TITLE: Probability measure on $\mathbb N$ such that $P(n \mathbb N) =1/n$ for all $n \ge 1$ cannot exist
QUESTION [8 upvotes]: How to prove that $\mathbb N$ cannot be endowed to a probablity space $(\mathbb N, \mathcal F, P)$ such that for all integer $n \ge 1$ we have $$P(n \mathbb N)=\frac{1}{n}$$
I imagine that divergence of the harmonic series and inclusion-exclusion principle are good ingredients to be used... But I don't know how up to now!

REPLY [16 votes]: Let $p_n$ be the $n$th prime. Then the events $p_n\mathbb N$ are pairwise independent:
$$P(p_n\mathbb N \cap p_m\mathbb N)=P(p_np_m\mathbb N)=\frac 1{p_np_m}=P(p_n\mathbb N)P(p_m\mathbb N).$$
The sum of the reciprocals of the primes
$$\sum_n \frac 1{p_n}$$
famously diverges. So, by the second Borel-Cantelli lemma, the event that infinitely many of the events $p_n\mathbb N$ occur has probability one. But this cannot be satisfied, since no natural number divides by infinitely many primes. (Apart from $0$, but setting $P(0)=1$ is not a solution either.)