Interpreting tail events and their "dependence" on finite r.v.'s Let $X_1,X_2...$ be random variables defined on the same sample space. Let $$\mathcal{F}_n = \sigma(X_n,X_{n+1}...), \mathcal{F} = \bigcap_n\mathcal{F_n}$$ where the latter is the tail $\sigma$-field.

I am trying to understand what we mean when we say that a tail event $A \in \mathcal{F}$ does not depend on any finite set $\{X_1...X_n\}$. From a previous answer, this is my "intuitive" understanding:
If $A \in \mathcal{F}$, then for all $n \in \mathbb{N}$, $A \in \mathcal{F_n}$. But since $\mathcal{F_n} \subset \mathcal{F_1}$, we see that "omitting" the first $n-1$ $X_i$'s does not affect our event being in $\mathcal{F}$. Thus, if we think about the $\sigma$-fields as "current knowledge" about our space, we see that this "current knowledge" is not "dependent" on $X_1 ... X_{n-1}$.
However, I am not sure how this can be made rigorous. In particular,
(i) is the term "dependent" being used here comparable to how we speak of independence of events and random variables? What is the mathematical definition of dependent here? Or is just "dependent" in an intuitive sense?
(ii) What justifies seeing the $\sigma$-field as our "current knowledge" about the space? I know that $\sigma(X_n,X_{n+1}...)$ is the collection of sets $[\omega \in \Omega : X(\omega) \in B]$, where $X = (X_n,X_{n+1}...)$ and $B$ is a Borel set in $\mathbb{R^{\mathbb{N}}} $. I suppose this is linked to the view of a $\sigma-$field as a set of events, but what do we mean by "current knowledge"
(iii) Finally, how can I view (i) and (ii) together? In math, what does it mean for our current knowledge -the $\sigma$-field $\sigma(X)$ - to not "depend" on $X_1...X_n$?

If it helps for you to explain, I suppose a helpful and simple example to think might be the event $\{\omega \in \Omega: \sum X_i(\omega) < \infty\}$  - I understand why this does not "depend" on the value that finitely many $X_i$ take from an analysis perspective, but now I am trying to understand it from this probability theory perspective.
 A: One way of conceptualising events that foreground the "information" heuristic is thinking of a sample space $\Omega$ as all possible outcomes of a random experiment, and an event $A$ as a yes/no question about a realised outcome $\omega \in \Omega$.
Informallly, the question would be "is $\omega \in A$?"
A $\sigma$-algebra $\mathcal{F}$ can then be thought of as a set of yes/no questions about $\omega$ that we know the answer to assuming that we can synthesise a countable number of answers.
You can reconcile the $\sigma$-algebra axioms with this this intuitive description.
In this way, $\sigma(X_n)$ is the set of yes/no questions that we can answer about $\omega$ if we only know the value of $X_n(\omega)$.
Now, random processes $(X_1, X_2, \dotsc)$ are linearly structured.
That is, we usually think of observing a process not all at once, but at a specific time $n$, at which point we can observe "past" values $(X_1, \dotsc, X_n)$ but not "future" values $(X_{n+1}, X_{n+2}, \dotsc)$.
In this way, the $\sigma$-algebra $\mathcal{F}_n = \sigma(X_1, \dotsc, X_n)$ is the set of yes/no questions that can be answered if we know the values of the process $X$ up to time $n$.
But this is exactly what we would want in a definition of "current knowledge about $X$ at time $n$."
In this framework, phrases like "$\mathcal{F}_n$ doesn't depend on $X_{n+1}$" mean that the value of $X_{n+1}$ never influences the answer to any question in $\mathcal{F}_n$.
This has nothing to do with statistical dependence, as its perfectly possible (and  is the case in most non-trivial examples) that $X_{n+1}$ is not statistically independent of $\mathcal{F}_n$ - it's just that the word "depends" is unfortunately overloaded in statistics.
In the case of the tail $\sigma$-algebra, $\mathcal{F}$ is "the set of questions about the process $X$ whose answers are not influenced by any finite set of values $(X_1, \dotsc, X_n)$".
Another way of thinking about $\mathcal{F}$ is that it is the set of questions that we would be able to answer if we knew all of $X$, but can never answer by stepping forward linearly in time and observing the current information $\mathcal{F}_n$ at each time point $n$.
