Equidistributed sequence in all $\Bbb R $?

Question

Does there exist a real sequence $(x_n)$ such that

$$\forall \ a < b,\ c < d \in \Bbb R$$

$$ \lim_{n\to \infty}\frac{\operatorname{Card}\{0 ≤ i ≤n\ :\ a≤x_i≤b\}}{\operatorname{Card}\{0 ≤ i ≤n\ :\ c≤x_i≤d\}} = \frac{b-a}{d-c}$$

Answer 1

A sequence that goes through all the rational numbers carefully should do the trick.

Let's define the sequence $(x_n)$ in blocks, so that after the $k^{\text{th}}$ block the sequence has included all the (finitely many) fractions $\frac pq$ with $-k \le \frac pq \le k$ and $|q| \le k$. To simplify analysis, we won't simplify the fractions: we will visit $\frac12$ in the second block, $\frac24$ in the fourth block, and $\frac36$ in the sixth block.

Choose $n$ large enough so that $x_1, x_2, \dots, x_n$ included $k$ complete blocks and is partway through the next, and assume $k > \max\{|a|,|b|,|c|,|d|\}$. For each $j \le k$, there are approximately $j(b-a)$ fractions with denominator $j$ in $[a,b]$: at least $j(b-a)-1$ and no more than $j(b-a)+1$. So in the first $k$ blocks, we have $(1 + 2 + \dots + k)(b-a) + O(k)$ elements of $[a,b]$. The $(k+1)^{\text{th}}$ block also includes only $O(k)$ elements of $[a,b]$. So the fraction in the limit is $$\frac{\binom{k+1}{2}(b-a) + O(k)}{\binom{k+1}{2}(d-c) + O(k)}$$ which converges to $\frac{b-a}{d-c}$ as $k \to \infty$.