A sequence $\omega=(x_n)$ is called ud mod $1$ if $\forall a,b\in \mathbb R$, $0\le a<b\le 1$, we have $$\lim_{n\to \infty} \frac{A([a,b),N,\omega)}{N}=b-a$$ where $A(E,N,\omega)$ represents the number of terms $x_n$, $1\le n\le N$ for which $\{x_n\}\in E$ where $\{x\}=x-[x]$ is the fractional part of $x$.
It is clear that if we leave out finitely many terms from a sequence that is ud mod $1$, the resulting sequence is still ud mod $1$. What additional condition is needed if "finitely" is replaced by "infinitely"? What should be an analogous result if instead of leaving out the terms, we decide to change them in an arbitrary manner?
My guess is that the infinite subsequence we are deleting (or changing) also needs to be ud mod $1$. But, I can't see how to prove it. I am not even sure that my guess is right. Please help me to crack it.
As Reuns pointed out, a subsequence of density $0$ may be deleted without any difficulties. But, I need a proof of that.
You can remove any subsequence of density $0$. Obviously it may fail if you remove a subsequence of density $> 0$.