[proofplan] We prove the result by viewing a sum-indexed empirical process as the image of the empirical process on $\mathcal F\cup\mathcal G$ under two continuous operations: restriction to the two sub-classes and addition of the two restricted coordinates. The continuous mapping principle transfers the Donsker convergence through these maps. The only point requiring care is that a function $h\in\mathcal H$ can have several decompositions $h=f+g$; the Brownian bridge covariance semimetric shows that the limiting Gaussian process is constant on the quotient classes whenever two decompositions agree $P$-a.s. The assumed measurable separable quotient version then gives exactly the Donsker convergence for $\mathcal H$. [/proofplan]

[step:Restrict the Donsker convergence from $\mathcal F\cup\mathcal G$ to the two summand classes]Let \begin{align*} \mathcal E:=\mathcal F\cup\mathcal G. \end{align*} For $n\in\mathbb N$, let $\alpha_{n,\mathcal E}:\Omega\to\ell^\infty(\mathcal E)$ denote the empirical process indexed by $\mathcal E$, defined by \begin{align*} \alpha_{n,\mathcal E}(h):=\sqrt n\{P_nh-Ph\} \end{align*} for $h\in\mathcal E$, where the sample maps $X_i:(\Omega,\mathcal A)\to(S,\mathcal S)$ are those in the theorem statement, \begin{align*} P_nh:=\frac{1}{n}\sum_{i=1}^n h(X_i) \end{align*} and \begin{align*} Ph:=\int_S h(x)\,dP(x). \end{align*} Since $\mathcal E$ is $P$-Donsker, there is a tight $P$-Brownian bridge $G_{P,\mathcal E}$ in $\ell^\infty(\mathcal E)$ such that \begin{align*} \alpha_{n,\mathcal E}\xrightarrow{d}G_{P,\mathcal E} \end{align*} in $\ell^\infty(\mathcal E)$. Define the restriction map \begin{align*} R:\ell^\infty(\mathcal E)&\to\ell^\infty(\mathcal F)\times\ell^\infty(\mathcal G) \end{align*} by \begin{align*} R(z):=(z|_{\mathcal F},z|_{\mathcal G}) \end{align*} for $z\in\ell^\infty(\mathcal E)$, where the product space is equipped with the norm \begin{align*} \|(u,v)\|_{\times}:=\max\{\|u\|_{\ell^\infty(\mathcal F)},\|v\|_{\ell^\infty(\mathcal G)}\}. \end{align*} For $z,w\in\ell^\infty(\mathcal E)$, \begin{align*} \|R(z)-R(w)\|_{\times}\le \|z-w\|_{\ell^\infty(\mathcal E)}. \end{align*} Thus $R$ is Lipschitz, hence continuous. The theorem statement assumes the $\mathcal E$-indexed empirical process is taken in a measurable separable version, so the random elements to which the continuous mapping principle is applied are measurable in the required version. By the [citetheorem:9834] applied to the continuous map $R$, \begin{align*} R(\alpha_{n,\mathcal E})\xrightarrow{d}R(G_{P,\mathcal E}) \end{align*} in $\ell^\infty(\mathcal F)\times\ell^\infty(\mathcal G)$.[/step]

