If $A$ and $B$ both have defined asymptotic density, and $A\cap B=\emptyset$, does $A \cup B$? Let $A \subseteq \mathbb N$. Define the asymptotic density of $A$ as $$d(A) = \lim_{n\to \infty}\frac {|A \cap \{1,...,n\}|}{n}$$
If $A\cap B=\emptyset$, $d(A)$ and $d(B)$ are defined, is $d(A\cup B)$ defined too?
I know someone has asked a similar question here: If two sets have a natural (asymptotic) density, does their union?
But in that thread, $A$ and $B$ might not be disjoint.

Note that
$$
\begin{eqnarray}d(A\cup B)&=&\lim_{n\rightarrow\infty}\frac{\left|(A\cup B)\cap [n]\right|}{n}\\&=&\lim_{n\rightarrow\infty}\frac{\left|A\cap[n]\right|+\left|B\cap[n]\right|-\left|(A\cap B)\cap[n]\right|}{n}\\
&=&\lim_{n\rightarrow\infty}\frac{\left|A\cap[n]\right|}{n}+\lim_{n\rightarrow\infty}\frac{\left|B\cap[n]\right|}{n}-\lim_{n\rightarrow\infty}\frac{\left|A\cap B \cap[n]\right|}{n}\\
&=&d(A)+d(B)-d(A\cap B)
\end{eqnarray}
$$
(where $[n]\equiv\{1,2,\ldots,n\}$), assuming that the limits on the right-hand side converge.
So if $A\cap B=\emptyset$ (or even under the weaker condition that $d(A\cap B)=0$), then $$d(A\cup B)=d(A)+d(B).$$
