[proofplan] The $P$-Donsker hypothesis means that the empirical process indexed by $\mathcal F$ converges in $\ell^\infty(\mathcal F)$ to a tight Brownian bridge $B_P$. A standard asymptotic equicontinuity characterization of Donsker classes then forces the index space, after quotienting by the zero-distance relation of the canonical semimetric, to be [totally bounded](/page/Totally%20Bounded). The covariance formula for the Brownian bridge identifies $d_P$ with the canonical Gaussian semimetric of the limit. Finally, the standard [regularity theorem](/theorems/2750) for tight Gaussian processes gives a separable uniformly continuous modification, and [uniform continuity](/page/Uniform%20Continuity) extends each sample path uniquely to the metric completion. [/proofplan]

[step:Use the Donsker hypothesis to obtain a tight Brownian bridge limit] The [probability space](/page/Probability%20Space) $(\Omega_X,\mathcal A_X,\mathbb P_X)$, the independent identically distributed sequence $X_1,X_2,\dots$ with law $P$, the empirical means $P_nh$, the empirical processes $G_n$, and the Brownian bridge $B_P$ on $(\Omega_B,\mathcal A_B,\mathbb P_B)$ are part of the theorem statement. By the definition of the $P$-Donsker assumption used there, the random elements $G_n$ converge weakly in $\ell^\infty(\mathcal F)$ to the tight Borel measurable centred Gaussian process $B_P$. Its covariance is the Brownian bridge covariance \begin{align*} \mathbb E_B[B_P(f)B_P(g)]=P(fg)-Pf\,Pg \end{align*} for all $f,g\in\mathcal F$, where $\mathbb E_B$ denotes expectation with respect to $\mathbb P_B$. The uniform boundedness assumption ensures that $f,g,(f-g)^2$, and $fg$ are $P$-integrable, so all displayed quantities are finite. [/step]

[step:Identify the canonical Gaussian semimetric with $d_P$]Define the Gaussian canonical semimetric $\rho:\mathcal F\times\mathcal F\to[0,\infty)$ of $B_P$ by \begin{align*} \rho(f,g)^2:=\mathbb E_B[(B_P(f)-B_P(g))^2]. \end{align*} Using centredness and the covariance formula, \begin{align*} \rho(f,g)^2=\mathbb E_B[B_P(f)^2]-2\mathbb E_B[B_P(f)B_P(g)]+\mathbb E_B[B_P(g)^2]. \end{align*} Substituting the Brownian bridge covariance gives \begin{align*} \rho(f,g)^2=P(f^2)-(Pf)^2-2(P(fg)-Pf\,Pg)+P(g^2)-(Pg)^2. \end{align*} Collecting terms, \begin{align*} \rho(f,g)^2=P(f-g)^2-(Pf-Pg)^2=d_P(f,g)^2. \end{align*} Thus $\rho=d_P$ on $\mathcal F$.[/step]

[guided]We need to connect the metric appearing in the statement with the metric naturally attached to the Gaussian limit. Define the Gaussian canonical semimetric $\rho:\mathcal F\times\mathcal F\to[0,\infty)$ by \begin{align*} \rho(f,g)^2:=\mathbb E_B[(B_P(f)-B_P(g))^2]. \end{align*} This is the correct semimetric for sample-path regularity of a Gaussian process: two indices are close when the corresponding Gaussian coordinates have small $L^2(\mathbb P)$ distance. Because $B_P$ is centred, expanding the square gives \begin{align*} \rho(f,g)^2=\mathbb E_B[B_P(f)^2]-2\mathbb E_B[B_P(f)B_P(g)]+\mathbb E_B[B_P(g)^2]. \end{align*} The Brownian bridge covariance formula says that, for every $u,v\in\mathcal F$, \begin{align*} \mathbb E_B[B_P(u)B_P(v)]=P(uv)-Pu\,Pv. \end{align*} Applying this formula with $(u,v)=(f,f)$, $(f,g)$, and $(g,g)$ yields \begin{align*} \rho(f,g)^2=P(f^2)-(Pf)^2-2(P(fg)-Pf\,Pg)+P(g^2)-(Pg)^2. \end{align*} The right-hand side is exactly \begin{align*} P(f-g)^2-(Pf-Pg)^2. \end{align*} Therefore \begin{align*} \rho(f,g)=d_P(f,g) \end{align*} for all $f,g\in\mathcal F$. This calculation is the reason the quotient in the theorem is taken with respect to $d_P$: it is precisely the zero-variance relation for the limiting Gaussian process.[/guided]

[step:Apply the Gaussian regularity hypothesis for the tight Brownian bridge]We apply the Gaussian regularity hypothesis stated in the theorem to the tight Borel measurable centred Gaussian process $B_P$ indexed by $\mathcal F$. The preceding step identifies the canonical Gaussian semimetric of $B_P$ with $d_P$. Therefore the quotient of $\mathcal F$ by the relation $d_P(f,g)=0$ is totally bounded under the induced metric, and there is a separable version of the Brownian bridge, still denoted $B_P$, whose sample paths factor through this quotient and are uniformly continuous on it. Equivalently, if $q:\mathcal F\to\mathcal F_0$ denotes the quotient map, there exist a process \begin{align*} \widetilde B_P:\Omega_B\times\mathcal F_0\to\mathbb R \end{align*} and an event $N\in\mathcal A_B$ with $\mathbb P_B(N)=0$ such that $\widetilde B_P(\omega,q(f))=B_P(\omega,f)$ for every $\omega\in\Omega_B\setminus N$ and every $f\in\mathcal F$, and such that, for every $\omega\in\Omega_B\setminus N$, \begin{align*} \sup\{|\widetilde B_P(\omega,a)-\widetilde B_P(\omega,b)|:a,b\in\mathcal F_0,\ d_P(a,b)<\delta\}\to 0 \end{align*} as $\delta\downarrow 0$.[/step]

