[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]