[proofplan] The proof first passes from convergence of the processes in $\ell^\infty(\mathcal F)$ to convergence of their suprema, using the continuity of the norm map. Conditional convergence gives convergence in probability of the bootstrap conditional distribution functions at every continuity point of the limiting distribution function $F$. The regular quantile condition provides deterministic continuity points $a_\eta<c_p<b_\eta$ whose $F$-values bracket $p$ up to an error of size $\eta$, and the bootstrap quantile is trapped between those points with probability tending to one. Finally, convergence of $T_n$ at these deterministic continuity points sandwiches the target coverage between $p-\eta$ and $p+\eta$, and then $\eta\downarrow0$. [/proofplan]

[step:Pass from process convergence to convergence of the supremum statistics]Define the norm map \begin{align*} \Phi:\ell^\infty(\mathcal F)&\to\mathbb R \end{align*} \begin{align*} z&\mapsto \|z\|_{\mathcal F}. \end{align*} For $z,w\in\ell^\infty(\mathcal F)$, the [reverse triangle inequality](/theorems/2300) gives \begin{align*} |\Phi(z)-\Phi(w)|\le \|z-w\|_{\mathcal F}. \end{align*} Thus $\Phi$ is $1$-Lipschitz, hence continuous. Applying the [citetheorem:9834] to $Z_n\rightsquigarrow Z$ gives \begin{align*} T_n=\Phi(Z_n)\rightsquigarrow \Phi(Z)=T. \end{align*} The same continuity, applied under the stated conditional [weak convergence](/page/Weak%20Convergence) convention for $Z_n^*$, gives \begin{align*} T_n^*=\Phi(Z_n^*)\rightsquigarrow T \end{align*} conditionally in probability. Consequently, for every continuity point $t\in\mathbb R$ of $F$, \begin{align*} F_n^*(t\mid X_1,\dots,X_n)\xrightarrow{\mathbb P}F(t). \end{align*}[/step]

[guided]The first reduction is to replace the empirical processes by the scalar statistics that actually appear in the theorem. We define the map \begin{align*} \Phi:\ell^\infty(\mathcal F)&\to\mathbb R \end{align*} \begin{align*} z&\mapsto \|z\|_{\mathcal F}. \end{align*} This map is continuous because for every $z,w\in\ell^\infty(\mathcal F)$ the reverse triangle inequality for the supremum norm yields \begin{align*} |\Phi(z)-\Phi(w)|=|\|z\|_{\mathcal F}-\|w\|_{\mathcal F}|\le \|z-w\|_{\mathcal F}. \end{align*} So $\Phi$ is in fact $1$-Lipschitz. Since $Z_n\rightsquigarrow Z$ in $\ell^\infty(\mathcal F)$ and $\Phi$ is continuous, the [citetheorem:9834] gives \begin{align*} \Phi(Z_n)\rightsquigarrow \Phi(Z). \end{align*} By the definitions of $T_n$ and $T$, this is exactly \begin{align*} T_n\rightsquigarrow T. \end{align*} For the bootstrap sequence, the hypothesis says that $Z_n^*$ converges conditionally in probability to the same tight limit $Z$ in $\ell^\infty(\mathcal F)$. The same continuous map $\Phi$ may be applied conditionally, because the stated convention for conditional weak convergence includes the conditional continuous mapping principle. Therefore \begin{align*} T_n^*=\Phi(Z_n^*)\rightsquigarrow \Phi(Z)=T \end{align*} conditionally in probability. The scalar consequence we need is convergence of conditional distribution functions at continuity points. Let $t\in\mathbb R$ be a continuity point of \begin{align*} F:\mathbb R&\to[0,1] \end{align*} \begin{align*} s&\mapsto \mathbb P(T\le s). \end{align*} The set $(-\infty,t]$ is then a continuity set for the law of $T$, because its boundary is $\{t\}$ and $\mathbb P(T=t)=0$. By the conditional portmanteau implication included in the hypothesis, \begin{align*} \mathbb P(T_n^*\le t\mid X_1,\dots,X_n)\xrightarrow{\mathbb P}\mathbb P(T\le t). \end{align*} Using the definition of $F_n^*$ and $F$, this becomes \begin{align*} F_n^*(t\mid X_1,\dots,X_n)\xrightarrow{\mathbb P}F(t). \end{align*}[/guided]