A: I think @Damian Pavlyshyn's answer is quite good already, but maybe I can rephrase things a little and add some more examples to help. Like Damian said, $\Omega$ can be thought of as all the outcomes of a certain experiment you're doing. For example, $\Omega$ is the set of all points on a dartboard, and the experiment you are performing is throwing a dart at the dartboard. Then, every $\omega\in \Omega$ can be identified with one trial of the experiment, namely the trial in which you get the outcome $\omega\in \Omega$.
Then, a random variable $X: \Omega \to \mathbb R$ can be thought of as some instrument that can measure the outcome of your experiment. For example, the dartboard could have rings of scores, like $100$ points for the center, $90$ points for close to the center, etc. and $X(\omega)$ could be this score. The $\sigma$-algebra $\mathscr B$ (normally Borel sets of course) on $\mathbb R$ represents all the questions you can ask about the measurement you receive from $X$. For example if $\mathscr B$ is $\{\varnothing, (-\infty, 0), [0,\infty), \mathbb R\}$, maybe you can think of this as the user-interface of the instrument $X$ is really bad and can only gives very coarse readings.
(Normally for the Borel sets $\mathscr B$, basically any question you can ask about the reading $X$ gives you is valid --- though you can't ask cursed questions like "is $X(\omega)$ in some Vitali set", for technical reasons. In fact actually if you abandon the Axiom of Choice you don't need to worry about these technicalities, and in fact are allowed to ask "any question".)
Then the pullback $\sigma$-algebra on $\Omega$, namely $\sigma(X)$, is just pulling back all questions "is $X(\omega) \in B$" to "did $X$ measure the outcome of the trial $\omega$ of the experiment to be in $B$". In other words, using $X$ to measure some facet of the outcomes of your experiment, you gained some information on the outcome of each trial of your experiment, and $\sigma(X)$ represents exactly all this information you've gained. For example, you know know the answer to "did this particular trial $\omega_0$ result in $X$ measuring at least $>90$ points", or "which trials of the experiment resulted in $X$ measuring $0$ points".
Now thinking about a random process, pretend like you are again running an experiment, with $\omega\in \Omega$ denoting each trial. Every day, you find a new instrument $X_i$ to measure your experiment with. So the first day, you now know the answers corresponding to $\sigma(X_1)$, and on the second day, you now know the answers corresponding to $\sigma(X_1,X_2)$, and so on. For example, by day 3, you know the answer to "which trials of the experiment result in all three measurements having $>90$ points", which corresponds to the set $\{\omega\in \Omega: X_1(\omega),X_2(\omega),X_3(\omega)\} \in \sigma(X_1,X_2,X_3)$. And then the probability $P(\{\omega\in \Omega: (X_1,X_2,X_3)(\omega) \in (B_1,B_2,B_3)\})$ is exactly the answer to "what fraction of the trials of my experiments resulted in measurements $X_i$ landing in $B_i$, for $i=1,2,3$" (the measure-theoretic approach to probability is indeed a "frequentist" approach).
In contrast, $\{\omega\in \Omega : X_4(\omega)=100\}$ is not $\sigma(X_1,X_2,X_3)$-measurable, because we literally have not measured $X_4$ yet, i.e. that information is not accessible from the day 3 information $\sigma(X_1,X_2,X_3)$, unless of course $X_4$ is completely determined by the results of $X_1,X_2,X_3$, i.e. $X_4=f(X_1,X_2,X_3)$ for some (measurable) function $f: \mathbb R^3 \to \mathbb R$.
So we see that $\mathscr F_N:=\mathscr F_{\leq N}:=\sigma(X_1,\ldots, X_N)$ is all the information we now know about all the trials of our experiment by day $N$. Similarly, $\mathscr F_{> N}:=\sigma(X_{N+1},\ldots)$ is all the information we know about the trials of our experiment if we only know the results of days $N+1, N+2$, through infinity. But we really only like using the words "current knowledge" to talk about "forward-time" processes like $\mathscr F_1,\mathscr F_2,\ldots$, since that matches our intuition of "learning new information as time passes".
