Characterization of tail sigma algebra (and its complement)

Question

I am not able to extract the concept of a tail $\sigma$-algebra from its definition. So, given a probability space $(\Omega,\mathcal{A},P)$ and sub-$\sigma$-algebrae $\mathcal{A}_1,\mathcal{A}_2,\dots$ (e.g. $\mathcal{A}_j:=\sigma(X_j)$ for r.v. $X_1,X_2,\dots$ on $\Omega$) I call $$\mathcal{C}_\infty:=\bigcap_{j=1}^\infty\sigma\left(\bigcup_{k=j}^\infty \mathcal{A}_k\right)$$ the tail $\sigma$-algebra.

So how would I extract from this defintion the property that allows people say "this event is not part of $\mathcal{C}_\infty$ because it depends on $X_1$" as I often read?

There must be some assumptions for this statement to hold, e.g. $(X_j=X_1,\,j\in\mathbb{N})$ would disallow the statement (right?); I feel the $\mathcal{A}_1,\mathcal{A}_2,\dots$ must somehow "separate" the space?

Can someone give me a criterion with proof that allows me talk conceptually about tail $\sigma$-algebrae? Thank you very much. (I don't have trouble to see why e.g. $\limsup$'s are in the tail $\sigma$-algebra, I rather have trouble with giving negative answers)

Answer 1

Here is one result that captures the intuition of the tail algebra, at least for me.

Define two sequences $x = (x_1, x_2, \ldots)$ and $y = (y_1, y_2, \ldots)$ to be tail equivalent if there exists an index $N$ such that for all $n \geq N$, $x_n = y_n$. Then for any measurable set $S$ in the tail algebra, $x \in S$ if and only if $y \in S$. So from the point of view of probability, $x$ and $y$ are indistinguishable.

This follows from this elementary result. Let $Y$ be a probability space. Consider the function $\pi: X \times Y \to Y$, $\pi( (x, y) ) = y$, and give $X \times Y$ the measure $\sigma(\pi)$. Then for a measurable set $S$, $(x, y) \in S$ iff $(x', y) \in S$. (This in turn follows from the definition of $\sigma(\pi)$.)

Answer 2

How about this as a simple heuristic, suppose there exists some $B\in \mathcal{B} = \sigma(\mathbb{R})$ (the Borel $\sigma$-algebra on the reals) such that $(X_{1}\in B)\setminus\left\{\bigcup_{k=2}^\infty (X_k \in B)\right\} \ne \emptyset$, so that there is some distinction between $X_1$ and what follows (they are certainly not identically the same as in your comment above, nor for that matter is each $X_j$ contained in its successor for example), then certainly: $$\sigma(X_1)\setminus(\bigcap_{j=2}^\infty\sigma(X_j))\ne\emptyset$$ Since there will definitely be events depending purely on $X_1$ and not on any of its successors. So in full generality then, $unless$ we have special conditions on the $X_j$ (eg inclusion in successors), we must have that the tail $\sigma$-algebra is $strictly$ smaller than $\sigma(X_1)$. Does that get to what you are asking or is it too simplistic? (And agreed we don't need a measure to define this). Cheers