[step:Show that the regular quantile is a continuity point with value $p$] Since $F$ is a distribution function and $c_p$ is finite, right-continuity of $F$ and the definition of $c_p$ imply \begin{align*} F(c_p)\ge p. \end{align*} Let $\eta>0$. By regularity, choose a continuity point $b_\eta>c_p$ such that \begin{align*} F(b_\eta)-p<\eta. \end{align*} Monotonicity gives \begin{align*} F(c_p)\le F(b_\eta)<p+\eta. \end{align*} Since $\eta>0$ was arbitrary, \begin{align*} F(c_p)\le p. \end{align*} Thus \begin{align*} F(c_p)=p. \end{align*} It remains to identify the left limit. Let $F(c_p-):=\lim_{s\uparrow c_p}F(s)$ denote the left limit of $F$ at $c_p$. For every $\eta>0$, regularity gives a continuity point $a_\eta<c_p$ such that \begin{align*} p-F(a_\eta)<\eta. \end{align*} Because $a_\eta<c_p$, monotonicity gives \begin{align*} F(a_\eta)\le F(c_p-)\le F(c_p). \end{align*} Therefore \begin{align*} p-\eta<F(c_p-)\le p. \end{align*} Letting $\eta\downarrow0$ gives \begin{align*} F(c_p-)=p. \end{align*} Since $F(c_p)=p$, the function $F$ is continuous at $c_p$. [/step]

[step:Sandwich the bootstrap quantile between deterministic continuity points]Fix $\eta>0$, and choose continuity points $a_\eta,b_\eta\in\mathbb R$ as in the regularity assumption. Define the random distribution function \begin{align*} G_n:\mathbb R&\to[0,1] \end{align*} \begin{align*} t&\mapsto F_n^*(t\mid X_1,\dots,X_n). \end{align*} Then $G_n$ is nondecreasing, and \begin{align*} c_{n,p}^*=\inf\{t\in\mathbb R:G_n(t)\ge p\}. \end{align*} If $c_{n,p}^*<a_\eta$, then by the definition of the infimum there exists $s<a_\eta$ such that $G_n(s)\ge p$. Since $G_n$ is nondecreasing, this implies $G_n(a_\eta)\ge p$. Hence \begin{align*} \{c_{n,p}^*<a_\eta\}\subseteq \{G_n(a_\eta)\ge p\}. \end{align*} Because $a_\eta$ is a continuity point of $F$ and $G_n(a_\eta)\xrightarrow{\mathbb P}F(a_\eta)$, while $F(a_\eta)<p$, we obtain \begin{align*} \mathbb P_{\mathrm{out}}(c_{n,p}^*<a_\eta)\to0. \end{align*} Similarly, if $G_n(b_\eta)\ge p$, then the set $\{t\in\mathbb R:G_n(t)\ge p\}$ intersects $(-\infty,b_\eta]$, so $c_{n,p}^*\le b_\eta$. Therefore \begin{align*} \{c_{n,p}^*>b_\eta\}\subseteq \{G_n(b_\eta)<p\}. \end{align*} Because $b_\eta$ is a continuity point of $F$ and $G_n(b_\eta)\xrightarrow{\mathbb P}F(b_\eta)$, while $F(b_\eta)>p$, we obtain \begin{align*} \mathbb P_{\mathrm{out}}(c_{n,p}^*>b_\eta)\to0. \end{align*}[/step]

