Question in proof of Glivenko-Cantelli Theorem

Question

A question came up when I was working through the proof of the Glivenko-Cantelli theorem in Durrett's Probability: Theory and Examples (pg. 76 in the fourth edition).

The Glivenko Cantelli Theorem: Let $X_{1}, X_{2}, \ldots$ be i.i.d. random variables and let \begin{align} F_{n}(x) = n^{-1} \sum_{i=1}^{n} 1(X_{i}\leq x) \end{align} Then as $n\to \infty$ \begin{align} \sup_{x} | F_{n}(x) - F(x)| \to 0 \quad\text{ a.s.} \end{align}

The proof starts like this:

Define $Y_{i} = 1(X_{i}\leq x)$. Then we have that $\mathbb{E} Y_{i} = P(X_{i} \leq x) = F(x)$. Finally we know by the strong law of large numbers that \begin{align} F_{n}(x) = n^{-1} \sum_{i=1}^{n} Y_{i} \to F(x) \text{ a.s.} \end{align} FAIR ENOUGH. Can't argue with Durrett there. However, in the proof Durrett then starts talking about how we have proved that $F_{n}(x)$ converges pointwise. A quote:

For $1\leq j\leq k-1$ let $x_{j,k} = \inf\{y : F(y)\geq j/k\}$. The pointwise convergence of $F_{n}(x)$ and $F_{n}(x^{-})$ imply that we can pick $N_{k}(\omega)$ so that, if $n\geq N_{k}(\omega)$, then \begin{align*} |F_{n}(x_{j,k}) - F(x_{j,k})| \leq k^{-1} &&\text{ and } &&|F_{n}(x_{j,k}^{-}) - F(x_{j,k}^{-})| \leq k^{-1} \end{align*}

Two questions:

1. Does $F_{n}(x)\to F(x)$ almost surely for every $x\in \mathbb{R}$ imply that $F_{n}(x) \to F(x)$ pointwise for every $x\in \mathbb{R}$? Is there a proof of this?
2. If $F_{n}(x) \to F(x)$ pointwise, why is the $N_{k}(\omega)$ a function of $\omega\in\Omega$ (element of sample space) and not $x$ in the above quote?

Answer 1

In order to understand the details, you need to explicitly write the $\omega$ variable. For example, the law of large numbers guarantees that if we define $$\Omega^\prime=\bigcap_{k\geq 2, 1\leq j<k}\left\{\omega: F_n(x_{j,k}-,\omega)\to F(x_{j,k}-) \mbox{ and }F_n(x_{j,k},\omega)\to F(x_{j,k})\right\},$$ then $\mathbb{P}(\Omega^\prime)=1.$ For each $\omega\in\Omega^\prime$ we have the convergence $F_n(x_{j,k}-,\omega)\to F(x_{j,k}-)$ and $F_n(x_{j,k},\omega)\to F(x_{j,k})$, so the value $N_k(\omega)$ exists as claimed.

For $n\geq N_k(\omega)$, Durrett's argument now gives $\sup_x|F_n(x,\omega)-F(x)|\leq 2/k.$