[proofplan] The forward implication from discrepancy to uniform distribution is immediate because the discrepancy is the supremum of all interval-counting errors. For the converse, assume the interval counts converge for every half-open interval and first obtain uniform control over a finite mesh of intervals. Then approximate an arbitrary interval from inside and outside by mesh intervals; the mesh endpoint error is controlled by the mesh size, while the counting error on mesh intervals is uniformly small because there are only finitely many of them. Letting first $N\to\infty$ and then the mesh size tend to $0$ gives $D_N(x_1,\dots,x_N)\to 0$. [/proofplan]

[step:Bound each fixed interval count by the discrepancy] Assume that \begin{align*} \lim_{N\to\infty}D_N(x_1,\dots,x_N)=0. \end{align*} Let $[a,b)\subset [0,1)$ be a half-open interval. For each $N\in\mathbb N$, define \begin{align*} A_N([a,b)):=\#\{1\le n\le N:\{x_n\}\in [a,b)\}. \end{align*} By the definition of $D_N(x_1,\dots,x_N)$ as a supremum over all half-open intervals in $[0,1)$, we have \begin{align*} \left|\frac{A_N([a,b))}{N}-(b-a)\right|\le D_N(x_1,\dots,x_N). \end{align*} The right-hand side tends to $0$, so \begin{align*} \lim_{N\to\infty}\frac{A_N([a,b))}{N}=b-a. \end{align*} Since $[a,b)$ was arbitrary, $(x_n)_{n\ge 1}$ is uniformly distributed modulo $1$. [/step]

[step:Obtain uniform counting control on a finite mesh]Conversely, assume that $(x_n)_{n\ge 1}$ is uniformly distributed modulo $1$. Fix an integer $M\ge 1$. For integers $0\le r\le s\le M$, define the mesh interval \begin{align*} I_{r,s}:=\left[\frac{r}{M},\frac{s}{M}\right). \end{align*} There are finitely many such intervals. By uniform distribution modulo $1$, for every pair $0\le r\le s\le M$, \begin{align*} \lim_{N\to\infty}\frac{1}{N}\#\{1\le n\le N:\{x_n\}\in I_{r,s}\}=\frac{s-r}{M}. \end{align*} Therefore, for every $\varepsilon>0$ there exists $N_0=N_0(M,\varepsilon)\in\mathbb N$ such that, for all $N\ge N_0$ and all $0\le r\le s\le M$, \begin{align*} \left|\frac{1}{N}\#\{1\le n\le N:\{x_n\}\in I_{r,s}\}-\frac{s-r}{M}\right|\le \varepsilon. \end{align*}[/step]

[guided]We need a statement that is stronger than convergence on one interval at a time. For a fixed mesh size $M$, however, there are only finitely many mesh intervals \begin{align*} I_{r,s}:=\left[\frac{r}{M},\frac{s}{M}\right) \end{align*} with $0\le r\le s\le M$. Uniform distribution modulo $1$ says that each one satisfies \begin{align*} \lim_{N\to\infty}\frac{1}{N}\#\{1\le n\le N:\{x_n\}\in I_{r,s}\}=\frac{s-r}{M}. \end{align*} The reason this becomes uniform over the mesh is finiteness. Given $\varepsilon>0$, each fixed pair $(r,s)$ has some threshold $N_{r,s}$ after which its counting error is at most $\varepsilon$. Since the set of pairs $(r,s)$ with $0\le r\le s\le M$ is finite, define $N_0$ to be the maximum of these finitely many thresholds. Then for every $N\ge N_0$ and every mesh interval $I_{r,s}$, \begin{align*} \left|\frac{1}{N}\#\{1\le n\le N:\{x_n\}\in I_{r,s}\}-\frac{s-r}{M}\right|\le \varepsilon. \end{align*} This is the key compactness-in-finite-form step: pointwise convergence over finitely many mesh intervals becomes simultaneous convergence over all of them.[/guided]

