Chaining and Uniform Limits — Examples

Fix a finite set $A\subseteq [0,1]$. If $A=\mathrm{\varnothing }$, then $P(A)=0$. Otherwise write $m=\mathrm{\mid }A\mathrm{\mid }\ge 1$. Given $ϵ>0$, for each $x\in A$ let $I_x=\{y\in [0,1]:\mathrm{\mid }y-x\mathrm{\mid }<ϵ\mathrm{/}(2m)\}$. These $m$ intervals cover $A$, and each has length at most $ϵ\mathrm{/}m$. By [A1] and finite subadditivity [F2], $0\le P(A)\le {\sum }_{x\in A}P(I_x)\le ϵ$. Since this holds for every $ϵ>0$, $P(A)=0$. By [F1], each event $\{X_i\in A\}$ has probability zero; countable subadditivity [F2] gives $P(\mathrm{\exists }i\ge 1:X_i\in A)=0$. Hence with probability one $P_n(A)=0=P(A)$ for every $n$. This proves the pointwise strong law for this fixed $A$, without asserting one common event for all finite sets.

For any realized finite sample $x_1,\dots ,x_n$, the set $A_n=\{x_1,\dots ,x_n\}$ is finite, so $A_n\in \mathcal{A}$, $P_n(A_n)=1$, and $P(A_n)=0$. Every indicator deviation is at most one, so the supremum over $\mathcal{A}$ equals one for every sample. This supremum is the constant measurable function one; by [F4] it does not converge to zero in probability.

The class is pointwise measurable. Since $1\in \mathbb{Q}$, the rational thresholds together with $1$ are indexed by $\mathbb{Q}\cap [0,1]$, a countable set by [F5]–[F6]. For each fixed $t<1$, Countable Choice [A2] selects $q_k\in \mathbb{Q}\cap (t,\min (1,t+1\mathrm{/}k))$, which exist by [F7]. Then $q_k\to t$ from above, so ${\mathbf{1}}_{[0,q_k]}(x)\to {\mathbf{1}}_{[0,t]}(x)$ for every $x\in [0,1]$.

For each integer $m\ge 1$, the $m$ brackets $[f_{j\mathrm{/}m},f_{(j+1)\mathrm{/}m}]$, $0\le j<m$, cover $\mathcal{F}$ and have $L^1(P)$ width $(j+1)\mathrm{/}m-j\mathrm{/}m=1\mathrm{/}m$. Thus for every $ϵ>0$, taking an integer $m>1\mathrm{/}ϵ$ gives $N_{[]}(ϵ,\mathcal{F},L^1(P))\le m<\mathrm{\infty }$. The envelope condition is [A1].

By [F4] under [A2], the hypotheses verified in steps 1.1 and 1.2 imply ${\sup }_{t\in [0,1]}\mathrm{\mid }{P}_n{f}_t-P{f}_t\mathrm{\mid }\to 0$ almost surely, which is the claimed strong Glivenko–Cantelli property by [F1].

The class $\mathcal{F}=\{f_t:t\in [0,1]\}$ is pointwise measurable. Since $1\in \mathbb{Q}$, the rational thresholds together with $1$ are indexed by $\mathbb{Q}\cap [0,1]$, which is countable by [F5]–[F6]. For each fixed $t<1$, [A2] and [F7] give rationals $q_k\in \mathbb{Q}\cap (t,\min (1,t+1\mathrm{/}(k+1)))$ for every integer $k\ge 0$, so $q_k\to t$ from above and $f_{q_k}(x)\to f_t(x)$ for every $x\in [0,1]$. For $t=1$, use the constant sequence $q_k=1$. The envelope $H\equiv 1$ is measurable and square-integrable.

For an integer $m\ge 1$, the $m$ brackets $[f_{j\mathrm{/}m},f_{(j+1)\mathrm{/}m}]$ have $L^2(P)$ width $\sqrt{1\mathrm{/}m}$, since their endpoint difference is the indicator of an interval of length $1\mathrm{/}m$. Thus for $0<ϵ\le 1$, choosing $m=\lceil {ϵ}^{-2}\rceil$ gives $N_{[]}(ϵ,\mathcal{F},L^2(P))\le m$.

Consequently $\log N_{[]}(ϵ,\mathcal{F},L^2(P))\le \log (1+{ϵ}^{-2})\le \log 2+2\log (1\mathrm{/}ϵ)$ for $0<ϵ\le 1$. Since $\log (1\mathrm{/}ϵ)\le 1\mathrm{/}ϵ$, the square-root entropy is bounded by $\sqrt{\log 2}+\sqrt{2}{ϵ}^{-1\mathrm{/}2}$, an integrable function on $(0,1]$. All hypotheses of [F4] hold.

Applying [F4] gives a tight centered Gaussian limit. For $s,t\in [0,1]$, $P(f_s{f}_t)=P([0,\min (s,t)])=\min (s,t)$ and $P{f}_{s}P{f}_t=st$, so [F2] gives covariance $\min (s,t)-st$. This is precisely the defining covariance of the standard Brownian bridge stated in the example, proving the claimed limit. AC is carried from [F4].