[guided]We now turn convergence of the conditional distribution functions into convergence of their quantiles. Fix $\eta>0$. The regularity assumption gives continuity points $a_\eta,b_\eta\in\mathbb R$ satisfying \begin{align*} a_\eta<c_p<b_\eta, \end{align*} \begin{align*} F(a_\eta)<p<F(b_\eta), \end{align*} \begin{align*} p-F(a_\eta)<\eta, \end{align*} and \begin{align*} F(b_\eta)-p<\eta. \end{align*} Define \begin{align*} G_n:\mathbb R&\to[0,1] \end{align*} \begin{align*} t&\mapsto F_n^*(t\mid X_1,\dots,X_n). \end{align*} This is a conditional distribution function, so it is nondecreasing in $t$. The bootstrap quantile is \begin{align*} c_{n,p}^*=\inf\{t\in\mathbb R:G_n(t)\ge p\}. \end{align*} First consider the event that the bootstrap quantile falls too far to the left. If $c_{n,p}^*<a_\eta$, then the infimum of the set $\{t\in\mathbb R:G_n(t)\ge p\}$ is strictly below $a_\eta$. Hence there is some $s<a_\eta$ with $G_n(s)\ge p$. Since $G_n$ is nondecreasing, \begin{align*} G_n(a_\eta)\ge G_n(s)\ge p. \end{align*} Thus \begin{align*} \{c_{n,p}^*<a_\eta\}\subseteq \{G_n(a_\eta)\ge p\}. \end{align*} The point $a_\eta$ is a continuity point of $F$, so the previous step gives \begin{align*} G_n(a_\eta)=F_n^*(a_\eta\mid X_1,\dots,X_n)\xrightarrow{\mathbb P}F(a_\eta). \end{align*} Since $F(a_\eta)<p$, choose the positive gap \begin{align*} \delta_a:=p-F(a_\eta)>0. \end{align*} Then \begin{align*} \{G_n(a_\eta)\ge p\}\subseteq \{|G_n(a_\eta)-F(a_\eta)|\ge \delta_a\}, \end{align*} and the probability of the right-hand event tends to $0$. Therefore \begin{align*} \mathbb P_{\mathrm{out}}(c_{n,p}^*<a_\eta)\to0. \end{align*} The right tail is analogous but uses the opposite implication. If $G_n(b_\eta)\ge p$, then $b_\eta$ itself belongs to the set whose infimum defines $c_{n,p}^*$. Hence \begin{align*} c_{n,p}^*\le b_\eta. \end{align*} Taking the contrapositive gives \begin{align*} \{c_{n,p}^*>b_\eta\}\subseteq \{G_n(b_\eta)<p\}. \end{align*} Again $b_\eta$ is a continuity point of $F$, so \begin{align*} G_n(b_\eta)=F_n^*(b_\eta\mid X_1,\dots,X_n)\xrightarrow{\mathbb P}F(b_\eta). \end{align*} Since $F(b_\eta)>p$, define \begin{align*} \delta_b:=F(b_\eta)-p>0. \end{align*} Then \begin{align*} \{G_n(b_\eta)<p\}\subseteq \{|G_n(b_\eta)-F(b_\eta)|\ge \delta_b\}, \end{align*} whose probability tends to $0$. Therefore \begin{align*} \mathbb P_{\mathrm{out}}(c_{n,p}^*>b_\eta)\to0. \end{align*}[/guided]

[step:Use deterministic quantile brackets rather than location convergence] For every $\eta>0$, let $a_\eta,b_\eta\in\mathbb R$ be the continuity points supplied by the regularity assumption. The preceding step gives \begin{align*} \mathbb P_{\mathrm{out}}(c_{n,p}^*<a_\eta)\to0 \end{align*} and \begin{align*} \mathbb P_{\mathrm{out}}(c_{n,p}^*>b_\eta)\to0. \end{align*} We do not need, and the hypotheses do not imply, convergence of $c_{n,p}^*$ to $c_p$ in location. The only information needed later is that the random threshold lies in the deterministic interval $[a_\eta,b_\eta]$ with outer probability tending to one, while the endpoint probabilities satisfy \begin{align*} p-\eta<F(a_\eta)<p<F(b_\eta)<p+\eta. \end{align*} [/step]

[step:Sandwich the coverage probability by the deterministic bracket values]Fix $\eta>0$, and let $a_\eta,b_\eta\in\mathbb R$ be the continuity points supplied by regularity. On the event $\{c_{n,p}^*\ge a_\eta\}$, the implication $T_n\le a_\eta\implies T_n\le c_{n,p}^*$ holds. Hence \begin{align*} \{T_n\le a_\eta\}\setminus\{c_{n,p}^*<a_\eta\}\subseteq \{T_n\le c_{n,p}^*\}. \end{align*} Taking outer probabilities and using subadditivity of outer probability gives \begin{align*} \mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*)\ge \mathbb P_{\mathrm{out}}(T_n\le a_\eta)-\mathbb P_{\mathrm{out}}(c_{n,p}^*<a_\eta). \end{align*} On the event $\{c_{n,p}^*\le b_\eta\}$, the implication $T_n\le c_{n,p}^*\implies T_n\le b_\eta$ holds. Therefore \begin{align*} \{T_n\le c_{n,p}^*\}\subseteq \{T_n\le b_\eta\}\cup\{c_{n,p}^*>b_\eta\}, \end{align*} and subadditivity gives \begin{align*} \mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*)\le \mathbb P_{\mathrm{out}}(T_n\le b_\eta)+\mathbb P_{\mathrm{out}}(c_{n,p}^*>b_\eta). \end{align*} Because $a_\eta$ and $b_\eta$ are continuity points of $F$, the outer-probability portmanteau implication in the stated weak-convergence convention for $T_n\rightsquigarrow T$ gives \begin{align*} \mathbb P_{\mathrm{out}}(T_n\le a_\eta)\to F(a_\eta) \end{align*} and \begin{align*} \mathbb P_{\mathrm{out}}(T_n\le b_\eta)\to F(b_\eta). \end{align*} Combining these limits with the preceding step yields \begin{align*} F(a_\eta)\le \liminf_{n\to\infty}\mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*) \end{align*} and \begin{align*} \limsup_{n\to\infty}\mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*)\le F(b_\eta). \end{align*} Regularity gives $p-\eta<F(a_\eta)<p<F(b_\eta)<p+\eta$, so \begin{align*} p-\eta\le \liminf_{n\to\infty}\mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*) \end{align*} and \begin{align*} \limsup_{n\to\infty}\mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*)\le p+\eta. \end{align*} Letting $\eta\downarrow0$ gives \begin{align*} \mathbb P_{\mathrm{out}}\left(T_n\le c_{n,p}^*\right)\to p. \end{align*} This is the asserted bootstrap validity for the supremum statistic.[/step]

