SOLVED
Let $A=\{a_1<a_2<\cdots\}\subseteq \mathbb{N}$ be infinite such that $a_{i+1}/a_i\to 1$. For any $x\geq a_1$ let
\[f(x) = \frac{x-a_i}{a_{i+1}-a_i}\in [0,1),\]
where $x\in [a_i,a_{i+1})$. Is it true that, for almost all $\alpha$, the sequence $f(\alpha n)$ is uniformly distributed in $[0,1)$?
For example if $A=\mathbb{N}$ then $f(x)=\{x\}$ is the usual fractional part operator.
A problem due to Le Veque [LV53], who proved it in some special cases.
This is false is general, as shown by Schmidt [Sc69].
