reduce to the case $[0,1]$ in equidistribution modulo $1$ I am trying to prove a theorem using an optimal solution. Given $(y_n)_{n \in \mathbb{N^{*}}}$ such that $\exists \delta > 0, \forall n \in \mathbb{N^{*}} y_{n+1} - y_n \geq \delta$ then for almost every $\xi \in \mathbb{R}$ (Lebesgue) the sequence $(\xi y_n)_{n \in \mathbb{N^{*}}}$ is equidistributed modulo $1$.
I found a proof considering an arbitrary interval $[a,b]$ but some people told me it's sufficient to prove the result for almost every $\xi \in [0,1]$ but I am not convinced. In fact, if we prove the result just for almost every $\xi \in [0,1]$ the sequence $\xi y_n$ does not cover all the interval $\xi \in [0,1]$.
I am trying to find a counterexample, if we have a sequence $ \frac{1}{16} \leq y_n \leq \frac{1}{2}$ then multiplying by $\xi \in [0,1]$ gives us  $ 0 \leq \xi y_n \leq \frac{1}{2}$ and if we take $\xi \in [1,2]$ we got $ 0 \leq \xi y_n \leq 1$. Do you see what I mean? I think that the argument reducing to $[0,1]$ is not always true.
Thanks in advance !

I discovered that such an example does not exist. In fact, we have $y_{n+1} - y_{n} \geq \delta$ so $\lim\limits_{n  \longrightarrow +\infty}^{}$. If we suppose that we have proved the result for almost every $\xi \in [0,1]$. Then the result is true for almost every $\xi \in \mathbb{R}$. In fact, if we take $a > 0$, then we got $a y_{n+1} - a y_{n} \geq a \delta > 0 $   so $a \xi y_n$ is equidistributed for almost every $\xi \in [0,1]$ then $\xi y_n$ is equidistributed for almost every $\xi \in \mathbb{R}^{+}$. Then we deduce the result (because if ${x_n}$ is equidistributed modulo $1$ then ${-x_n}$ is equidistributed modulo $1$.