[guided]The key point is that we should not try to prove $c_{n,p}^*\to c_p$ as a real number. The regularity assumption controls the distribution values near level $p$, not the physical distance of the bracketing points from $c_p$. So we keep the deterministic bracket $a_\eta<c_p<b_\eta$ and compare probabilities at those two endpoints. Fix $\eta>0$, and let $a_\eta,b_\eta\in\mathbb R$ be the continuity points from the regularity assumption. The previous step proved \begin{align*} \mathbb P_{\mathrm{out}}(c_{n,p}^*<a_\eta)\to0 \end{align*} and \begin{align*} \mathbb P_{\mathrm{out}}(c_{n,p}^*>b_\eta)\to0. \end{align*} On the event $\{c_{n,p}^*\ge a_\eta\}$, every outcome satisfying $T_n\le a_\eta$ also satisfies $T_n\le c_{n,p}^*$. This gives the set inclusion \begin{align*} \{T_n\le a_\eta\}\setminus\{c_{n,p}^*<a_\eta\}\subseteq \{T_n\le c_{n,p}^*\}. \end{align*} By [monotonicity and subadditivity](/theorems/1081) of outer probability, \begin{align*} \mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*)\ge \mathbb P_{\mathrm{out}}(T_n\le a_\eta)-\mathbb P_{\mathrm{out}}(c_{n,p}^*<a_\eta). \end{align*} Similarly, on the event $\{c_{n,p}^*\le b_\eta\}$, the implication $T_n\le c_{n,p}^*\implies T_n\le b_\eta$ holds. Hence \begin{align*} \{T_n\le c_{n,p}^*\}\subseteq \{T_n\le b_\eta\}\cup\{c_{n,p}^*>b_\eta\}, \end{align*} and therefore \begin{align*} \mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*)\le \mathbb P_{\mathrm{out}}(T_n\le b_\eta)+\mathbb P_{\mathrm{out}}(c_{n,p}^*>b_\eta). \end{align*} Now we use weak convergence of $T_n$ only at deterministic continuity points. The first step gave $T_n\rightsquigarrow T$. The theorem statement explicitly places nonmeasurable maps under the usual outer-probability convention, so the relevant portmanteau consequence is: if $t$ is a continuity point of the distribution function $F$ of $T$, then \begin{align*} \mathbb P_{\mathrm{out}}(T_n\le t)\to \mathbb P(T\le t)=F(t). \end{align*} Both $a_\eta$ and $b_\eta$ were chosen to be continuity points of $F$, so \begin{align*} \mathbb P_{\mathrm{out}}(T_n\le a_\eta)\to F(a_\eta) \end{align*} and \begin{align*} \mathbb P_{\mathrm{out}}(T_n\le b_\eta)\to F(b_\eta). \end{align*} Combining these deterministic convergence statements with the two vanishing bootstrap-tail probabilities yields \begin{align*} F(a_\eta)\le \liminf_{n\to\infty}\mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*) \end{align*} and \begin{align*} \limsup_{n\to\infty}\mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*)\le F(b_\eta). \end{align*} Finally, the regularity assumption says that the endpoint distribution values are within $\eta$ of $p$: \begin{align*} p-\eta<F(a_\eta)<p<F(b_\eta)<p+\eta. \end{align*} Therefore \begin{align*} p-\eta\le \liminf_{n\to\infty}\mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*) \end{align*} and \begin{align*} \limsup_{n\to\infty}\mathbb P_{\mathrm{out}}(T_n\le c_{n,p}^*)\le p+\eta. \end{align*} Since $\eta>0$ was arbitrary, the lower and upper limits are both equal to $p$. Thus \begin{align*} \mathbb P_{\mathrm{out}}\left(T_n\le c_{n,p}^*\right)\to p. \end{align*} This proves the asserted bootstrap validity.[/guided]