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 $\mu(\{nk: k\in \mathbb N\})=\dfrac 1{n \log n},\forall n>1$ ?
1 Answer
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 $1/\zeta(x)\asymp x1$ 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.

1$\begingroup$ Nice. WolframAlpha says $$\mu(\mathbb N)\mu(\{1\})= \sum_{n=2}^\infty \int_1^\infty \frac{n^{x}}{\zeta(x)}\,dx=\int_1^\infty \frac{\sum_{n=2}^\infty n^{x}}{\zeta(x)}\,dx=\int_1^\infty1\frac{1}{\zeta(x)}\,dx = 1.1239120333264 $$ $\endgroup$ Dec 22, 2018 at 7:11