Probability measure on subset of natural numbers... How one would define a probability measure on all subsets of natural numbers, which is finite-additive and such that the variables $\chi_p(n)=\left\{\begin{matrix}
1 & p|n \\ 
0 & \text{esle}
\end{matrix}\right.$ are independent. 
Any help you can give will be greatly appreciated.
 A: Zeta distributions provide a family of examples, where
$$P(X=n):=\frac{1}{\zeta(s)n^s}$$
for any parameter $s>1$ (and $\zeta(s)$ is the Riemann zeta function).
Qiaochu's blog post discusses information related to this topic, including haar measure on profinite integers and conditions (monotonicity, entropy etc) which restrict what sort of distributions can exist with the property that $p\mid X$ are all mutually independent events, in particular when the zeta distributions are the only such ones.
A: It is natural to ask for probability of divisibility to be independent for relatively prime integers, which amounts to choosing an integer by independently selecting the power of every prime dividing it.  
To achieve this, 


*

*for every prime $q$ take a constant $c_q \in (0,1]$, such that the infinite product of all constants $\prod c_q$ is nonzero; 

*then take (for every prime $q$) any probability distribution $P_q$ on the non-negative integer powers of $q$ such that $P_q[1]=c_q$.  


Define $\pi(n) = \prod P_{q}[q^{e_q(n)}]$ where $q$ runs over all primes and $e_q(n)$ is the power of $q$ dividing $n$.  This is the product distribution on (infinite) prime factorizations restricted to its probability $1$ subset, the products of finitely many primes.  Condition (1) ensures that the complement of the finite factorizations has measure $0$.  
One can check that $\sum \pi(n) = 1$ without any reliance on probability arguments. The sum is at most $1$, and it can be extended past $1 - \epsilon$ by taking enough terms  to include the product of [sum of the first $k$ terms] for the first $k$ prime distributions $P_q$, for sufficiently large  $k$.
This is a countably additive probability satisfying the conditions of the problem.  Examples can be had by taking any Dirichlet series with non-negative coefficients and a non-negative Euler product at a value of $s$ where both converge.  If $$ F(s) = \sum a_n n^{-s} =  \prod f_q(s) $$ the distribution is $\pi(k) = \frac{a_k}{k^{-s}F(s)}$ and the distribution $P_q$ is proportional to the coefficients of $f_q(s)$, with $c_q = \frac{1}{f_q(s)}$.  
Amusingly, $P_q$ is the determined from the terms of the Dirichlet series $\frac{f_q(s)}{f_q(s)}$ where the numerator is a formal series and the denominator is a real number equal to the sum of the same series.
