Is or when is $\mathbb{P}(X_1 \in A_1, ... X_n \in A_n)$ equivalent to $\mathbb{P}(A_1 \bigcap ... \bigcap A_n)$? In the context of independence:
Is $\mathbb{P}(X_1 \in A_1, ... X_n \in A_n)$ equivalent to $\mathbb{P}(A_1 \bigcap ... \bigcap A_n)$?
$X_i$s are random variables, $A_i \subset \Omega$ (the sample space) 
I read (Independence) that this could be possible using indicator functions $\mathbb{1}_{A_i}$, which is claimed to turn
$$\mathbb{P}(X_1 \in A_1, ... X_n \in A_n)=\mathbb{P}(X_1 \in A_1)...\mathbb{P}(X_n \in A_n)$$
into
$$\mathbb{P}(A_1 \bigcap ... \bigcap A_n)=\mathbb{P}(A_1)...\mathbb{P}(A_n)$$
True?
For example:
Can one e.g. construct the argument (the one with $S = \{1,2,3,4\}$) on p. 13 here using $X_i \in A_i$ rather than $ A_i \bigcap A_j$.
 A: Not in general.
Consider, for example, the one index case, with probability space the interval $[-1, 1]$ holding the uniform measure.
Let $Z$ be the coordinate variable, $X = Z^2$, and $A = [0.5, 1]$.  Then:
$$ P(X \in A) = P([-1, -\sqrt{0.5}] \cup [\sqrt{0.5}, 1]) \approx 0.293$$
while
$ P(A) = 0.25$

I wonder why I'm seeing two definitions for independence. Some using intersections and some the inclusions. Or are these different variations of independence (I've heard the term "mutually independent")?

I have some time to elaborate more on my comment now.
You must distinguish between independents of events and independence of random variables.
Give an probability space $\Omega$ (a.k.a sample space, if that's your preferred term), a subset $A$ which is assignable a probability (a.k.a. a measurable set) is called an event.  Two events $A_1, A_2 \subset \Omega$ are said to be independent when
$$ P(A_1 \cap A_2) = P(A_1) P(A_2) $$
Now, if we have a random variable on the probability space $X: \Omega \rightarrow R$, then any subset of the range of $X$ determines an event, we may call it the pre-image event.  To be specific, for an interval $[y_1, y_2]$ in the range, we may associate the event
$$ \{\omega \in \Omega : X(\omega) \in [y_1, y_2] \} $$
To save keystrokes, we can abbreviate this event
$$ X \in [y_1, y_2] $$
or for a general $A \subset R$
$$X \in A$$
Now we can define independence of two random variables $X_1, X_2$ in terms of these pre-image events.  Two random variables are independent if, for any two subsets $A_1, A_2$ of $R$ (notice the switch in what space the $A$s are a subset of, which @Chaconne alluded to in e's comment)
$$ P(X_1 \in A_1, X_2 \in A_2) = P(X_1 \in A_1) P(X_2 \in A_2) $$
The comma here is an abbreviation for and, it looks better on paper than the possibly more pedantic
$$ P(X_1 \in A_1 \cap X_2 \in A_2) = P(X_1 \in A_1) P(X_2 \in A_2) $$

I read (Independence) that this could be possible using indicator functions $1_A$.

We can make sense of this now.  To any event $A \subset \Omega$ we may associate the indicator function $1_A: \Omega \rightarrow \{0, 1\} \subset R$.  It is defined by the two relations $\omega \in A \Rightarrow 1_A(\omega)=1$, and the complementary $\omega \notin A \Rightarrow 1_A(\omega)=0$.  Since this indicator function is defined on a  probability space, it is a random variable.
If you have two such events, giving indicators $1_{A_1}$ and $1_{A_2}$, you will see, upon working through the definitions above, that independence of these random variables is equivalent to independence of the underlying events.
A: It is not true that $P(\bigcap_n X_n \in A_n) = P(\bigcap_n A_n)$ even if we include the fact that $X_n$'s or $A_n$'s are independent. I'm not even quite sure that such equation is meaningful.
Perhaps what you meant to do is define the events $B_n := (X_n \in A_n)$. Then we can say that:
$$P(\bigcap_n X_n \in A_n) = P(\bigcap_n B_n)$$
Note that the $A_n$'s are not events but Borel sets while the $B_n$'s are events.
Consider a coin toss that pays off 1 if you flip heads and -1 o/w. We can denote the payoff as a random variable $X$.
Note that the event $(X \in (0,2))$ is equal to the event $H$.
We do not have that $P(X \in (0,2)) = P(0,2)$ because in the first place $P(0,2)$ is not defined because $(0,2)$ is not an event.

I guess we can consider the case where the Borel sets are also measurable. It's still not true.
Consider the probability space $(\mathbb R, \mathscr B(\mathbb R), \lambda)$.
Thus, the events $B \in \mathscr B(\mathbb R)$ are also Borel sets.
Consider the random variable $X := 2 \times 1_{(0,1)}$, the event and Borel set $B = (0,1)$ and $\omega = 0.5$. Then $X(\omega) = 2$. It is clear that we do not have that
$$X(\omega) \in B \iff \omega \in B$$

On independence:



*

*Two events $E$ and $F$ are said to be independent if $P(E\cap F)=P(E)P(F)$ holds. More generally, the events $E_1,E_2,\dots$ are said to be independent if, for every subset $E_{n_1},E_{n_2},\dots,E_{n_r},r \in \mathbb{N}$, of these events, $$P\left(\bigcap_{i=1}^rE_{n_i}\right) =\prod_{i=1}^rP(E_{n_i})$$





*Two collections of events (eg $\sigma$-algebras) $\mathscr F$ and $\mathscr E$ are said to be independent if $\forall F \in \mathscr F$ and $\forall E \in \mathscr E$, $P(E\cap F)=P(E)P(F)$. More generally, the collections $\mathscr E_1,\mathscr E_2,\dots$ are said to be independent if, $\forall E_{n_1} \in \mathscr E_{n_1}, \forall E_{n_2} \in \mathscr E_{n_2}, \dots, \forall E_{n_r} \in \mathscr E_{n_r}, r \in \mathbb{N}$, $$P\left(\bigcap_{i=1}^rE_{n_i}\right) =\prod_{i=1}^rP(E_{n_i})$$





*Two random variables $X$ and $Y$ are said to be independent if their $\sigma$-algebras, $\sigma(X)$ and $\sigma(Y)$, are independent. More generally, the random variables $X_1,X_2,\dots$ are said to be independent if, for every distinct indices $n_1,n_2,\dots,n_r,r \in \mathbb{N}$, $$\sigma(X_{n_1}), \sigma(X_{n_2}), \dots, \sigma(X_{n_r}) \ \text{are independent}$$





*Mutually independent is the same as independent. Pairwise independent means for some events, collections of events or random variables denoted by $A_1, A_2, ...$ and distinct indices $m, n \in \mathbb N$, $A_m$ and $A_n$ are independent.





*The events $E_1, E_2, ...$ are independent if and only if the random variables $1_{E_1}, 1_{E_2}, ...$ are independent because $\sigma(1_{E_i}) = \sigma(E_i)$