[step:Approximate an arbitrary interval by mesh intervals] Fix $M\ge 1$, $\varepsilon>0$, and $N\ge N_0(M,\varepsilon)$ as in the previous step. Let $[a,b)\subset [0,1)$ be arbitrary. Choose integers $r_-,r_+,s_-,s_+$ with \begin{align*} \frac{r_-}{M}\le a<\frac{r_-+1}{M},\qquad \frac{r_+ -1}{M}<a\le \frac{r_+}{M}, \end{align*} and \begin{align*} \frac{s_-}{M}\le b<\frac{s_-+1}{M},\qquad \frac{s_+ -1}{M}<b\le \frac{s_+}{M}. \end{align*} Equivalently, $r_-=\lfloor Ma\rfloor$, $r_+=\lceil Ma\rceil$, $s_-=\lfloor Mb\rfloor$, and $s_+=\lceil Mb\rceil$. Define the outer mesh interval \begin{align*} I^+:=\left[\frac{r_-}{M},\frac{s_+}{M}\right) \end{align*} and define the inner mesh interval \begin{align*} I^-:=\left[\frac{r_+}{M},\frac{s_-}{M}\right) \end{align*} if $r_+\le s_-$, while if $r_+>s_-$ we set $I^-:=\varnothing$ and regard its length and count as $0$. Then \begin{align*} I^-\subset [a,b)\subset I^+. \end{align*} For a half-open interval $J=[u,v)\subset [0,1)$, write $|J|:=v-u$ for its length, and set $|\varnothing|:=0$. Moreover, \begin{align*} |I^+|-(b-a)\le \frac{2}{M} \end{align*} and \begin{align*} (b-a)-|I^-|\le \frac{2}{M}. \end{align*} Let \begin{align*} A_N(J):=\#\{1\le n\le N:\{x_n\}\in J\} \end{align*} for every half-open interval $J\subset [0,1)$, and set $A_N(\varnothing):=0$. Since $I^-\subset [a,b)\subset I^+$, monotonicity of cardinality gives \begin{align*} A_N(I^-)\le A_N([a,b))\le A_N(I^+). \end{align*} The finite-mesh estimate gives \begin{align*} \frac{A_N(I^+)}{N}\le |I^+|+\varepsilon \end{align*} and, because either $I^-$ is a mesh interval or $I^-=\varnothing$, \begin{align*} \frac{A_N(I^-)}{N}\ge |I^-|-\varepsilon. \end{align*} Hence \begin{align*} \frac{A_N([a,b))}{N}-(b-a)\le \varepsilon+\frac{2}{M} \end{align*} and \begin{align*} (b-a)-\frac{A_N([a,b))}{N}\le \varepsilon+\frac{2}{M}. \end{align*} Therefore \begin{align*} \left|\frac{A_N([a,b))}{N}-(b-a)\right|\le \varepsilon+\frac{2}{M}. \end{align*} [/step]

[step:Take the supremum over intervals and send the mesh size to infinity] The estimate in the previous step holds for every half-open interval $[a,b)\subset [0,1)$ whenever $N\ge N_0(M,\varepsilon)$. Taking the supremum over all $0\le a\le b\le 1$ gives \begin{align*} D_N(x_1,\dots,x_N)\le \varepsilon+\frac{2}{M} \end{align*} for all $N\ge N_0(M,\varepsilon)$. Thus \begin{align*} \limsup_{N\to\infty}D_N(x_1,\dots,x_N)\le \varepsilon+\frac{2}{M}. \end{align*} Since $\varepsilon>0$ was arbitrary, we obtain \begin{align*} \limsup_{N\to\infty}D_N(x_1,\dots,x_N)\le \frac{2}{M}. \end{align*} Letting $M\to\infty$ gives \begin{align*} \limsup_{N\to\infty}D_N(x_1,\dots,x_N)\le 0. \end{align*} Because $D_N(x_1,\dots,x_N)\ge 0$ by definition, this implies \begin{align*} \lim_{N\to\infty}D_N(x_1,\dots,x_N)=0. \end{align*} This proves the converse implication and completes the proof. [/step]