Let $\mu^\star$ be a real-valued function defined on the power set of the positive integers $\mathbf{N}^+$ such that for all $X,Y\subseteq \mathbf{N}^+$ the following axioms hold:


(F1) $\mu^\star(\mathbf{N}^+)=1$;


(F2) $\mu^\star(X) \le \mu^\star(Y)$ if $X\subseteq Y$;


(F3) $\mu^\star(X\cup Y) \le \mu^\star(X)+\mu^\star(Y)$;


(F4) $\mu^\star(\{kx:x \in X\})=\mu^\star(X)/k$ for all $k \in \mathbf{N}^+$;


(F5) $\mu^\star(\{x+h: x \in X\})=\mu^\star(X)$ for all $h \in \mathbf{N}^+$.


A function of this type is said to be an *(arithmetic) upper density*. The set of these functions include the upper asymptotic, Banach, logarithmic, analytic, Polya, Buck densities and many others. Related questions on these type of functions can be found [here](https://mathoverflow.net/questions/214878/additivity-of-upper-densities-with-respect-to-arithmetic-progressions-of-integer) and [here](https://mathoverflow.net/questions/220268/does-a-monotone-subadditive-f-mathcalp-bf-n-to-0-1-admit-a-finite-part).


At this point we can defined its associated lower density $\mu\_\star$ for all $X\subseteq \mathbf{N}^+$ by 
$$\mu\_\star(X)=1-\mu^\star(X^c).$$
Now, all examples I have in mind are superadditive functions, namely
$$
\mu\_\star(X\cup Y) \ge \mu\_\star(X)+\mu\_\star(Y)
$$
whenever $X$ and $Y$ are disjoint subsets of $\mathbf{N}^+$. Does this property hold in general?


After an easy manipulation, the question turns out to be equivalent to the following:



> 
> **Question.** Let $\mu^\star$ be an upper density on $\mathbf{N}^+$, that is, a function satisfying axioms (F1)-(F5). Is it true that if $X,Y$ are subsets of $\mathbf{N}^+$ such that $X\cup Y=\mathbf{N}^+$ then
>  $$
> 1+\mu^\star(X\cap Y) \le \mu^\star(X)+\mu^\star(Y)?
> $$
> 
> 
> 




---


In turn, this can be viewed as a strenghtening of (F3) above; moreover, it would imply that the induced density $\mu$, which can be seen as the restriction of $\mu^\star$ on $\{X\subseteq \mathbf{N}^+:\mu^\star(X)=\mu\_\star(X)\}$, is additive.



Superadditivity of the lower density
I will reuse the same trick as in the [answer](https://mathoverflow.net/a/226461/16537) to Paolo's other question to show that the answer to this is still in the negative.


Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function
$$h := ((1-\alpha) f^q + \alpha g^q)^{\frac{1}{q}}$$
is an upper density too (most notably, condition (F3) follows from Minkowski's inequality, which is where we need $q \ge 1$). So in particular, assume from now on that $f$ is the upper asymptotic (or natural) density (on $\mathbf N^+$), namely the function
$$
\mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto \limsup\_{n \to \infty} \frac{|X \cap [1,n]|}{n},
$$
and $g$ the upper Banach (or uniform) density, viz. the function
$$
\mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto \lim\_{n \to \infty} \max\_{h \ge 0} \frac{|X \cap [1+h,n+h]|}{n}
$$
(the latter limit exists as a consequence of [Fekete's lemma](https://mathoverflow.net/questions/99122/generalizations-and-relative-applications-of-feketes-subadditive-lemma)); both of these functions are upper densities, as was mentioned in the OP.


Next, fix $a \in {]0,1]}$, and set
$$
V\_a := \bigcup\_{n \ge 1} [\![a(2n-1)!+(1-a) (2n)!, (2n)!]\!]
$$
and
$$
X := V\_a \cup (V\_a^c \cap (2\cdot \mathbf N))
\quad\text{and}\quad
Y := V\_a \cup (V\_a^c \cap (2 \cdot \mathbf N+1)),
$$
where $V\_a^c := \mathbf N^+ \setminus V\_a$. It is clear that $X \cup Y = \mathbf N^+$ and $X \cap Y = V\_a$. Moreover, we get from [here](https://mathoverflow.net/questions/207522/reference-for-a-lemma-on-the-asymptotic-upper-density-of-special-sets-with-large) that
$$
f(X) \le f(V\_a) + f(V\_a^c \cap (2 \cdot \mathbf N)) \le a +\frac{1}{2},
$$
and similarly $f(Y) \le a + \frac{1}{2}$ and $f(V\_a) = a$. On the other hand, $V\_a$ contains arbitrarily large intervals of consecutive integers, hence $g(X) = g(Y) = g(V\_a) = 1$. It follows that
$$
1+h(X \cap Y) = 1 + (\alpha +a^q(1-\alpha))^{\frac{1}{q}}
$$
and
$$
h(X) + h(Y) \le 2 \cdot (\alpha +(0.5+a)^q(1-\alpha))^{\frac{1}{q}}.
$$
With this in hand, suppose to a contradiction that $1+h(A \cup B) \le h(A) + h(B)$ for all $A, B \subseteq \mathbf N^+$ such that $A \cup B = \mathbf N^+$, regardless of the actual values of the parameters $a$, $\alpha$ and $q$. Then, we have from the above that
$$
1 + \alpha^{\frac{1}{q}} \le 2 \cdot (\alpha +\left(0.5+a\right)^q(1-\alpha))^{\frac{1}{q}},
$$
which ultimately implies, in the limit as $a \to 0^+$, that
$$
1 + \alpha^{\frac{1}{q}} \le (2^q\alpha + 1-\alpha)^{\frac{1}{q}}
$$
for all $\alpha \in [0,1]$ and $q \in [1,\infty[$. This, however, is quite false (e.g., let $\alpha = \frac{1}{2}$ and $q = 2$).


