Measure on the positive integers respecting independence of prime divisors

Question

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...

Answer 1

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.

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

Answer 2

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.

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