[guided]The reason for passing through the union class is that the Donsker assumption gives convergence for one process indexed by every element of $\mathcal F\cup\mathcal G$. We first extract from that process its two coordinate restrictions. Define \begin{align*} \mathcal E:=\mathcal F\cup\mathcal G. \end{align*} For every $n\in\mathbb N$, define the empirical process \begin{align*} \alpha_{n,\mathcal E}:\Omega&\to\ell^\infty(\mathcal E) \end{align*} by \begin{align*} \alpha_{n,\mathcal E}(h):=\sqrt n\{P_nh-Ph\} \end{align*} for $h\in\mathcal E$, where the sample maps $X_i:(\Omega,\mathcal A)\to(S,\mathcal S)$ are the independent identically distributed maps from the theorem statement, \begin{align*} P_nh:=\frac{1}{n}\sum_{i=1}^n h(X_i) \end{align*} and \begin{align*} Ph:=\int_S h(x)\,dP(x). \end{align*} The hypothesis that $\mathcal E$ is $P$-Donsker means precisely that, for a measurable separable version, there is a tight $P$-Brownian bridge $G_{P,\mathcal E}$ in $\ell^\infty(\mathcal E)$ such that \begin{align*} \alpha_{n,\mathcal E}\xrightarrow{d}G_{P,\mathcal E} \end{align*} in $\ell^\infty(\mathcal E)$. Now define \begin{align*} R:\ell^\infty(\mathcal E)&\to\ell^\infty(\mathcal F)\times\ell^\infty(\mathcal G) \end{align*} by \begin{align*} R(z):=(z|_{\mathcal F},z|_{\mathcal G}). \end{align*} We equip the product with the norm \begin{align*} \|(u,v)\|_{\times}:=\max\{\|u\|_{\ell^\infty(\mathcal F)},\|v\|_{\ell^\infty(\mathcal G)}\}. \end{align*} For $z,w\in\ell^\infty(\mathcal E)$, the two restrictions cannot increase the supremum norm, so \begin{align*} \|R(z)-R(w)\|_{\times}\le \|z-w\|_{\ell^\infty(\mathcal E)}. \end{align*} Thus $R$ is a continuous map. The continuous mapping principle [citetheorem:9834] therefore transfers the convergence of $\alpha_{n,\mathcal E}$ to the joint convergence of the two restricted empirical processes: \begin{align*} R(\alpha_{n,\mathcal E})\xrightarrow{d}R(G_{P,\mathcal E}) \end{align*} in $\ell^\infty(\mathcal F)\times\ell^\infty(\mathcal G)$.[/guided]

[step:Add the two restricted coordinates by a continuous linear map] Define \begin{align*} A:\ell^\infty(\mathcal F)\times\ell^\infty(\mathcal G)&\to\ell^\infty(\mathcal F\times\mathcal G) \end{align*} by \begin{align*} A(u,v)(f,g):=u(f)+v(g) \end{align*} for $(u,v)\in\ell^\infty(\mathcal F)\times\ell^\infty(\mathcal G)$ and $(f,g)\in\mathcal F\times\mathcal G$. This map is well defined because \begin{align*} \|A(u,v)\|_{\ell^\infty(\mathcal F\times\mathcal G)}\le \|u\|_{\ell^\infty(\mathcal F)}+\|v\|_{\ell^\infty(\mathcal G)}\le 2\|(u,v)\|_{\times}. \end{align*} Therefore $A$ is continuous. Applying the continuous mapping principle to $A$ gives \begin{align*} A(R(\alpha_{n,\mathcal E}))\xrightarrow{d}A(R(G_{P,\mathcal E})) \end{align*} in $\ell^\infty(\mathcal F\times\mathcal G)$. For $(f,g)\in\mathcal F\times\mathcal G$, \begin{align*} A(R(\alpha_{n,\mathcal E}))(f,g)=\alpha_{n,\mathcal E}(f)+\alpha_{n,\mathcal E}(g). \end{align*} By linearity of $P_n$ and $P$, \begin{align*} \alpha_{n,\mathcal E}(f)+\alpha_{n,\mathcal E}(g)=\sqrt n\{P_n(f+g)-P(f+g)\}. \end{align*} Thus $A(R(\alpha_{n,\mathcal E}))$ is the empirical process indexed by decompositions $(f,g)$ of functions in $\mathcal H$. [/step]

