# On the existence of a particular type of finite measure on $\mathbb N$

Let $$\mathbb N$$ denote the set of all positive integers. Does there exist a countably additive measure $$\mu : \mathcal P(\mathbb N) \to [0,\infty)$$ such that $$\mu(\mathbb N)<\infty$$ and $$\mu(\{nk: k\in \mathbb N\})=\dfrac 1{n \log n},\forall n>1$$ ?

Other than the fact that the measure's value on $$\{1\}$$ is completely unspecified by the given constraint, there is only one possibility for such a measure (call it $$\lambda$$) if it exists: the Möbius inversion formula for infinite series means that $$\frac1{n\log n} = \sum_{k\in\mathbb N} \lambda(\{nk\}) \implies \lambda(\{n\}) = \sum_{k\in\mathbb N} \mu(k) \frac1{nk \log nk} = \frac1n \sum_{k\in\mathbb N} \frac{\mu(k)}{k \log nk} \quad(n\ge2),$$ where $$\mu$$ is the Möbius function. Using Dirichlet series one can further show that $$\lambda(\{n\}) = \int_1^\infty \frac{n^{-x}}{\zeta(x)} \,dx \quad(n\ge2),$$ where $$\zeta(s)$$ is the Riemann zeta function. Since $$1/\zeta(x)\asymp x-1$$ for $$x\in[1,2]$$ while $$1/\zeta(x) \asymp 1$$ for $$x\in[2,\infty)$$, one can show from this integral representation that $$\lambda(\{n\})$$ has order of magnitude $$1/(n\log^2 n)$$, so that this really does represent a finite (positive) measure on $$\mathbb N$$. And indeed \begin{align*} \lambda\big( \{nk\colon k\in\mathbb N\}\big) &= \sum_{k=1}^\infty \int_1^\infty \frac{(nk)^{-x}}{\zeta(x)} \,dx \\ &= \int_1^\infty \frac{n^{-x}}{\zeta(x)} \sum_{k=1}^\infty \int_1^\infty k^{-x} \,dx \\ &= \int_1^\infty \frac{n^{-x}}{\zeta(x)} \zeta(x) \,dx = \int_1^\infty n^{-x} \,dx = \frac1{n\log n} \quad(n\ge2) \end{align*} as desired.
• Nice. WolframAlpha says $$\mu(\mathbb N)-\mu(\{1\})= \sum_{n=2}^\infty \int_1^\infty \frac{n^{-x}}{\zeta(x)}\,dx=\int_1^\infty \frac{\sum_{n=2}^\infty n^{-x}}{\zeta(x)}\,dx=\int_1^\infty1-\frac{1}{\zeta(x)}\,dx = 1.1239120333264$$ – Bjørn Kjos-Hanssen Dec 22 '18 at 7:11