[proofplan] We prove convergence in $\ell^\infty([0,1])$ by the standard two-part criterion for [weak convergence](/page/Weak%20Convergence) of stochastic processes: convergence of all finite-dimensional distributions and tightness in the sup norm. The finite-dimensional distributions reduce to the multivariate [central limit theorem](/theorems/521) applied to multinomial interval counts. Tightness is obtained from dyadic control of empirical increments, which gives asymptotic equicontinuity and hence upgrades finite-dimensional convergence to process-level convergence. The limiting covariance is exactly the covariance of the Brownian bridge. [/proofplan] [step:Declare the empirical process and compute its covariance] Let $(\Omega,\mathcal{F},\mathbb{P})$ denote the underlying probability space on which the random variables $U_1,U_2,\dots$ are defined. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $[0,1]$. For $n \in \mathbb{N}$, define the empirical distribution function $G_n: [0,1] \to [0,1]$ by \begin{align*} G_n(t) := \frac{1}{n}\sum_{k=1}^{n} \mathbb{1}_{[0,t]}(U_k), \qquad t \in [0,1]. \end{align*} Define the centred empirical process $\alpha_n: [0,1] \to \mathbb{R}$ by \begin{align*} \alpha_n(t) := \sqrt{n}\,(G_n(t)-t), \qquad t \in [0,1]. \end{align*} For $s,t \in [0,1]$, since $U_1 \sim \operatorname{Unif}(0,1)$, we have \begin{align*} \mathbb{E}[\mathbb{1}_{[0,s]}(U_1)] = s. \end{align*} Moreover, \begin{align*} \mathbb{E}[\mathbb{1}_{[0,s]}(U_1)\mathbb{1}_{[0,t]}(U_1)] = \mathbb{P}(U_1 \in [0,s] \cap [0,t]) = \min\{s,t\}. \end{align*} Using independence of $U_1,\dots,U_n$, the covariance of $\alpha_n$ is therefore \begin{align*} \operatorname{Cov}(\alpha_n(s),\alpha_n(t)) = \operatorname{Cov}(\mathbb{1}_{[0,s]}(U_1),\mathbb{1}_{[0,t]}(U_1)) = \min\{s,t\}-st. \end{align*} This is the covariance function of the limiting centred Gaussian process $B$. More precisely, $B$ denotes a stochastic process on some probability space, regarded through its law as an $\ell^\infty([0,1])$-valued random element, whose sample paths are continuous and whose covariance satisfies \begin{align*} \operatorname{Cov}(B(s),B(t)) = \min\{s,t\}-st, \qquad s,t\in[0,1]. \end{align*} [/step] [step:Prove convergence of finite-dimensional distributions by multinomial counts] Fix $m \in \mathbb{N}$ and points $0 \leq t_1 < \cdots < t_m \leq 1$. Define $t_0 := 0$ and $t_{m+1} := 1$. Define measurable intervals $A_1 := [0,t_1]$ and $A_j := (t_{j-1},t_j]$ for $2 \leq j \leq m+1$. Define the interval-count vector $N_n \in \mathbb{N}^{m+1}$ by \begin{align*} (N_n)_j := \sum_{k=1}^{n} \mathbb{1}_{A_j}(U_k), \qquad 1 \leq j \leq m+1. \end{align*} The vector $N_n$ has multinomial distribution with parameters $n$ and probabilities $p_j := \mathbb{P}(U_1\in A_j)=t_j-t_{j-1}$ for $1 \leq j \leq m+1$; the equality for $j=1$ uses $\mathbb{P}(U_1=0)=0$. For each $k \in \mathbb{N}$, define the random vector $Y_k: \Omega \to \mathbb{R}^{m+1}$ by \begin{align*} Y_k = \bigl(\mathbb{1}_{A_1}(U_k),\dots,\mathbb{1}_{A_{m+1}}(U_k)\bigr). \end{align*} They are i.i.d. and bounded, hence have finite second moments. By the [Multivariate Central Limit Theorem](/theorems/1854) applied to the i.i.d. bounded random vectors $(Y_k)_{k\in\mathbb{N}}$, we obtain \begin{align*} \frac{N_n-np}{\sqrt{n}} \xrightarrow{d} Z \end{align*} in $\mathbb{R}^{m+1}$, where $p := (p_1,\dots,p_{m+1})$ and $Z$ is centred Gaussian with covariance matrix \begin{align*} \operatorname{Cov}(Z_i,Z_j) = p_i\mathbb{1}_{\{i=j\}}-p_i p_j. \end{align*} For $1 \leq r \leq m$, define the [linear map](/page/Linear%20Map) $T: \mathbb{R}^{m+1} \to \mathbb{R}^{m}$ by \begin{align*} (Tx)_r := \sum_{j=1}^{r} x_j, \qquad 1 \leq r \leq m. \end{align*} Then \begin{align*} (\alpha_n(t_1),\dots,\alpha_n(t_m)) = T\left(\frac{N_n-np}{\sqrt{n}}\right). \end{align*} The map $T$ is continuous because it is linear between finite-dimensional Euclidean spaces. By the [Continuous Mapping Theorem](/theorems/1847) applied to this continuous map, we get \begin{align*} (\alpha_n(t_1),\dots,\alpha_n(t_m)) \xrightarrow{d} T Z. \end{align*} The limiting vector $TZ$ is centred Gaussian and has covariance \begin{align*} \operatorname{Cov}((TZ)_r,(TZ)_q) = \sum_{i=1}^{r}\sum_{j=1}^{q} \bigl(p_i\mathbb{1}_{\{i=j\}}-p_i p_j\bigr) = \sum_{i=1}^{\min\{r,q\}}p_i - \left(\sum_{i=1}^{r}p_i\right)\left(\sum_{j=1}^{q}p_j\right) = \min\{t_r,t_q\}-t_r t_q. \end{align*} Thus the finite-dimensional distributions of $\alpha_n$ converge to those of the Brownian bridge $B$. [/step] [step:Use bracketing of short intervals to obtain asymptotic equicontinuity] Define the empirical increment map $\Delta_n: \{(a,b) \in [0,1]^2 : a < b\} \to \mathbb{R}$ by \begin{align*} \Delta_n(a,b) = \alpha_n(b)-\alpha_n(a) = \frac{1}{\sqrt{n}}\sum_{k=1}^{n}\left(\mathbb{1}_{(a,b]}(U_k)-(b-a)\right). \end{align*} Let $\mathcal{I}_\delta$ denote the class of intervals $(a,b] \subset [0,1]$ with $0 \leq a < b \leq 1$ and $b-a \leq \delta$. Define the centred empirical measure map $\nu_n: \mathcal{I}_\delta \to \mathbb{R}$ by \begin{align*} \nu_n(I) = \frac{1}{\sqrt{n}}\sum_{k=1}^{n}\left(\mathbb{1}_I(U_k)-\mathbb{P}(U_k\in I)\right). \end{align*} Then $\Delta_n(a,b)=\nu_n((a,b])$. For a class $\mathcal{F}$ of [measurable functions](/page/Measurable%20Functions) and $u>0$, define $N_{[]}(u,\mathcal{F},L^2(\mathbb{P}))$ to be the least cardinality of a family of brackets $[\ell_r,h_r]$, where $\ell_r,h_r\in L^2(\mathbb{P})$, $\ell_r\leq h_r$ pointwise, $\|h_r-\ell_r\|_{L^2(\mathbb{P})}\leq u$, and every $f\in\mathcal{F}$ satisfies $\ell_r\leq f\leq h_r$ for some bracket in the family. We use the following fully stated external result, the localized bracketing maximal inequality for empirical processes, Theorem 2.14.2 in van der Vaart and Wellner, *Weak Convergence and Empirical Processes*. Let $\mathcal{F}$ be a pointwise measurable class of measurable functions $f: [0,1] \to \mathbb{R}$ with measurable envelope $F: [0,1] \to [0,\infty)$ satisfying $|f|\leq F\leq 1$ for every $f\in\mathcal{F}$. Define \begin{align*} \rho := \sup_{f\in\mathcal{F}}\|f\|_{L^2(\mathbb{P})}. \end{align*} Assume that $\rho\leq1$ and that the bracketing entropy satisfies $N_{[]}(u,\mathcal{F},L^2(\mathbb{P}))\leq A u^{-v}$ for $0<u\leq1$, with constants $A,v$ independent of $n$. Then there is a constant $K=K(A,v)>0$ such that \begin{align*} \limsup_{n\to\infty}\mathbb{E}\left[\sup_{f\in\mathcal{F}}\left|\frac{1}{\sqrt n}\sum_{k=1}^n(f(U_k)-\mathbb{E}[f(U_k)])\right|\right] \leq K\int_0^\rho \sqrt{1+\log N_{[]}(u,\mathcal{F},L^2(\mathbb{P}))}\,d\mathcal{L}^1(u). \end{align*} Apply this theorem with $\mathcal{F}_\delta := \{\mathbb{1}_I:I\in\mathcal{I}_\delta\}$. This class is pointwise measurable because intervals with rational endpoints form a countable