[step:Show that the Gaussian limit is constant on quotient classes]Define \begin{align*} Z:\mathcal F\times\mathcal G&\to\mathbb R \end{align*} by \begin{align*} Z(f,g):=G_{P,\mathcal E}(f)+G_{P,\mathcal E}(g). \end{align*} Then $Z=A(R(G_{P,\mathcal E}))$. Suppose $(f,g),(f',g')\in\mathcal F\times\mathcal G$ satisfy \begin{align*} f+g=f'+g'\quad P\text{-a.s.} \end{align*} Define \begin{align*} q:S&\to\mathbb R \end{align*} by \begin{align*} q:=f+g-f'-g'. \end{align*} Then $q=0$ $P$-a.s. Since $\mathcal E$ is $P$-Donsker, every index in $\mathcal E$ has finite second moment under $P$; hence $f,g,f',g'\in L^2(P)$ and $q\in L^2(P)$. Since the covariance semimetric of the $P$-Brownian bridge is \begin{align*} \rho_P(a,b)^2:=P(a-b)^2-\{P(a-b)\}^2 \end{align*} for square-integrable indices $a,b$, the variance of the Gaussian [random variable](/page/Random%20Variable) \begin{align*} Z(f,g)-Z(f',g') \end{align*} is \begin{align*} Pq^2-(Pq)^2. \end{align*} Because $q=0$ $P$-a.s., \begin{align*} Pq^2=\int_S q(x)^2\,dP(x)=0 \end{align*} and \begin{align*} Pq=\int_S q(x)\,dP(x)=0. \end{align*} Hence \begin{align*} \operatorname{Var}(Z(f,g)-Z(f',g'))=0. \end{align*} A centred Gaussian random variable with variance $0$ is equal to $0$ almost surely, so \begin{align*} Z(f,g)=Z(f',g')\quad\text{a.s.} \end{align*} Thus the limiting Gaussian process is compatible with the quotient [equivalence relation](/page/Equivalence%20Relation) in the canonical semimetric: equivalent decompositions have distance $0$ for the centred Gaussian increment $Z(f,g)-Z(f',g')$. By the theorem statement, the Brownian-bridge limit on $Q$ is taken in the corresponding measurable separable quotient version. Denote this quotient-indexed version by \begin{align*} \beta:Q\to\mathbb R. \end{align*} It is characterized by the pullback identity \begin{align*} \beta(\pi(f,g))=Z(f,g) \end{align*} for the representatives in the separating version, and separability with respect to $d_Q$ makes this identity determine the process on all quotient indices. Therefore the limiting Gaussian process descends to the quotient $(\mathcal F\times\mathcal G)/\sim$ as a single quotient-indexed separable modification.[/step]

[guided]The delicate point is that equality for one fixed pair of equivalent decompositions is not by itself the same as having one process that is constant on all equivalence classes. We first prove the fixed-pair assertion with all integrability hypotheses visible. Let $(f,g),(f',g')\in\mathcal F\times\mathcal G$ satisfy \begin{align*} f+g=f'+g'\quad P\text{-a.s.} \end{align*} and define the [measurable function](/page/Measurable%20Function) \begin{align*} q:S&\to\mathbb R \end{align*} by \begin{align*} q:=f+g-f'-g'. \end{align*} The $P$-Donsker hypothesis on $\mathcal E=\mathcal F\cup\mathcal G$ implies that each index in $\mathcal E$ has finite second moment under $P$. Therefore $f,g,f',g'\in L^2(P)$, and the linear combination $q$ also belongs to $L^2(P)$. The covariance formula for the $P$-Brownian bridge gives the variance of \begin{align*} Z(f,g)-Z(f',g') \end{align*} as \begin{align*} Pq^2-(Pq)^2. \end{align*} Since $q=0$ $P$-a.s., both terms vanish: \begin{align*} Pq^2=\int_S q(x)^2\,dP(x)=0 \end{align*} and \begin{align*} Pq=\int_S q(x)\,dP(x)=0. \end{align*} Thus $Z(f,g)-Z(f',g')$ is a centred Gaussian random variable with variance $0$, so it is equal to $0$ almost surely. Now we turn the fixed-pair conclusion into a quotient-indexed process. The repaired theorem statement specifies the quotient semimetric \begin{align*} d_Q(\pi(f,g),\pi(f',g')):=\left(P\{(f+g)-(f'+g')\}^2-\left(P\{(f+g)-(f'+g')\}\right)^2\right)^{1/2}. \end{align*} The calculation above says exactly that $d_Q(\pi(f,g),\pi(f',g'))=0$ whenever two decompositions represent the same quotient index. The Brownian-bridge limit is therefore taken in the measurable separable quotient version specified in the statement. Writing this quotient-indexed version as $\beta:Q\to\mathbb R$, its pullback satisfies \begin{align*} \beta(\pi(f,g))=Z(f,g) \end{align*} on the separating version, and separability with respect to $d_Q$ determines the same identity on all quotient indices. Hence the Gaussian limit is not merely pointwise compatible with the quotient; it is represented by a single separable quotient-indexed modification.[/guided]

