A function $f\colon \mathcal{P}(\mathbf{N})\to [0,1]$ is said to have the *Darboux property* whenever for all $X \subseteq \mathbf{N}$ and $y \in [0,f(X)]$, there exists $Y \subseteq X$ such that $f(Y)=y$. 


Moreover, $f$ is said to be *monotone* if $f(X)\le f(Y)$ whenever $X\subseteq Y$ and *subadditive* if $f(X\cup Y) \le f(X)+f(Y)$ for all $X,Y \subseteq \mathbf{N}$.



> 
> **Question.** Let $f\colon \mathcal{P}(\mathbf{N})\to [0,1]$ be a monotone subadditive function with the Darboux property such that $f(\mathbf{N})=1$. Does there exist necessarily a *finite* partition $\{A\_1,\ldots,A\_k\}$ of $\mathbf{N}$ for which
>  $$
> f(A\_1), \ldots, f(A\_k) \in (0,1)?
> $$
> 
> 
> 




---


A simpler version of this question has been posted on MSE one week ago [here](https://math.stackexchange.com/questions/1457731/a-monotone-subadditive-non-atomic-f-on-the-power-of-bf-n-hasa-partition-a), showing that in general the answer is negative if $k$ is bounded. 


The question comes from the study of properties of a class of functions satisfying certain axioms, which can be called as "upper densities" (with this respect, you can see this MO [question](https://mathoverflow.net/questions/214878/additivity-of-upper-densities-with-respect-to-arithmetic-progressions-of-integer)). [Here](https://mathoverflow.net/questions/219274/reference-request-darboux-properties-of-real-valued-set-functions-measures-de), instead, there is MO question related to the definition of Darboux property. 


A related construction (which does not ask for subadditivity) by [Salvo Tringali](https://mathoverflow.net/users/16537/salvo-tringali?tab=profile) has been provided [here](https://math.stackexchange.com/questions/1404968/does-darboux-property-imply-strong-darboux-property).



Does a monotone subadditive $f: \mathcal{P}(\bf N)\to [0,1]$ admit a finite partition with values in $(0,1)$?
$\let\eps\varepsilon$It seems that the following function fits:
$$
 f(A)=\inf\left\{\alpha\colon \quad \sum\_{x\in A}x^{-\alpha}<\infty\right\}.
$$
(I assume that $\mathbb N$ starts with 1.)
Clearly, it is monotone, and $f(\mathbb N)=1$. Moreover, it is more than subadditive: we have $f(A\cup B)=\max\{f(A),f(B)\}$. This also implies that if $A\_1\cup A\_2\cup\dots\cup A\_k=\mathbb N$, then $f(A\_i)=1$ for some $i$.


It remains to show that $f$ has the Darboux property. Let $f(A)=\alpha$ and $\beta\in[0,\alpha)$; we need to find $B\subseteq A$ with $f(B)=\beta$. We have $\sum\_{x\in A}x^{-\beta}=\infty$. Set $X\_n=[2^n,2^{n+1})\cap \mathbb N$, $A\_n=A\cap X\_n$, and $a\_n=\sum\_{x\in A\_n}x^{-\beta}$. Then the series $\sum\_n a\_n$ diverges.


Set $b\_n=\min\{a\_n,1\}$. The series $\sum\_n b\_n$ diverges as well (this is trivial if $b\_i=1$ infinitely many times, as well as if it happens only finitely many times). Now, choose $B\_n$ to be the subset of $A\_n$ such that $S\_n=\sum\_{x\in B\_n}x^{-\beta}\leq b\_n$, and $S\_n$ is maximal subject to this property. Then $S\_n\geq b\_n/2$ for all $n\geq 1$. 


Set $B=\cup\_n B\_n$; then $\sum\_{x\in B}x^{-\beta}\geq \sum\_n b\_n/2=\infty$. On the other hand, for every $\eps>0$ we have
$$
 \sum\_{x\in B}=
 \sum\_n\sum\_{x\in B\_n}x^{-\beta-\eps}
 \leq \sum\_n 2^{-\eps n}\sum\_{x\in B\_n}x^{-\beta}
 \leq \sum\_n2^{-\eps n}<\infty.
$$
Thus $f(B)=\beta$.


