Let $\mathsf{d}^\star$ be the asymptotic upper density, defined on the power set of positive integers $\mathbf{N}$, so that
$$
\mathsf{d}^\star\colon \mathcal{P}(\mathbf{N}) \to\mathbf{R}\colon X\mapsto \limsup_{n\to \infty} \frac{|X\cap [1,n]|}{n}.
$$
It is easy to verify that if $k\cdot \mathbf{N}+h:=\{kx+h\colon x \in \mathbf{N}\}$ is an arithmetic progression of $\mathbf{N}$, and $X$ is a set of integers having no elements in common with $k\cdot \mathbf{N}+h$, then
$$
\mathsf{d}^\star(X\cup (k\cdot \mathbf{N}+h))=\mathsf{d}^\star(X)+\frac{1}{k}.
$$ 
The same reasoning can be extended to the upper analytic, upper logarithmic, upper Banach and upper Buck densities, at least. That's why one may ask if this property holds in general:

We say that a set function $\mu^\star\colon \mathcal{P}(\mathbf{N})\to \mathbf{R}$ is an "*upper density*" whenever it satisfies the following axioms:

(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) $\mu^\star(k\cdot X+h)=\mu^\star(X)/k$

for all positive integers $k,h$ and sets $X,Y \subseteq \mathbf{N}$. Then, is it true that if $X \cap (k\cdot \mathbf{N}+h)=\emptyset$ then
$$
\mu^\star(X \cup (k\cdot \mathbf{N}+h))=\mu^\star(X)+\frac{1}{k}\,\,?
$$