[step:Pass from decompositions to the quotient-indexed sum class]Let \begin{align*} \pi:\mathcal F\times\mathcal G&\to(\mathcal F\times\mathcal G)/\sim \end{align*} be the quotient map, and write $Q:=(\mathcal F\times\mathcal G)/\sim$. Define the isometric pullback map \begin{align*} J:\ell^\infty(Q)&\to\ell^\infty(\mathcal F\times\mathcal G) \end{align*} by \begin{align*} Jb:=b\circ\pi \end{align*} for $b\in\ell^\infty(Q)$. Its range is precisely the closed subspace of bounded functions on $\mathcal F\times\mathcal G$ that are constant on $\sim$-equivalence classes. Let \begin{align*} \beta_n:Q&\to\mathbb R \end{align*} be the quotient-indexed empirical process defined by \begin{align*} \beta_n(\pi(f,g)):=\sqrt n\{P_n(f+g)-P(f+g)\}. \end{align*} This is well defined under the stated quotient representation assumption, and the decomposition-indexed process satisfies \begin{align*} J\beta_n=A(R(\alpha_{n,\mathcal E})). \end{align*} The preceding steps give \begin{align*} J\beta_n\xrightarrow{d}J\beta \end{align*} in the closed subspace $J\ell^\infty(Q)\subset\ell^\infty(\mathcal F\times\mathcal G)$, where $\beta:Q\to\mathbb R$ is the quotient-indexed separable Brownian-bridge version specified in the theorem statement and identified above by $J\beta=A(R(G_{P,\mathcal E}))$. Since $J$ is an isometry from $\ell^\infty(Q)$ onto this closed subspace, convergence of $J\beta_n$ to $J\beta$ is equivalent to convergence of $\beta_n$ to $\beta$ in $\ell^\infty(Q)$. Under the identification of $Q$ with the $P$-a.s. equivalence classes of sums $f+g\in\mathcal H$, the covariance of $\beta$ is \begin{align*} \operatorname{Cov}(\beta(\pi(f,g)),\beta(\pi(f',g')))=P\{(f+g)(f'+g')\}-P(f+g)P(f'+g'), \end{align*} which is exactly the $P$-Brownian bridge covariance for the sum functions in $\mathcal H$. Therefore the empirical process indexed by $\mathcal H$ converges in distribution to its $P$-Brownian bridge in the assumed measurable separable quotient representation. Hence $\mathcal H$ is $P$-Donsker.[/step]

[guided]The decomposition-indexed convergence lives first in $\ell^\infty(\mathcal F\times\mathcal G)$, but the theorem is about the quotient index set $Q=(\mathcal F\times\mathcal G)/\sim$. To pass between them, define \begin{align*} J:\ell^\infty(Q)&\to\ell^\infty(\mathcal F\times\mathcal G) \end{align*} by \begin{align*} Jb:=b\circ\pi. \end{align*} For every $b\in\ell^\infty(Q)$, \begin{align*} \|Jb\|_{\ell^\infty(\mathcal F\times\mathcal G)}=\|b\|_{\ell^\infty(Q)}, \end{align*} so $J$ is an isometry. Its image consists exactly of those bounded functions on $\mathcal F\times\mathcal G$ that are constant on equivalence classes, because such a function has a unique value on each quotient class. The quotient-indexed empirical process is \begin{align*} \beta_n:Q&\to\mathbb R \end{align*} with \begin{align*} \beta_n(\pi(f,g)):=\sqrt n\{P_n(f+g)-P(f+g)\}. \end{align*} The statement assumes this is the measurable separable quotient representation. Pulling it back gives \begin{align*} J\beta_n(f,g)=\sqrt n\{P_n(f+g)-P(f+g)\}=A(R(\alpha_{n,\mathcal E}))(f,g). \end{align*} Thus the process obtained by continuous mapping is exactly the pullback of the quotient process, not merely a related process. The previous step constructed a separable quotient-indexed Gaussian process $\beta$ satisfying \begin{align*} \beta(\pi(f,g))=Z(f,g). \end{align*} Consequently \begin{align*} J\beta=A(R(G_{P,\mathcal E})). \end{align*} Since $J$ is an isometry onto its closed range, convergence of $J\beta_n$ to $J\beta$ in the range is the same as convergence of $\beta_n$ to $\beta$ in $\ell^\infty(Q)$. Finally, identifying $Q$ with the $P$-a.s. equivalence classes of sums in $\mathcal H$, the covariance of $\beta$ is the Brownian bridge covariance of those sum functions. This is precisely the definition of the $P$-Donsker property for $\mathcal H$ in the assumed measurable separable version.[/guided]