The following is an excerpt from John Walsh's Knowing the Odds and it concerns the Glivenko-Cantlli theorem.
The proof first derives the convergence for all rationals and then appeals to Dini's theorem for the uniform convergence part, given that $F^0$ and $F_n^{0}$ are nondecreasing. My question is: how is Dini's theorem applicable here? The version of which I am aware, see for example wikipedia, requires that the functions $F_n^{0}$ are continuous and satisfy $F_n^{0}(x) \leq F_{n+1}^{0}(x)$, neither of which is satisfied here. Am I misunderstanding something?
You are correct, this is not an application of Dini's theorem. Here is a correct way to finish the proof whenever the distribution function $F(x)$ is continuous over a particular compact set $[a,b]$ of interest. It uses another lemma that I state and prove below.
Lemma:
Fix $a,b$ real numbers with $a<b$. Let $g_n:[a,b]\rightarrow\mathbb{R}$ be (possibly discontinuous) functions defined for $n \in \{1, 2, 3,...\}$ that converge pointwise to a continuous function $g:[a,b]\rightarrow\mathbb{R}$ as $n\rightarrow \infty$. Further assume that for each positive integer $n$ we have $g_n(x)$ is nondecreasing over $x \in [a,b]$. Then $g_n\rightarrow g$ uniformly.
Proof: Fix $\epsilon>0$. Since $g(x)$ is continuous over the compact interval $[a,b]$, it must be uniformly continuous. Choose an integer $k>0$ such that $$|x-y|\leq (b-a)/k \implies |g(x)-g(y)|\leq \epsilon$$ Divide the interval $[a,b]$ into $k$ disjoint subintervals, each of size $(b-a)/k$. Let $\{x_0, x_1, ..., x_k\}$ be the gridpoints, where $x_0=a, x_k=b$. We know $g_n(x_i)\rightarrow g(x_i)$ for all $i \in \{0, 1, ..., k\}$ and so we can choose $m$ such that $n\geq m$ implies $|g_n(x_i)-g(x_i)|\leq \epsilon$ for all $i \in \{0, ..., k\}$.
Fix $x \in [a,b]$ and fix $n \geq m$. The value $x$ must be in between two gridpoints, so there is a $j \in \{0, ..., k-1\}$ such that $$ x_j \leq x \leq x_{j+1} $$ Since each function $g_n(x)$ is nondecreasing in $x$ we get: $$ g_n(x_j) \leq g_n(x) \leq g_n(x_{j+1}) $$ Thus \begin{align} &g(x)-\underbrace{|g(x)-g(x_j)|}_{\leq \epsilon} - \underbrace{|g(x_j)-g_n(x_j)|}_{\leq \epsilon} \\ &\leq g_n(x) \\ &\leq g(x) + \underbrace{|g(x)-g(x_{j+1})|}_{\leq \epsilon} + \underbrace{|g(x_{j+1})-g_n(x_{j+1})|}_{\leq \epsilon} \end{align} Thus $$ g(x) - 2\epsilon \leq g_n(x) \leq g(x) + 2\epsilon$$ That is, $|g(x)-g_n(x)|\leq 2\epsilon$. This holds for all $x \in [a,b]$. So convergence is uniform. $\Box$
The rest of the proof of Glivenko-Cantelli (with $F(x)$ continuous over an interval $[a,b]$ of interest) uses this lemma together with my first comment that simultaneous convergence over the rationals (with $F_n(x)$ nondecreasing in $x$ and continuity of $F(x)$) implies that we get simultaneous convergence over the reals.