Uniform convergence of empirical risk-set fractions

Each $f_t$ is measurable. The subclass with rational thresholds in $[0,\tau ]$ is countable and pointwise dense: for $t>0$ choose rationals $q_k\uparrow t$ with $0\le q_k<t$, while for $t=0$ use the member $f_0$ itself. If $x\ge t$, every $f_{q_k}(x)=1$; if $x<t$, then $q_k>x$ eventually and $f_{q_k}(x)=0$. Hence $f_{q_k}(x)\to f_t(x)$ for every $x$. The envelope is the constant function $1$, which is measurable and has $P_g$-integral $1$.

Put $q(t):=P_g(X_{g1}\ge t)$. For $s<t$, the indicators are nested and $P_g\mathrm{\mid }{f}_s-f_t\mathrm{\mid }=P_g(s\le X_{g1}<t)=q(s)-q(t)$. The function $q$ is nonincreasing with range in $[0,1]$. Fix $ϵ>0$ and set $\delta =ϵ\mathrm{/}3$. Partition $[0,1]$ into finitely many intervals of length at most $\delta$ and group thresholds whose $q$-values lie in the same interval. Each threshold class in one group is nested; its pointwise infimum and supremum are measurable threshold indicators (possibly their one-sided endpoint limits), and they bracket the group. The $L^1(P_g)$ width is the supremum of $P_g\mathrm{\mid }{f}_s-f_t\mathrm{\mid }$ over thresholds in that group, at most $\delta$, by the displayed identity. Thus at most $\lceil 1\mathrm{/}\delta \rceil +1$ brackets cover $\mathcal{F}$, so $N_{[]}(ϵ,\mathcal{F},L^1(P_g))<\mathrm{\infty }$. This also handles atoms: a jump in $q$ larger than a bin width places the two threshold sides in different bins, while a smaller jump contributes at most the bin's total $q$-width.

Steps 1.1 and 1.2 verify pointwise measurability, the integrable envelope, and every finite-bracketing hypothesis in [F6]. Its convergence-in-probability conclusion gives ${\sup }_{t\in [0,\tau ]}\mathrm{\mid }{P}_{g,n}{f}_t-P_g{f}_t\mathrm{\mid }{\to }_{P}0$. Since $P_{g,n}{f}_t=Y_{g,n}(t)\mathrm{/}{n}_g$ and $P_g{f}_t=s_g(t)$ by their definitions, this is the asserted uniform risk-set convergence. No Countable Choice or strong Glivenko--Cantelli conclusion is used.