[proofplan]
Fix a probability measure $P$ on $(S,\mathcal A)$. The uniform polynomial $L^2(Q)$ covering bound for $\mathcal F$, together with the bound $|f|\le 1$, is converted by the convex-hull bracketing entropy theorem into a finite bracketing entropy integral for $\operatorname{sconv}(\mathcal F)$. Since every element of the symmetric convex hull is still bounded in absolute value by $1$, the class has envelope $1\in L^2(P)$. The [bracketing Donsker theorem](/theorems/9836) then gives that $\operatorname{sconv}(\mathcal F)$ is $P$-Donsker, and the ordinary convex hull is handled by restricting the limiting empirical process to the subclass $\operatorname{conv}(\mathcal F)$.
[/proofplan]
[step:Fix the probability measure and record the entropy consequence for the symmetric convex hull]
Let $P$ be an arbitrary probability measure on $(S,\mathcal A)$. For a class $\mathcal H$ of real-valued [measurable functions](/page/Measurable%20Functions) on $S$ and a probability measure $Q$ on $(S,\mathcal A)$, let $N_{[]}(\varepsilon,\mathcal H,L^2(Q))$ denote the least cardinality of a family of measurable brackets $[l,u]$ such that $\|u-l\|_{L^2(Q)}\le\varepsilon$ and every $h\in\mathcal H$ satisfies $l\le h\le u$ pointwise. We use the following standard convex-hull bracketing entropy theorem: if a uniformly bounded class $\mathcal F$ satisfies
\begin{align*}
N(\varepsilon,\mathcal F,L^2(Q))\le A_0\varepsilon^{-v}
\end{align*}
for every finitely supported probability measure $Q$ and every $\varepsilon\in(0,1)$, then for every probability measure $P$ there exist constants $C\in(0,\infty)$ and $\alpha\in(0,2)$, depending only on $A_0$ and $v$, such that
\begin{align*}
\log N_{[]}(\varepsilon,\operatorname{sconv}(\mathcal F),L^2(P))\le C\varepsilon^{-\alpha}
\end{align*}
for every $\varepsilon\in(0,1)$. This is the convex-hull entropy bound for uniformly bounded classes.
It follows that the bracketing entropy integral is finite:
\begin{align*}
\int_0^1\sqrt{\log N_{[]}(\varepsilon,\operatorname{sconv}(\mathcal F),L^2(P))}\,d\mathcal L^1(\varepsilon)
\le \sqrt C\int_0^1\varepsilon^{-\alpha/2}\,d\mathcal L^1(\varepsilon).
\end{align*}
Since $\alpha<2$, the last integral is finite and equals
\begin{align*}
\int_0^1\varepsilon^{-\alpha/2}\,d\mathcal L^1(\varepsilon)=\frac{1}{1-\alpha/2}.
\end{align*}
Therefore
\begin{align*}
\int_0^1\sqrt{\log N_{[]}(\varepsilon,\operatorname{sconv}(\mathcal F),L^2(P))}\,d\mathcal L^1(\varepsilon)<\infty.
\end{align*}
[guided]
We first isolate the only non-elementary entropy input. The hypothesis controls ordinary $L^2(Q)$ covering numbers of the original class $\mathcal F$, uniformly over all finitely supported probability measures $Q$. The conclusion needed for the bracketing Donsker theorem is different: it is an $L^2(P)$ bracketing entropy integral for the larger class $\operatorname{sconv}(\mathcal F)$. The bridge between these two statements is the standard convex-hull entropy bound for uniformly bounded classes.
Precisely, the theorem says that under the assumptions
\begin{align*}
|f(s)|\le 1
\end{align*}
for all $f\in\mathcal F$ and $s\in S$, and
\begin{align*}
N(\varepsilon,\mathcal F,L^2(Q))\le A_0\varepsilon^{-v}
\end{align*}
for every finitely supported probability measure $Q$ and every $\varepsilon\in(0,1)$, there are constants $C\in(0,\infty)$ and $\alpha\in(0,2)$, depending only on $A_0$ and $v$, such that for the fixed probability measure $P$,
\begin{align*}
\log N_{[]}(\varepsilon,\operatorname{sconv}(\mathcal F),L^2(P))\le C\varepsilon^{-\alpha}
\end{align*}
for every $\varepsilon\in(0,1)$. This is the convex-hull entropy bound for uniformly bounded classes.
Now we check that this bound gives the exact integrability condition required later. Taking square roots gives
\begin{align*}
\sqrt{\log N_{[]}(\varepsilon,\operatorname{sconv}(\mathcal F),L^2(P))}\le \sqrt C\,\varepsilon^{-\alpha/2}
\end{align*}
for $\varepsilon\in(0,1)$. Integrating with respect to one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal L^1$ on $(0,1)$ gives
\begin{align*}
\int_0^1\sqrt{\log N_{[]}(\varepsilon,\operatorname{sconv}(\mathcal F),L^2(P))}\,d\mathcal L^1(\varepsilon)
\le \sqrt C\int_0^1\varepsilon^{-\alpha/2}\,d\mathcal L^1(\varepsilon).
