Finite measure on positive integers

Question

Disclaimer: I am sure that this idea is not at all new, but I have had trouble locating content directly related. I humbly accept that this question may be the result of a brain fart.

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Suppose that there is a measure $\mu$ on some $\sigma$-algebra of the positive integers such that $\mu(n\mathbb{Z^+})=\frac{1}{n}$ for every $n\in\mathbb{Z^+}$. Heuristically, the inclusion/exclusion principle shows that $$\mu(\{1\})=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{6}-\cdots=\frac{1}{\zeta(1)}=0,$$ and similarly other points have measure $0$. Since $\mu$ is countably additive, $1=\mu(\mathbb{Z^+})=0$.

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On the other hand, we can look at a (possibly non-existent) family of measures $\mu_{\epsilon}:\Omega(\mathbb{Z^+})\rightarrow[0,1]$ where $\mu_{\epsilon}(n\mathbb{Z^+})=\frac{1}{n^{1+\epsilon}}$ for $\epsilon>0$ and $n\in\mathbb{Z^+}$.

I have three questions:

• Is it known whether measures like these exist (rather than being an unrealizable fantasy)?
• If they do exist, are they useful or are there some common barriers to using them.
• If they are useful, would you point me to some literature utilizing them.

Answer 1

I assume you mean $\mathbb N$ rather than $\mathbb Z$: you'd have trouble otherwise with $n\le 0$.

If you define $\mu_\epsilon(\{m\}) = c/m^{1+\epsilon}$ for each positive integer $m$, where $c$ is constant, then $$\mu_\epsilon(n \mathbb N) = \sum_{k=1}^\infty \dfrac{c}{(nk)^{1+\epsilon}} = \dfrac{c}{n^{1+\epsilon}} \zeta(1+\epsilon)$$ so all you need is $c = 1/\zeta(1+\epsilon)$ to have a perfectly good probability measure satisfying your requirements when $\epsilon > 0$.