[guided]The decisive point is that the regularity conclusion is being applied to the Gaussian limit, not to the raw function class with the uncentred $L^2(P)$ distance. The theorem statement includes the exact Gaussian regularity hypothesis needed here: for a tight Borel Gaussian element of $\ell^\infty(\mathcal F)$, its canonical Gaussian semimetric has a totally bounded zero-distance quotient, and the process has a separable version whose sample paths factor through that quotient and are uniformly continuous for the induced metric. We verify the only identification needed to use that hypothesis in the present notation. The previous step defined the canonical Gaussian semimetric $\rho$ of $B_P$ by \begin{align*} \rho(f,g)^2:=\mathbb E_B[(B_P(f)-B_P(g))^2] \end{align*} for $f,g\in\mathcal F$, and proved that $\rho=d_P$ on $\mathcal F$. Thus the zero-distance quotient in the Gaussian regularity hypothesis is exactly the quotient $\mathcal F_0$ appearing in the theorem statement. The conclusion of the hypothesis therefore gives [total boundedness](/page/Total%20Boundedness) of $\mathcal F_0$ and gives a separable version of $B_P$ whose paths are well-defined on equivalence classes. This last phrase is important: it avoids taking an uncountable union of null events over all pairs $f,g$ with $d_P(f,g)=0$. Instead, the factorisation through $q:\mathcal F\to\mathcal F_0$ is part of the regularity conclusion. In explicit event notation, there are a process $\widetilde B_P:\Omega_B\times\mathcal F_0\to\mathbb R$ and a null event $N\in\mathcal A_B$ such that $\widetilde B_P(\omega,q(f))=B_P(\omega,f)$ for every $\omega\in\Omega_B\setminus N$ and every $f\in\mathcal F$, and such that, for every $\omega\in\Omega_B\setminus N$, \begin{align*} \sup\{|\widetilde B_P(\omega,a)-\widetilde B_P(\omega,b)|:a,b\in\mathcal F_0,\ d_P(a,b)<\delta\}\to 0 \end{align*} as $\delta\downarrow 0$.[/guided]

[step:Pass from the quotient to the metric completion] Let $q:\mathcal F\to\mathcal F_0$ be the quotient map. From the previous step, we already have a version of the Brownian bridge that factors through $q$ on a probability-one event; denote the induced process by \begin{align*} \widetilde B_P:\Omega_B\times\mathcal F_0\to\mathbb R. \end{align*} On the event $\Omega_B\setminus N$, the map $\widetilde B_P(\omega,\cdot):\mathcal F_0\to\mathbb R$ is uniformly continuous. Therefore, for every $\omega\in\Omega_B\setminus N$, it has a unique uniformly continuous extension \begin{align*} \overline B_P(\omega):\overline{\mathcal F_0}\to\mathbb R. \end{align*} For $\omega\in N$, define $\overline B_P(\omega,a):=0$ for every $a\in\overline{\mathcal F_0}$. For each $a\in\overline{\mathcal F_0}$, choose a sequence $(a_k)_{k\ge 1}$ in $\mathcal F_0$ with $a_k\to a$ in the completed metric. Then \begin{align*} \overline B_P(\omega,a)=\lim_{k\to\infty}\widetilde B_P(\omega,a_k) \end{align*} for every $\omega\in\Omega_B\setminus N$. Since each coordinate map $\omega\mapsto\widetilde B_P(\omega,a_k)$ is measurable and $N\in\mathcal A_B$, the coordinate map $\omega\mapsto\overline B_P(\omega,a)$ is measurable as a pointwise limit after the harmless definition on $N$. Since $\mathcal F_0$ is totally bounded, its completion $\overline{\mathcal F_0}$ is also totally bounded. [/step]

[step:Record separability and identify the extension with the original Brownian bridge] The separability of the version on $\mathcal F_0$ is part of the Gaussian regularity hypothesis applied above. Since $\mathcal F_0$ is totally bounded, choose a countable [dense subset](/page/Dense%20Subset) $D\subset\mathcal F_0$. The extension $\overline B_P(\omega,\cdot)$ is continuous on $\overline{\mathcal F_0}$ for every $\omega\in\Omega_B\setminus N$, so the same [countable set](/page/Countable%20Set) $D$, viewed inside $\overline{\mathcal F_0}$, is a separating set for the extended paths. Thus the extended process is separable on the completed index space. The process $\overline B_P$ constructed above is a version of the original Brownian bridge on $\mathcal F$ because, for every finite collection $f_1,\dots,f_m\in\mathcal F$, the random vector $(\overline B_P(q(f_1)),\dots,\overline B_P(q(f_m)))$ agrees almost surely with $(B_P(f_1),\dots,B_P(f_m))$. Thus it has the same finite-dimensional distributions on the dense image $q(\mathcal F)\subset\overline{\mathcal F_0}$ and is a modification of the Brownian bridge indexed by the original class. Its sample paths are uniformly continuous for the completed canonical metric. Hence the effective completed index space is totally bounded and the limiting Brownian bridge has the asserted separable uniformly continuous version. [/step]