I would like to have the following proof of the Glivenko-Cantelli theorem checked, which states that if ${X_1,X_2,\dots}$ are iid copies of a real random variable ${X}$, then almost surely, one has
$\displaystyle \frac{1}{n} | \{ 1 \leq i \leq n: X_i \leq t \}| \rightarrow {\bf P}( X \leq t )$
uniformly in ${t}$ as ${n \rightarrow \infty}$.
Hint: For any natural number ${m}$, let ${f_m(x)}$ denote the largest integer multiple of ${1/m}$ less than ${x}$. Show first that ${f_m( \frac{1}{n} | \{ 1 \leq i \leq n: X_i \leq t \}| )}$ is within ${O(1/m)}$ of ${f_m( {\bf P}( X \leq t ) )}$ for all ${t}$ when ${n}$ is sufficiently large. Once one has uniformity in $t$ for the $f_m$-approximate version of the problem, the usual analysis arguments (splitting an epsilon into say three smaller epsilons, etc.) will then give uniformity in $t$ for the original problem.
Proof: Let $F_n(t) := \frac{1}{n}|\{ 1 \leq i \leq n: X_i \leq t \}|$ denote the empirical distribution function and $F(t) := {\bf P}(X \leq t)$ the CDF. For any natural number ${m}$, let ${f_m(x)}$ denote the largest integer multiple of ${1/m}$ less than ${x}$.
By construction, $f_m$ is a piecewise constant function on the unit interval, divided at $0, 1/m, 2/m, \dots, 1$ (we let $f_m(0) := 0$). Subsequently ${\bf R}$ is partitioned into $m$ intervals, divided at $t_0 := F^{-1}(0), t_1 := F^{-1}(1/m), \dots, t_m := F^{-1}(1)$, here we let $t_0 = F^{-1}(0) := -\infty$ and $t_m = F^{-1}(1) := \infty$.
For any $t \in {\bf R}$, $t \in [t_{j-1}, t_j]$ for some $j$, and correspondingly $F(t) \in [\frac{j-1}{m}, \frac{j}{m}]$, as all of $f_m, F, F_n$ are non-decreasing, we have \begin{align*} f_m(F_n(t)) - f_m(F(t)) &\leq f_m(F_n(t_j)) - f_m(F(t_{j-1})) \\ &= f_m(F_n(t_j)) - f_m(F(t_j)) + 1/m. \end{align*} Therefore, \begin{align*} \|f_m(F_n) - f_m(F)\|_\infty &= \sup_{t \in \bf R} |f_m(F_n(t)) - f_m(F(t))| \\ &\leq \max_{j \in \{1, \dots, m\}} |f_m(F_n(t_j)) - f_m(F(t_j))| + 1/m. \end{align*} By the strong law of large numbers (and the definition of $f_m$), we see that $\forall t \in {\bf R}, |f_m(F_n(t)) - f_m(F(t))| = O(1/m)$ almost surely for sufficiently large $n$. By the triangle inequality, we thus have \begin{align*} |F_n(t) - F(t)| &\leq |F_n(t) - f_m(F_n(t))| + |F(t) - f_m(F(t))| + |f_m(F_n(t)) - f_m(F(t))| \\ &= O(1/m) \end{align*} almost surely for sufficiently large $n$, as $1/m$ can be arbitrarily small, this proves the claim.
