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. $\text{mod }1$) 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 the fractional part of $x_n \in E.$
Let $x_1,\cdots x_N$ be a finite sequence of real numbers.
The number $$D_N = D_N(x_1,\cdots x_N)= \sup_{0 \leq \alpha < \beta \leq 1}\Bigg|\frac{A([\alpha,\beta);N)}{N}-(\beta - \alpha)\Bigg|$$
is called the discrepancy of the given sequence.
Now main problem.
Prove that $$D_N(x_1,\cdots, x_N)= \sup_{0 \leq \alpha \leq \beta \leq 1} \Bigg|\frac{A([\alpha,\beta];N)}{N} - (\beta - \alpha)\Bigg|$$
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 partially 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?
Some notations.
Without loss of relevance or generality, assume $N$ is fixed and all $x_i$ are in $[0,1)$.
Let $$D_N(\alpha,\beta)=\left|\frac{A([\alpha,\beta);N)}{N} - (\beta - \alpha)\right|.$$ So $D_N=\displaystyle\sup_{0 \leq \alpha \leq \beta \leq 1} D_N(\alpha, \beta).$
Similarly, let $$\Delta_N(\alpha,\beta)=\left|\frac{A([\alpha,\beta];N)}{N} - (\beta - \alpha)\right|,$$ and $\Delta_N=\displaystyle\sup_{0 \leq \alpha \leq \beta \leq 1} \Delta_N(\alpha, \beta).$
We are asked to show $D_N=\Delta_N$.
Remarks on the approach in the question
Call $\Delta_N$ "the stronger formula" is misleading. While the extra case when $\alpha=\beta$ may help make $\Delta_N$ larger than $D_N$, the fact that $A([\alpha,\beta];N)$ may be strictly larger than $A([\alpha,\beta);N)$ does not necessarily help make $\Delta_N$ larger because of the absolute value. Although $0.36>0.2$, we have $|0.36-0.3| < |0.2-0.3|$.
It looks your (idea of) proof is insufficient or lacking the details.
A complete proof
We will prove each of $D_N$ and $\Delta_N$ is $\le$ the other.
Prove $\Delta_N \le D_N$
Consider $\Delta_N(\alpha, \beta)$, where $0\le\alpha\le\beta\le1$. There are two cases.
Since all $x_i<1$, $\ A([\alpha, 1];N)=A([\alpha, 1);N)$. Hence $\Delta_N(\alpha,\beta)=D_N(\alpha,\beta)\le D_N$
Since $$\left|D_N(\alpha, \gamma)-\Delta_N(\alpha,\beta)\right| \le\left|\frac{A([\alpha, \gamma);N)-A([\alpha, \beta];N)}N\right|+\left|\beta-\gamma\right|$$ for all $\beta<\gamma\le 1$ and $$\lim_{\gamma\to\beta^+}A([\alpha, \gamma);N)=A([\alpha, \beta];N)$$ We have $$\lim_{\gamma\to\beta^+}\left|D_N(\alpha, \gamma)-\Delta_N(\alpha,\beta)\right|=0 $$ Hence $\Delta_N(\alpha,\beta)\le D_N$.
So in all cases, $\Delta_N(\alpha,\beta)\le D_N$. Since $\alpha,\beta$ are arbitrary, $\Delta_N\le D_N$.
Prove $D_N \le \Delta_N$
Consider $D_N(\alpha, \beta)$, where $0\le\alpha<\beta\le1$. Since $$\left|\Delta_N(\alpha, \gamma)-D_N(\alpha,\beta)\right| \le\left|\frac{A([\alpha, \gamma];N)-A([\alpha, \beta);N)}N\right|+\left|\beta-\gamma\right|$$ for all $\alpha\le\gamma\le 1$ and $$\lim_{\gamma\to\beta^-}A([\alpha, \gamma];N)=A([\alpha, \beta);N)$$ We have $$\lim_{\gamma\to\beta^-}\left|\Delta_N(\alpha, \gamma)-D_N(\alpha,\beta)\right|=0 $$ Hence $D_N(\alpha,\beta)\le \Delta_N$. Since $\alpha,\beta$ are arbitrary, $D_N\le\Delta_N$.