[proofplan]
We first record the transformed class and verify that it is a class of [measurable functions](/page/Measurable%20Functions) with a square-integrable envelope. When $\phi(0)=0$, the envelope is obtained directly from the Lipschitz inequality; in the general case it is part of the hypotheses. The Donsker conclusion is then exactly the standard Lipschitz permanence principle for Donsker classes, applied after checking its measurability and envelope assumptions.
[/proofplan]
[step:Declare the transformed class and its Lipschitz control]
Let $L\in[0,\infty)$ be a Lipschitz constant for $\phi:\mathbb R\to\mathbb R$, so that for all $u,v\in\mathbb R$,
\begin{align*}
|\phi(u)-\phi(v)|\le L|u-v|.
\end{align*}
For each $f\in\mathcal F$, the composition
\begin{align*}
\phi\circ f:S&\to\mathbb R
\end{align*}
is measurable because $f:(S,\mathcal S)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable and every Lipschitz real-valued function is Borel measurable. Thus $\phi\circ\mathcal F$ is a class of measurable real-valued functions on $S$.
For all $f,g\in\mathcal F$ and $P$-a.e. $x\in S$, the Lipschitz bound gives
\begin{align*}
|(\phi\circ f)(x)-(\phi\circ g)(x)|\le L|f(x)-g(x)|.
\end{align*}
Consequently, whenever $f-g\in L^2(P)$,
\begin{align*}
\|\phi\circ f-\phi\circ g\|_{L^2(P)}\le L\|f-g\|_{L^2(P)}.
\end{align*}
[/step]
[step:Verify the square-integrable envelope condition for the transformed class]
If $\phi(0)=0$, then for every $f\in\mathcal F$ and $P$-a.e. $x\in S$,
\begin{align*}
|(\phi\circ f)(x)|=|\phi(f(x))-\phi(0)|\le L|f(x)|\le L F(x).
\end{align*}
Define the measurable map
\begin{align*}
G_0:S&\to[0,\infty]
\end{align*}
\begin{align*}
x&\mapsto L F(x).
\end{align*}
Then $G_0$ is an envelope for $\phi\circ\mathcal F$, and
\begin{align*}
\int_S G_0(x)^2\,dP(x)=L^2\int_S F(x)^2\,dP(x)<\infty.
\end{align*}
In the alternative case, the existence of a measurable envelope
\begin{align*}
G:S\to[0,\infty]
\end{align*}
for $\phi\circ\mathcal F$ satisfying
\begin{align*}
\int_S G(x)^2\,dP(x)<\infty
\end{align*}
is assumed directly.
[guided]
The transformed class needs a square-integrable envelope because the Donsker permanence theorem is a statement about empirical processes indexed by $L^2(P)$-controlled functions. In the special case $\phi(0)=0$, no additional envelope assumption is needed. Indeed, if $L$ is a Lipschitz constant for $\phi$, then for every $f\in\mathcal F$ and $P$-a.e. $x\in S$,
\begin{align*}
|(\phi\circ f)(x)|=|\phi(f(x))-\phi(0)|\le L|f(x)|.
\end{align*}
Since $F$ is an envelope for $\mathcal F$, we also have $|f(x)|\le F(x)$ for every $f\in\mathcal F$ and $P$-a.e. $x\in S$. Combining the two inequalities gives
\begin{align*}
|(\phi\circ f)(x)|\le L F(x).
\end{align*}
Thus the map $G_0:S\to[0,\infty]$ defined by $G_0(x)=L F(x)$ is an envelope for $\phi\circ\mathcal F$. It is square-integrable because
\begin{align*}
\int_S G_0(x)^2\,dP(x)=L^2\int_S F(x)^2\,dP(x)<\infty.
\end{align*}
If $\phi(0)$ is not assumed to be $0$, this direct envelope may be replaced by the stated hypothesis that $\phi\circ\mathcal F$ has some measurable envelope $G:S\to[0,\infty]$ with $G\in L^2(P)$.
[/guided]
[/step]
[step:Apply Lipschitz permanence for Donsker classes]
We now invoke the Lipschitz permanence theorem for Donsker classes: if a class $\mathcal F$ is $P$-Donsker, $\phi:\mathbb R\to\mathbb R$ is Lipschitz, the transformed class $\phi\circ\mathcal F$ has a square-integrable envelope, and the relevant empirical processes satisfy the standard measurable separability assumptions, then $\phi\circ\mathcal F$ is $P$-Donsker. This is a standard empirical-process permanence result (citing a result not yet in the wiki: Lipschitz Preservation Theorem for Donsker Classes).
The hypotheses of this permanence theorem are satisfied here. The class $\mathcal F$ is $P$-Donsker by assumption. The function $\phi$ is Lipschitz by assumption. The measurable separability assumptions are included in the statement. The square-integrable envelope condition for $\phi\circ\mathcal F$ was verified in the preceding step, either by $G_0=LF$ when $\phi(0)=0$, or by the assumed envelope $G$ in the general case. Therefore $\phi\circ\mathcal F$ is $P$-Donsker.
[/step]