subclass whose indicators approximate every $\mathbb{1}_{(a,b]}$ pointwise except at the two endpoints, a $\mathcal{L}^1$-null set. Each function in $\mathcal{F}_\delta$ is bounded by $1$, and its localized radius is \begin{align*} \rho_\delta:=\sup_{I\in\mathcal{I}_\delta}\|\mathbb{1}_I\|_{L^2(\mathbb{P})}\leq\sqrt{\delta}. \end{align*} Also the interval bracketing construction using grid points of mesh at most $u^2/2$ gives an absolute constant $C_0>0$ such that $N_{[]}(u,\mathcal{F}_\delta,L^2(\mathbb{P}))\leq C_0u^{-4}$ for $0<u\leq1$; the exponent $4$ comes from choosing both interval endpoints from a grid of order $u^{-2}$ points, and $C_0$ is independent of $n$ and $\delta$. Hence the maximal inequality gives a constant $C=C(C_0)>0$, independent of $n$ and $\delta$, such that \begin{align*} \limsup_{n\to\infty}\mathbb{E}\left[\sup_{I\in\mathcal{I}_\delta}|\nu_n(I)|\right] \leq C\int_0^{\sqrt{\delta}}\sqrt{1+\log(1/u)}\,d\mathcal{L}^1(u). \end{align*} The integral on the right tends to $0$ as $\delta \downarrow 0$. For the non-negative [random variable](/page/Random%20Variable) \begin{align*} X_{n,\delta}:=\sup_{I\in\mathcal{I}_\delta}|\nu_n(I)|, \end{align*} the estimate $\mathbb{P}(X_{n,\delta}>\varepsilon)\leq\varepsilon^{-1}\mathbb{E}[X_{n,\delta}]$ follows from $X_{n,\delta}\geq \varepsilon\mathbb{1}_{\{X_{n,\delta}>\varepsilon\}}$. Hence, for every $\varepsilon>0$, \begin{align*} \lim_{\delta\downarrow 0}\limsup_{n\to\infty}\mathbb{P}\left(\sup_{|s-t|\leq\delta}|\alpha_n(s)-\alpha_n(t)|>\varepsilon\right)=0. \end{align*} This is the required asymptotic equicontinuity in probability. Also, \begin{align*} \sup_{t\in[0,1]}|\alpha_n(t)-\alpha_n(t-)| \leq \frac{1}{\sqrt{n}}, \end{align*} where $\alpha_n(t-)$ denotes the left limit at $t$ and the bound follows because $G_n$ has jumps of size $1/n$. Hence the jumps vanish uniformly in probability, so the preceding equicontinuity condition is the correct large-$n$ tightness condition in $\ell^\infty([0,1])$. [guided] The issue is that controlling dyadic increments at a fixed scale does not automatically control every pair $s,t$ with $|s-t|\leq\delta$. We instead use the empirical-process maximal inequality for the whole class of short intervals at once. Define the empirical increment map $\Delta_n: \{(a,b) \in [0,1]^2 : a < b\} \to \mathbb{R}$ by \begin{align*} \Delta_n(a,b) = \alpha_n(b)-\alpha_n(a) = \frac{1}{\sqrt{n}}\sum_{k=1}^{n}\left(\mathbb{1}_{(a,b]}(U_k)-(b-a)\right). \end{align*} Let $\mathcal{I}_\delta$ be the set of all intervals $(a,b] \subset [0,1]$ with length at most $\delta$. Define the centred empirical measure map $\nu_n: \mathcal{I}_\delta \to \mathbb{R}$ by \begin{align*} \nu_n(I) = \frac{1}{\sqrt{n}}\sum_{k=1}^{n}\left(\mathbb{1}_I(U_k)-\mathbb{P}(U_k\in I)\right). \end{align*} Then $\Delta_n(a,b)=\nu_n((a,b])$, so the ordinary modulus of $\alpha_n$ over pairs with distance at most $\delta$ is bounded by the supremum of $|\nu_n(I)|$ over $I\in\mathcal{I}_\delta$. For $u>0$, the bracketing number $N_{[]}(u,\mathcal{F}_\delta,L^2(\mathbb{P}))$ means the least number of brackets $[\ell_r,h_r]$ needed to cover $\mathcal{F}_\delta$, with $\ell_r,h_r\in L^2(\mathbb{P})$, $\ell_r\leq h_r$ pointwise, and $\|h_r-\ell_r\|_{L^2(\mathbb{P})}\leq u$. We apply the localized bracketing maximal inequality for empirical processes, Theorem 2.14.2 in van der Vaart and Wellner, *Weak