If we define the morphism from the positive integers to the interval $(0,1]$ as follows:
$f(x)=\left(\frac{1}{x+1},\frac{1}{x}\right]$
Then we can measure the density of any set of positive integers as a proportion of the whole, e.g. the odd numbers as follows:
$\displaystyle\sum_{x = 1}^{\infty} \left(\frac{1}{2x-1}-\frac{1}{2x}\right)=\ln{2}$
Obviously the even numbers have density $1-\ln{2}$ and every set of multiples of some integer has measure less than $1$, with the set only being fully closed if we include the point at infinity in our sum.
I presume this is a known measure?
What is the density of the prime numbers by this measure?
$\displaystyle\sum_{p\in\text{prime}} \left(\frac{1}{p}-\frac{1}{p+1}\right)\approx0.3302$
It looks to simplify to $\displaystyle\sum_{p\in\text{prime}}\frac{1}{p(p+1)}$
It looks like this has something to do with the relationship of the Zeta function to the distribution of primes. I imagine this sum over primes can be translated into an Euler product, a Dirichlet series and ultimately into a value of the Zeta function but I'm struggling to see how to proceed further.
Use the geometric series $$\frac{1}{p+1} = \frac{1}{p} \frac{1}{1+\frac1p} = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{p^{k}}$$ thus $$\sum_p \frac{1}{p}-\frac{1}{p+1} = \sum_{k=2}^\infty (-1)^{k} P(k) \\= \sum_{k=2}^\infty (-1)^{k} \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(nk)=\sum_{m=1}^\infty (\frac{\mu(m)}{m}-a_m) \log \zeta(m)$$ where $a_m = \sum_{d | m} (-1)^{m/d+1} \frac{\mu(d)}{d}=\prod_{p^k \| m} a_{p^k}$ and $P(s) = \sum_p p^{-s} = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$