Making sense of the Glivenko-Cantelli theorem -- where are the $\omega$'s in the r.v.'s? I have two questions:
First let's define 
$$\mathbb G_n(t)=\frac 1n \sum_{i=1}^n 1_{[0,t]}(\xi_i)$$
where $\xi_1, \xi_2, \ldots$ are i.i.d. Uniform$(0,1)$. Alternatively, we see that $n\mathbb G_n(t) = \#\{i\in\{1,2,\ldots, n\} : \xi_i \leq t\}$ where $\#$ denotes the counting measure on a set.

Question 1: at what $\omega\in \Omega$ are we evaluating the $\xi_i$? Intuitvely I think we can interpret this as we take random values in $(0,1)$, and count how many of them are $\leq$ some given $t$. However, in the context of measure theory and measure spaces, I don't know how to interpret the above statements.

Second, if we have $X_1, X_2, \ldots$ i.i.d. with c.d.f. $F(x)$, we can define
$$\mathbb F_n(x)=\frac 1n \sum_{i=1}^n 1_{(-\infty, x]}(X_i) = \frac 1n \#\{i\in\{1,\ldots, n\} : X_i \in (-\infty, x]\}$$
Again, I don't know what $\omega$ we are evaluating at, but that's already Question 1. 

Question 2: my book says that $\mathbb F_n(x)=\mathbb G_n(F(x))$; however, I don't see why
  $$\#\{i\in\{1,2,\ldots, n\} : \xi_i \leq F(x)\} = \#\{i\in\{1,\ldots, n\} : X_i \in (-\infty, x]\}$$
  I mean I think I understand that $P([\xi_i \leq F(x)])=P([X_i \in (-\infty, x]])$, but I don't see how the above statement is true. (Perhaps this is also because I don't understand what $\omega$ to evaluate at!)

 A: Question 1: So, since most of these arguments are related to the distributions, the actual underlying $\omega$ does not matter. That is: to state something happens almost surely, you just need to know that the probability of it is 1. This shows up similarly in the strong law of large numbers, if you were comfortable with the statement of that.
If you really want to see it defined on some measure spaces, you can take this canonical construction. Let your probability space be $(\Omega, \mathcal{A}, P) = ([0,1], \mathcal{B}[0,1], \lambda)$ the unit interval equipped with the Borel $\sigma$-field and Lebesgue measure.
For each $\omega \in \Omega$, represent it as a binary sequence. (i.e. $\omega = 0.x_1 x_2 x_3 x_4 \dots$...) These $x_i$ are a countable sequence of independent Bernoulli fair coin flips. Since it's countably infinite, you can break it up again into countably many sequences of independent Bernoulli coin flips. (Partition $\mathbb{N}$ into countably many disjoint sets and let these be the indices for each sequence.) Let $\xi_i$ have the binary expansion created from each of these sequences. Voilà! You've created countably many independent uniform random variables.
Question 2: Likely something buried in this book would be the definition of how the $X_i$ are generated. The equality you stated will clearly hold in distribution regardless. 
However, if you want the stated equality to actually hold on $\Omega$, you can take $X_i(\omega) = F_X(\xi_i(\omega))$. This is essentially using this trick to generate samples with a given CDF from samples of a uniform distribution. For more details: (https://en.wikipedia.org/wiki/Inverse_transform_sampling)
