# Prove Sequences are Uniformly Distributed Modulo 1

Can anybody show me the following Proof?

Prove that if $X_n$ for $n=1,2,\ldots,\infty$ is a real sequence that is uniformly distributed modulo $1$, and if $Y_n$, $n=1,2,\ldots,\infty$ is a real sequence such that $\lim_{n\to\infty} X_n – Y_n = \alpha$ then $Y_n$ is uniformly distributed modulo $1$.

Remark: This includes the situation where $X_n – Y_n = \varepsilon$ tends to zero as $n$ tends to infinity.

Hint: Prove for zero case first. Show that $X_n$ and $Y_n$ are close.

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Hint: for any $b \in [0,1]$, consider how many $Y_n$ for $1 \le n \le N$ are in $[0,b]$ modulo $1$. If $|Y_n - X_n| < \epsilon$, what interval for $X_n$ would guarantee $Y_n$ is in this interval? What interval must $X_n$ be in if $Y_n$ is in this interval?
For large $N$, estimate the fraction of $n \in [1, N]$ such that $Y_n$ is in the interval $[0,b]$ modulo $1$, using what you know about the fractions where $X_n$ is in certain other intervals. – Robert Israel Apr 18 '12 at 22:03
As an alternative to Robert's answer, it can also be done with the Weyl criterion. Let $X_n-Y_n=\alpha+b_n$ with $b_n\to0$. Then $$e^{2\pi ihY_n}=e^{2\pi ihX_n}e^{-2\pi ih\alpha}e^{-2\pi ihb_n}$$ The second term on the right is a constant, so it can be pulled out of any sum. The third term, you should show it can be written as $1+c_n$ with $c_n\to0$.