I am reading famous book about uniform distribution of sequences by Kuipers and Niederreiter and have questions about solving below exercise from that book. Before going to main exercise I will write here some definitions that are needed for problem.
DEFINITION The sequence $w = (x_n), n = 1, 2, ... ,$ of real numbers is said to be uniformly distributed modulo 1 (abbreviated u.d. mod 1) if for every pair a, b of real numbers with $0 < a < b < 1$ we have
$$lim_{N \to \infty} \frac{A([a,b);N;w)}{N} = b-a$$
where for a positive integer $N$ and a subset $E$ of $[0,1]$, let the counting function $A(E; N; w)$ be defined as the number of terms $x_n, 1 \leq n \leq N$, for which fractional part of $x_n \in E.$
Let $x_1,\cdots x_N$ be finite sequence of real numbers.
The number $$D_N = D_N(x_1,\cdots x_N)= sup_{0 \leq \alpha < \beta \leq 1}|\frac{A([\alpha,\beta);N)}{N}-(\beta - \alpha)|$$
is called the discrepancy of the given sequence.
Now main problem.
Prove that $$D_N(x_1,\cdots, x_N)= sup_{0 \leq \alpha \leq \beta \leq 1} |\frac{A([\alpha,\beta];N)}{N} - (\beta - \alpha)|$$
So two points needs to be proven, the interval could be just one point and closed interval instead of partial closed interval.
For first case why we can write $\alpha = \beta $ below sup.
Let $\alpha = \beta $ and let all fractional parts of ${x_n}$ be $\alpha$.Second $D_N$ will be equal to $1$. For first one let $\beta = \alpha + \epsilon$
Then $D_N = lim_{\epsilon \to 0} 1 - \epsilon = 1$
Same thing we can do with replacing partialy closed interval to closed interval. We will look $[\alpha, \beta + \epsilon)$ intervals.
Seems trivial work from my side don't think solution would be this trivial. Anything I am doing wrong?
