Existence of a monotone subadditive function with a jump on its values Let $f$ be a nonnegative function defined on the power of the set of positive integers for which:


*

*$f(X) \le f(Y) \le f(\mathbf{N})=1$ if $X\subseteq Y$;

*$f(X\cup Y)\le f(X)+f(Y)$;

*for each $X$ and each $y \in [0,f(X)]$ there exists $Y \subseteq X$ such that $f(Y)=y$.


Fix a set $A$ such that $f(A)<1$. Does there exist a set $B$ containing $A$ such that 
$$
f(A)<f(B)<1 \,\,?
$$
 A: Let $f(A)<1/2$, there is $n\in \mathbb{N}$ such that $f(A)<1/2-\frac{1}{2n}<1$. Since $f(\mathbb{N})=1$ so from property $(3)$: there is a set $C$ such that $$f(C)= \frac{1}{n}+f(A)$$ set $B:=A\cup C$, so from property $(1,2)$: $f(A)\leq f(B)\leq f(C)+f(A)$ and $f(C)\leq f(B)$ i.e., 
$$f(A) <f(C)\leq f(B)\leq f(C)+f(A)\leq \frac{1}{n}+2f(A)<1.$$
A: Actually $\frac 12$ is the limit. A possible construction of a "bad" function goes like this.
First of all, there exists a function $f$ satisfying (1-3). For instance, $f(X)=\limsup_{n\to\infty}\frac{X\cap[1,n]}{n}$, as suggested. Note that verifying $3$ is an easy, but not completely trivial exercise.
Next, split $\mathbb N$ into countably many infinite subsets and view each subset as a separate copy of $\mathbb N$. Then we can view any set $X$ as an infinite sequence $X_1,X_2,\dots$ of subsets of $\mathbb N$. Put $G(X)=\sum_{k\ge 1}f(X_k)\in[0,+\infty]$ and $g(x)=\frac{G(X)}{G(X)+1}$ with the usual convention $\frac{\infty}{\infty+1}=1$. Then $g$ still satisfies (1-3) but has the additional property that $g(X\cup Y)=1$ implies that $g(X)=1$ or $g(Y)=1$.
Finally, split $\mathbb N$ into two copies of $\mathbb N$, so each subset of the original $\mathbb N$ is now a pair $X,Y$ of subsets of $\mathbb N$ and put
$h(X,Y)=\frac 12\max(f(X),g(Y))$ if $g(Y)<1$ and $h(X,Y)=\frac 12+\frac 12f(X)$ if $g(Y)=1$.
1) Let $X\subset X',Y\subset Y'$. If $g(Y)<1$, then $h(X,Y)=\frac 12\max(f(X),g(Y))\le\frac 12\max(f(X'),g(Y'))\le h(X',Y')$. If $g(Y)=1$, then $g(Y')=1$ as well, so $h(X,Y)=\frac 12+\frac 12f(X)\le \frac 12+\frac 12f(X')=h(X',Y')$.
2) Let $(X,Y)$ and $(X',Y')$ be two (pairs of) sets. If $g(Y\cup Y')<1$, then the subadditivity follows from the subadditivity of $f$, $g$, and $\max$. Otherwise, WLOG, $g(Y)=1$, so
$$
h(X\cup X',Y\cup Y')=\frac 12+\frac 12f(X\cup X')\le \frac 12+\frac 12f(X)+\frac 12f(X')=h(X,Y)+\frac 12f(X')\le h(X,Y)+h(X',Y')$$.
3) If $g(Y)<1$, just decrease $f(X)$ and $g(Y)$ proportionally. If $g(Y)=1$, start with decreasing $f(X)$ until it drops to $0$, then decrease $g(Y)$.
Finally, let $A=(\mathbb N,\varnothing)$    
