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 $\mu(\{nk: k\in \mathbb N\})=\dfrac 1{n \log n},\forall n>1$ ?



On the existence of a particular type of finite measure on $\mathbb N$
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](https://math.stackexchange.com/questions/2391847/m%C3%B6bius-inversion-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](https://en.wikipedia.org/wiki/M%C3%B6bius_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](https://en.wikipedia.org/wiki/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.


