An example satisfying Q1 is as follows:
\begin{align*}
& A=\left\{\bigcup_{n\in\mathbb{Z}}[3n,3n+2)\setminus D'\right\}\cup D, \\
& B=\left\{\bigcup_{n\in\mathbb{Z}}[3n-1,3n)\setminus D\right\}\cup D', \\
& D=\{m/2^n\mid n=1,2,\dots,\,m=\pm 1,\pm 2,\dots,\,m/2^n\notin\mathbb{Z}\}, \\
& D'=\{m/3^n\mid n=1,2,\dots,\,m=\pm 1,\pm 2,\dots,\,m/3^n\notin\mathbb{Z}\}.
\end{align*}
We can confirm:
- $A\cap B=\emptyset$ and $A\cup B=\mathbb{R}$.
- $A$ and $B$ are dense in any open interval $(a,b)$ since $D\subset A$, $\bar{D}=\mathbb{R}$ and $D'\subset B$, $\bar{D'}=\mathbb{R}$. Here, for $X\subset \mathbb{R}$, $\bar{X}$ represents the closure of $X$ under the usual topology of $\mathbb{R}$. Note that $D$ means binary approximation and $D'$ means ternary approximation, and $D$ and $D'$ do not intersect.
- We have
\begin{align*}
\frac{\lambda(A\cap [-3n,3n])}{2\cdot 3n}
&= \frac{2n\cdot 2}{6n}=\frac{2}{3}, \qquad n=1,2,\dots
\end{align*}
then it follows that
\begin{align*}
\lim_{t\to\infty}\frac{\lambda(A\cap [-t,t])}{2t}
&= \frac{2}{3}\neq \frac{1}{2}.
\end{align*}
By the same manner as above, we see also
\begin{align*}
\lim_{t\to\infty}\frac{\lambda(B\cap[-t,t])}{2t}=\frac{1}{3}\neq \frac{1}{2}.
\end{align*}
Q2: The way of giving meaning is up to you.
I'd consider it as normalization over $2ε=λ(B_\varepsilon(x))$, and I would answer that it's the probability of a number chosen from a uniform distribution on $B_\varepsilon(x)$ being contained in $A$ (or $B$).
I have no idea of another interpretation. Other people may have different answers.
Also, does the $A$ and $B$ in this question correspond to the post you linked or to the example you want in question (1)?
– 1mdlrjcmed Aug 09 '23 at 05:49