Finally, we are ready to talk about the tail $\sigma$-algebra $\mathscr F:= \bigcap_{n=1}^\infty \mathscr F_n$. The key idea is this: the complete information you get by looking "with a birds-eye-view" at the totality of your experiment, namely $\sigma(X_1,X_2,\ldots, )$ (i.e. the information/questions you could answer if you knew $X_i(\omega)$ for every trial $\omega\in \Omega$ and every measurement $i\in \mathbb N^+$), is DIFFERENT from union of the information you learn at each finite day, namely $\bigcup_{n=1}^\infty \sigma(X_1,\ldots, X_n)$! For example, only knowing the union of all information from each finite day, you can never be sure that the sequence $\{X_i(\omega)\}_{i=1}^\infty$ converges (for given fixed $\omega\in \Omega$). But you WOULD know this information if you had the total/complete information $\sigma(X_1,X_2,\ldots, )$. The tail $\sigma$-algebra is exactly the difference $\sigma(X_1,X_2,\ldots, ) \smallsetminus \bigcup_{n=1}^\infty \sigma(X_1,\ldots, X_n)$, i.e. all information/questions you could answer if you knew the totality of EVERY measurement/trial altogether, but that you would not know on any given day $n$.
Another way of phrasing this: in this form of the original intersection definition, $S \in \mathscr F \iff S \in \mathscr F_{>n}$ for every $n\in \mathbb N^+$, so we see that $\mathscr F$ is all the information I am sure I can learn about any given trial $\omega\in \Omega$ of your experiment, if I were to ask you a question, and require that you tell me all the readings $X_i(\omega)$ except perhaps finitely many (that you are free to choose to not tell me for whatever reason, where this choice can depend on what question I ask).
That's all the theory. Now for some examples: above I said $S:=\{\omega\in \Omega: \lim_{n\to\infty} X_n(\omega) \text{ exists}\}$ is in the tail $\sigma$-algebra $\scr F$ (verify that this makes sense with respect to both interpretations of $\scr F$ above). More explicitly, observe that for any $n\in \mathbb N^+$,  $S = \{\omega\in \Omega: \text{the limit of } X_{n+1}(\omega), X_{n+2}(\omega), \ldots \text{ exists}\}$, where the RHS is clearly in $\scr F_{>n}$. Similarly for $S:=\{\omega\in \Omega: \lim_{n\to\infty} X_n(\omega) \text{ exists and equals }a\}$ for $a\in \mathbb R$, or $\{\omega\in \Omega: \lim_{n\to\infty} X_n(\omega) \text{ exists and lies in }B\}$ for Borel $B \in \scr B$.
On the other hand, defining $S_n:= X_1 + \ldots + X_n$,  $S:=\{\omega\in \Omega: \lim_{n\to\infty} S_n(\omega) \text{ exists}\}$ is in the tail $\sigma$-algebra for the same reason, BUT  $S:=\{\omega\in \Omega: \lim_{n\to\infty} S_n(\omega) \text{ exists and equals }a\}$ for $a\in \mathbb R$, or $\{\omega\in \Omega: \lim_{n\to\infty} S_n(\omega) \text{ exists and lies in }B\}$ for Borel $B \in \scr B$ are NOT in the tail $\sigma$-algebra, since the precise value of $X_1(\omega)$ has a huge impact on the ultimate value of the limit $\lim_{n\to\infty} S_n(\omega)$. In other words, if you only tell me the readings $X_2(\omega_0), X_3(\omega_0),\ldots$, I'll be able to tell if $\lim_{n\to\infty} S_n(\omega_0)$ converges, but I won't be able to know what exactly it converges to!
Finally, $S:=\{\omega\in \Omega: \lim_{n\to\infty} \frac{S_n}{n}(\omega) \text{ exists}\}$ or $\{\omega\in \Omega: \lim_{n\to\infty} \frac{S_n}{n}(\omega) \text{ exists and equals }a\}$ for $a\in \mathbb R$, or $\{\omega\in \Omega: \lim_{n\to\infty} \frac{S_n}{n}(\omega) \text{ exists and lies in }B\}$ for Borel $B \in \scr B$ are all in the tail $\sigma$-algebra. Exercise for the reader :)
Remark: this last example is VERY important, since it is crucially used in (some) proofs of the strong law of large numbers (SLLN).
A: Here’s something that might help with intuition for (iii). The tail $\sigma$-field is characterized by a saturation property. An event $A\in \mathcal F$ is tail measurable if and only if for each $\omega\in A$, another element $\tilde\omega$ of the sample space lies in $A$ if and only if $X_n(\omega)=X_n(\tilde\omega)$ for all sufficiently large $n$ (i.e. for all $n\ge n_0(\omega,\tilde\omega)$).
