TITLE: Exact definition of weak convergence of empirical processes
QUESTION [2 upvotes]: Consider the i.i.d. random variables $X_1,...,X_n$ with $X_i:\Omega \rightarrow \mathcal{X}$ $\forall i$ and probability distribution $P$. 
Consider a class $\mathcal{F}$ of functions $f:\mathcal{X}\rightarrow \mathbb{R}$. 
Define the empirical process $G_n:=\{\sqrt{n}(\frac{1}{n}\sum_{i=1}^nf(X_i)-\int_{\mathcal{X}}f(x)dP(x)) \text{ s.t. } f\in \mathcal{F}\}$. 
A sample path for a fixed $\omega \in \Omega$ is a map $f\in \mathcal{F}\mapsto \sqrt{n}(\frac{1}{n}\sum_{i=1}^nf(X_i(\omega))-\int_{\mathcal{X}}f(x)dP(x))\in \mathbb{R}$. Let $\mathcal{A}$ be the collection of all sample paths.  
$G_n$ can be also seen as a random map $ \Omega\rightarrow\mathcal{A}$.
Let $l^{\infty}(\mathcal{F})$ be the set of functions $z:\mathcal{F}\rightarrow \mathbb{R}$ such that $\sup_{f\in \mathcal{F}}|z(f)|<\infty$. 
Suppose I'm told that $G_n$ weakly converges to a stochastic process $G_p$ "in $l^\infty(\mathcal{F})$" (e.g. here at the top of page 3). What's the exact definition of weak convergence of the empirical process?
My attempt: $G_n$ converges in distribution to a stochastic process $G_p$ in $l^\infty(\mathcal{F})$ means that $\lim_{n\rightarrow\infty} E^*_{X_1,...,X_n}h(G_n)=E_{X_1,...,X_n}h(G_p)$ $\forall$ functions $h: l^{\infty}(\mathcal{F})\rightarrow \mathbb{R}$ continuous and bounded. 
But, if this interpretation is correct, do we need $l^{\infty}(\mathcal{F}) \in \mathcal{A}$? And, similarly, if $G_p$ is a map from $\Omega$  $\mathcal{B}$ do we need $l^{\infty}(\mathcal{F}) \in \mathcal{B}$?

REPLY [2 votes]: Your interpretation is not quite right. See the discussion at the top of page 5 of your (Wellner) reference.
Weak convergence in $\ell^\infty(\cal F)$ means
$$
\lim_{n\to\infty} E^*h(G_n)= Eh(G_p)
$$ 
for all continuous bounded $h$ that map $\ell^\infty(\cal F)$ to $\mathbb R$. The expectation on the LHS is indeed with respect to the joint distribution of $X_1\ldots X_n$, since that's all that $G_n$ is based on. But the rhs cannot depend on $n$. 
As for the connection between $\cal A$ and $l^\infty(F)$, typically we expect the sample paths for $G_n$ and $G_p$ to be bounded almost surely, which means we can assume $\cal A$ and $\cal B$ are both subsets of $l^\infty(F)$. If you stare at the definitions you'll realize that a statement like $l^{\infty}(\mathcal{F}) \in \mathcal{A}$ can't be right.
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