Does a monotone subadditive $f: \mathcal{P}(\bf N)\to [0,1]$ admit a finite partition with values in $(0,1)$? 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, 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). Here, instead, there is MO question related to the definition of Darboux property. 
A related construction (which does not ask for subadditivity) by Salvo Tringali has been provided here.
 A: $\let\eps\varepsilon$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 $f(B)=\beta$.