\end{align*}
The exponent condition $\alpha<2$ is exactly what is needed here, because $\alpha/2<1$ and hence
\begin{align*}
\int_0^1\varepsilon^{-\alpha/2}\,d\mathcal L^1(\varepsilon)=\frac{1}{1-\alpha/2}<\infty.
\end{align*}
Thus
\begin{align*}
\int_0^1\sqrt{\log N_{[]}(\varepsilon,\operatorname{sconv}(\mathcal F),L^2(P))}\,d\mathcal L^1(\varepsilon)<\infty.
\end{align*}
[/guided]
[/step]
[step:Verify that the symmetric convex hull has a square-integrable envelope]
Let $G:S\to[0,\infty)$ be the constant function
\begin{align*}
G(s):=1.
\end{align*}
We claim that $G$ is an envelope for $\operatorname{sconv}(\mathcal F)$. Let $h\in\operatorname{sconv}(\mathcal F)$. By definition of finite convex hull, there exist $m\in\mathbb N$, functions $g_1,\dots,g_m\in\mathcal F\cup(-\mathcal F)$, and coefficients $\lambda_1,\dots,\lambda_m\in[0,1]$ satisfying
\begin{align*}
\sum_{j=1}^m\lambda_j=1
\end{align*}
such that
\begin{align*}
h=\sum_{j=1}^m\lambda_j g_j.
\end{align*}
For each $j\in\{1,\dots,m\}$ and each $s\in S$, the assumption on $\mathcal F$ gives $|g_j(s)|\le 1$. Hence, by the triangle inequality in $\mathbb R$,
\begin{align*}
|h(s)|\le \sum_{j=1}^m\lambda_j |g_j(s)|\le \sum_{j=1}^m\lambda_j=1=G(s).
\end{align*}
Thus $G$ is an envelope. Since $P$ is a probability measure,
\begin{align*}
\int_S G(s)^2\,dP(s)=\int_S 1\,dP(s)=1<\infty.
\end{align*}
Therefore $\operatorname{sconv}(\mathcal F)$ has a measurable envelope in $L^2(P)$.
[/step]
[step:Apply the bracketing Donsker theorem to the symmetric convex hull]
The class $\operatorname{sconv}(\mathcal F)$ consists of measurable functions under the measurability and separability hypotheses imposed in the statement. The preceding step gives a measurable envelope $G\in L^2(P)$, and the first step gives
\begin{align*}
\int_0^1\sqrt{\log N_{[]}(\varepsilon,\operatorname{sconv}(\mathcal F),L^2(P))}\,d\mathcal L^1(\varepsilon)<\infty.
\end{align*}
These are precisely the hypotheses of the [Bracketing Donsker Theorem][citetheorem:9836]. Therefore $\operatorname{sconv}(\mathcal F)$ is $P$-Donsker.
[/step]
[step:Restrict the Donsker convergence to the ordinary convex hull]
For every $f\in\mathcal F$, we have $f\in\mathcal F\cup(-\mathcal F)$, and hence every finite convex combination of elements of $\mathcal F$ is also a finite convex combination of elements of $\mathcal F\cup(-\mathcal F)$. Thus
\begin{align*}
\operatorname{conv}(\mathcal F)\subseteq\operatorname{sconv}(\mathcal F).
\end{align*}
Since $\operatorname{sconv}(\mathcal F)$ is $P$-Donsker, the empirical process indexed by $\operatorname{sconv}(\mathcal F)$ converges in distribution in $\ell^\infty(\operatorname{sconv}(\mathcal F))$ to the corresponding tight Gaussian limit. The restriction map
\begin{align*}
R:\ell^\infty(\operatorname{sconv}(\mathcal F))&\to\ell^\infty(\operatorname{conv}(\mathcal F))
\end{align*}
\begin{align*}
z&\mapsto z|_{\operatorname{conv}(\mathcal F)}
\end{align*}
is linear and satisfies
\begin{align*}
\|Rz-Rw\|_{\ell^\infty(\operatorname{conv}(\mathcal F))}
\le
\|z-w\|_{\ell^\infty(\operatorname{sconv}(\mathcal F))}
\end{align*}
for all $z,w\in\ell^\infty(\operatorname{sconv}(\mathcal F))$. Hence $R$ is continuous. Applying the [continuous mapping theorem](/theorems/1847) to this restriction map, and using the separability and measurability hypotheses from the statement for the restricted class, gives convergence of the empirical process indexed by $\operatorname{conv}(\mathcal F)$ in $\ell^\infty(\operatorname{conv}(\mathcal F))$. Therefore $\operatorname{conv}(\mathcal F)$ is $P$-Donsker.
Because the probability measure $P$ was arbitrary, both conclusions hold for every probability measure $P$ on $(S,\mathcal A)$.
[/step]