Convergence and Empirical Processes*, to the class $\mathcal{F}_\delta := \{\mathbb{1}_I:I\in\mathcal{I}_\delta\}$. In the form needed here, the inequality requires a pointwise measurable class of measurable functions, a measurable envelope bounded by $1$, the localized radius \begin{align*} \rho_\delta := \sup_{f\in\mathcal{F}_\delta}\|f\|_{L^2(\mathbb{P})}, \end{align*} and an entropy integral for the bracketing numbers $N_{[]}(u,\mathcal{F}_\delta,L^2(\mathbb{P}))$. The class is pointwise measurable because intervals with rational endpoints form a countable approximating subclass; endpoint discrepancies have probability zero since $U_1$ has distribution $\operatorname{Unif}(0,1)$. The constant function $F: [0,1] \to [0,\infty)$ defined by $F(x):=1$ is a measurable envelope for $\mathcal{F}_\delta$, and every function in $\mathcal{F}_\delta$ is bounded by $1$. The localized radius is \begin{align*} \rho_\delta = \sup_{I\in\mathcal{I}_\delta}\|\mathbb{1}_I\|_{L^2(\mathbb{P})} = \sup_{I\in\mathcal{I}_\delta}\mathbb{P}(U_1\in I)^{1/2} \leq \sqrt{\delta}. \end{align*} The entropy bound comes from a uniform grid. Choose grid points with mesh at most $u^2/2$. A bracket is obtained by moving the left and right endpoints of $(a,b]$ to neighbouring grid points; the symmetric difference between the lower and upper intervals has $\mathbb{P}$-measure at most $u^2$, so the $L^2(\mathbb{P})$ bracket width is at most $u$. Since there are order $u^{-2}$ choices for each endpoint, there is an absolute constant $C_0>0$ such that \begin{align*} N_{[]}(u,\mathcal{F}_\delta,L^2(\mathbb{P}))\leq C_0u^{-4}, \qquad 0<u\leq1. \end{align*} The constant $C_0$ is geometric and does not depend on $n$ or $\delta$. Therefore the maximal inequality gives a constant $C=C(C_0)>0$, again independent of $n$ and $\delta$, such that \begin{align*} \limsup_{n\to\infty}\mathbb{E}\left[\sup_{I\in\mathcal{I}_\delta}|\nu_n(I)|\right] \leq C\int_0^{\sqrt{\delta}}\sqrt{1+\log(1/u)}\,d\mathcal{L}^1(u). \end{align*} The integral tends to $0$ as $\delta\downarrow0$, because the integrand is integrable near $0$ after the substitution $u=e^{-v}$. Define the non-negative random variable \begin{align*} X_{n,\delta}:=\sup_{I\in\mathcal{I}_\delta}|\nu_n(I)|. \end{align*} The elementary estimate \begin{align*} \mathbb{P}(X_{n,\delta}>\varepsilon)\leq \frac{1}{\varepsilon}\mathbb{E}[X_{n,\delta}] \end{align*} follows because $X_{n,\delta}\geq\varepsilon\mathbb{1}_{\{X_{n,\delta}>\varepsilon\}}$ and expectation preserves order. Therefore, for every $\varepsilon>0$, \begin{align*} \lim_{\delta\downarrow 0}\limsup_{n\to\infty}\mathbb{P}\left(\sup_{|s-t|\leq\delta}|\alpha_n(s)-\alpha_n(t)|>\varepsilon\right)=0. \end{align*} The order of limits matters. We first let $n\to\infty$ and only then let $\delta\downarrow0$; this avoids the false assertion of uniform control over all $n$, which cannot hold because the empirical distribution function has jumps. Those jumps do vanish at the process scale, since each jump of $G_n$ has size $1/n$ and therefore each jump of $\alpha_n$ has size at most $1/\sqrt n$: \begin{align*} \sup_{t\in[0,1]}|\alpha_n(t)-\alpha_n(t-)| \leq \frac{1}{\sqrt{n}}. \end{align*} Thus the empirical processes are asymptotically equicontinuous in the large-$n$ sense needed for weak convergence in $\ell^\infty([0,1])$. [/guided] [/step] [step:Apply the weak convergence criterion in $\ell^\infty([0,1])$] The finite-dimensional distributions of $(\alpha_n)_{n\in\mathbb{N}}$ converge to those of $B$, as proved above. For a subset $A\subseteq\Omega$, define the outer probability $\mathbb{P}^*(A)$ by \begin{align*} \mathbb{P}^*(A):=\inf\{\mathbb{P}(E): A\subseteq E,\ E\in\mathcal{F}\}. \end{align*} The preceding step proves asymptotic equicontinuity in outer probability: \begin{align*} \lim_{\delta\downarrow 0}\limsup_{n\to\infty}\mathbb{P}^*\left(\sup_{|s-t|\leq\delta}|\alpha_n(s)-\alpha_n(t)|>\varepsilon\right)=0 \end{align*} for every $\varepsilon>0$. For fixed $n$ and $\delta$, the displayed supremum is measurable: since $G_n$ is right-continuous with left limits and has only the sample points $U_1,\dots,U_n$ as possible jump points, its modulus over $|s-t|\leq\delta$ is the same as the supremum over rational pairs after allowing distances $|s-t|<\delta+1/r$ and intersecting over $r\in\mathbb{N}$. Thus it is a countable combination of measurable random variables, so $\mathbb{P}^*$ agrees with $\mathbb{P}$ for this term. Define $C([0,1])$ to be the space of continuous maps $[0,1]\to\mathbb{R}$, equipped with the sup norm inherited from $\ell^\infty([0,1])$. The Brownian bridge $B$ has a continuous version, so its law is supported on the separable subspace $C([0,1])\subset\ell^\infty([0,1])$. Therefore we apply the following fully stated external weak convergence criterion in $\ell^\infty$, the asymptotic equicontinuity criterion of Theorems 1.5.4 and 1.5.7 in van der Vaart and Wellner, *Weak Convergence and Empirical Processes*. If $(X_n)_{n\in\mathbb{N}}$ is an asymptotically measurable sequence of $\ell^\infty(T)$-valued indexed processes, if the finite-dimensional distributions of $X_n$ converge on a [dense subset](/page/Dense%20Subset) of a semimetric space $(T,d)$ to those of a tight Borel process $X$ whose sample paths are uniformly $d$-continuous and whose law is supported on a separable subset of $\ell^\infty(T)$, and if \begin{align*} \lim_{\delta\downarrow0}\limsup_{n\to\infty}\mathbb{P}^*\left(\sup_{d(s,t)\leq\delta}|X_n(s)-X_n(t)|>\varepsilon\right)=0 \end{align*} for every $\varepsilon>0$, then $X_n\xrightarrow{d}X$ in $\ell^\infty(T)$. We verify these hypotheses with $T=[0,1]$, $d(s,t):=|s-t|$, $X_n:=\alpha_n$, and $X:=B$. The finite-dimensional convergence was proved above, and the dense subset may be all of $[0,1]$. The asymptotic equicontinuity hypothesis is exactly the displayed modulus estimate, together with the fact that the jumps of $\alpha_n$ vanish uniformly because each jump of $G_n$ has size $1/n$. The asymptotic measurability hypothesis is also part of the cited criterion: a pointwise measurable indexed empirical process is asymptotically measurable as an $\ell^\infty$-valued random element, meaning that for every bounded continuous functional $H:\ell^\infty([0,1])\to\mathbb{R}$ the difference between the outer and inner expectations of $H(\alpha_n)$ tends to $0$. This applies because the class $\{\mathbb{1}_{[0,t]}:t\in[0,1]\}$ is pointwise measurable by rational endpoint approximation and $\mathbb{P}(U_1=t)=0$ for every $t\in[0,1]$. Finally, the law of $B$ is supported on $C([0,1])$, which is separable under the sup norm, and its sample paths are uniformly continuous on the compact interval $[0,1]$. Hence \begin{align*} \alpha_n \xrightarrow{d} B \end{align*} in $\ell^\infty([0,1])$, as required. [/step]