I need to prove whether two sequences are equidistributed or not For each subset $S\subset [0,1]$ write $X_{s}$ for the "indicator function" as $X_{s}(t)=1$ when $t\in S$ and $X_{s}(t)=0 $ when $t\notin S$.
A sequence $\{x_{n}\}$ in $[0,1]$ is called equidistributed when for each subinterval $[a,b]\subset[0,1]$  we have $$\lim_{N\to\infty}{1\over N}\sum_{j=1}^{N}X_{[a,b]}(x_{j})=(b-a).$$
Then determine whether the sequence $1\over 2$, $1\over 3$, $2\over 3$, $1\over 4$, $2\over 4$, $3\over 4$, $...$ is equidistributed in $[0,1]$?
And how about 1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, 1/16, 3/16, 5/16, 7/16,9/16, 11/16, 13/16, 15/16,...?
I asked my professor and he let me prove the next 6 statements are equivalent:
(1) {$x_{n}$} is equidistributed in [0,1]
(2) for each Riemann-integrable function f:[0,1]->R we have  $$\lim_{N\to\infty}{1\over N}\sum_{j=1}^{N}f(x_{j})=\int_{0}^{1}fdx.$$
(3) For each continuous function f:[0,1]->R we have $$\lim_{N\to\infty}{1\over N}\sum_{j=1}^{N}f(x_{j})=\int_{0}^{1}fdx.$$
(4) For each continuous function $f:[0,1]\to R$ such that $f(0)=f(1)$ we have $$\lim_{N\to\infty}{1\over N}\sum_{j=1}^{N}f(x_{j})=\int_{0}^{1}fdx.$$
(5) For each integer $l\ge 1$ we have $$\lim_{N\to\infty}{1\over N}\sum_{j=1}^{N}(x_{j})^l=\int_{0}^{1}x^ldx={1\over {l+1}}.$$
(6) For each integer $l\ge 1$ we have
$$0=\lim_{N\to\infty}{1\over N}\sum_{j=1}^{N}\cos(2\pi lx_{j})= \lim_{N\to\infty}{1\over N}\sum_{j=1}^{N}\sin(2\pi lx_{j}).$$
I have proved these 6 statements are equivalent and now I need to use one of them to prove whether those two sequence are equidistributed or not? But I still don't know how to continue? Can someone help me?That's my general thoughts
 A: I think you can prove it directly from the definition. How many fractions with denominator $n$ will lie in the interval $[a,b]$? That interval has length $b-a$; the space between consecutive fractions with denominator $n$ is $1/n$; you can fit roughly $(b-a)\div(1/n)$ intervals of length $1/n$ into an interval of length $b-a$, so the number of fractions with denominator $n$ in the interval $[a,b]$ is roughly $(b-a)\div(1/n)$, that is, roughly $(b-a)n$. If you're a little careful, you can probably get that to be $(b-a)n+\theta$, where $\theta$ depends on $a$, $b$, and $n$ but satisfies $|\theta|\le2$. 
So, the contribution to the sum from terms with denominator $n$ will be $(b-a)n+\theta$. 
Now, the total number of fractions with denominator at most $n$ is $1+2+\cdots+(n-1)=n(n-1)/2$. A given value of $N$ is between $n(n-1)/2$ and $(n+1)n/2$ for some $n$. So the sum up to $N$ is the sum over all fractions with denominator at most $n$, plus something I'll call $\eta$ that is bounded by $n$ (the "something" coming from the terms with denominator $n+1$). So the whole thing amounts to $${1\over N}\left(\sum_k^n((b-a)k+\theta)+\eta\right)$$ and it's all downhill from there. 
An alternative is to use the Weyl criterion, which is like your (6) but stated for the complex exponential function instead of for sines and cosines. It says for each nonzero integer $h$, $$\lim_{N\to\infty}{1\over N}\sum_1^Ne^{2\pi ihx_k}=0$$ You show that the sum over fractions with a given denominator is small. 
