Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 28 / 2 / a

For each fixed $t\in [0,1]$, the variables $1_{\{X_k\le t\}}$ are independent and identically distributed random variables with Bernoulli distribution of parameter $t$. Their expected value is $t$, and the strong law of large numbers gives $F_n(t)\to t$ almost surely.

For a finite set $F$, intersect these probability-one events over its finitely many elements. On the resulting single event, every one of the finitely many errors tends to zero, so their maximum tends to zero:If $F$ is empty, take the displayed maximum to be zero. The important point is simultaneous convergence on one event; pointwise convergence at an uncountable collection of arguments would not alone imply the uniform claim in part (ii).

For each positive integer $m$, apply part (i) to the grid $D_m=\{j/m:0\le j\le m\}$. Intersect the resulting probability-one events over all $m$, a countable intersection. Work on this common event.

For $x\in [j/m,(j+1)/m]$, monotonicity of the empirical distribution function gives $F_n(j/m)\le F_n(x)\le F_n((j+1)/m)$. ConsequentlyandThe endpoint $x=1$ is itself on the grid. ThereforeFor each fixed $m$, take the upper limit as $n\to \infty$; part (i) makes the maximum vanish. Then let $m\to \infty$. This provesThis is the uniform-sample case of the Glivenko-Cantelli theorem, proved by a finite-grid approximation rather than assumed.

Let $G$ be the cumulative distribution function of the arbitrary Borel probability measure $\mu$. It may have atoms or flat intervals. Use its quantile functionThe limits of $G$ at the two infinities make $Q(u)$ finite. Its monotonicity and right continuity imply the exact equivalence $Q(u)\le x⟺u\le G(x)$. For the nontrivial direction, if $Q(u)\le x$, points with $G\ge u$ lie arbitrarily close above $Q(u)$; if the infimum itself is $x$, right continuity gives $G(x)\ge u$. The other direction follows directly from the defining infimum.

Take independent uniform variables $U_k$ and set $Y_k^{\prime }=Q(U_k)$. Thenso $(Y_k^{\prime })$ has the same entire sequence law as the original iid sample $(Y_k)$. On this construction, the empirical distribution function satisfiesThus part (ii) givesThe supremum is measurable: since both distribution functions are right-continuous, it equals the supremum over rational $x$. The convergence event consequently depends measurably on the coordinate sequence, and equality of the sequence laws transfers its probability one to $(Y_k)$. HenceThis is the Quantile proof of the Glivenko-Cantelli theorem, including arbitrary distributions with atoms. The original PDF supplies the full definition and absolute-value signs in this part, which the TeX conversion incompletely records.