \\ & \le \lim \limits_{n \to \infty} \prod \limits_{k \ge n} e^{-P(A_{p_k})} \end{align*}

because $1-x \le e^{-x}$ (convexity inequality). We deduce finally :

$$P \Big( \bigcup \limits_{n \in \mathbb{N}} \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) \le \lim \limits_{n \to \infty} e^{-\sum \limits_{k=1}^n P(A_{p_k})}$$

But $$\sum \limits_{k=1}^n P(A_k) = \sum \limits_{k=1}^n \frac{1}{p_k} \underset{n \to \infty}{\longrightarrow} +\infty$$

(see Divergence sum of reciprocal of primes), so we conclude

$$P \Big( \bigcup \limits_{n \in \mathbb{N}} \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) = 0$$

which is absurd. Hence, there is no probability satisfying your conditions.

See Zhoraster's post which provides a better and clearer answer to your question.

• There are some typos, but essentially it is correct. In fact, you show that for each $n$ the measure of numbers not dividing $p_n,p_{n+1}\dots$ must be zero; as a result, by letting $n\to\infty$, the measure of whole $\mathbb N$ must be zero. The assumption that $\mu$ is a probability measure is not significant, as one can modify the argument for the set of even numbers, not the whole $n$ (my proof is essentially in the very same spirit). – zhoraster Feb 18 '17 at 12:29