# Intuition behind the construction of a tail $\sigma$-field

Before starting I should note that I've already read this, this, this and this, and none of them quite provide the intuition I seem to be missing. I've tried to provide as many details related to my own thought process to see if we can diagnose this misunderstanding more specifically.

I just started learning about Kolmogorov's zero-one law, and (not unlike many people who first encounter it I suppose), I'm having trouble wrapping my head around the concept of a tail field. In particular, if we have a sequence of events $$A_n\in\mathcal{F}$$, then the tail field $$\tau$$ is defined as $$\tau = \bigcap_{n=1}^\infty\sigma(A_n, A_{n+1},\cdots)$$ First, I'm slightly confused as to how $$\tau$$ is a $$\sigma$$-algebra at all (though I think I can check this myself).

What I'm really struggling with is simply how to interpret this definition. To me, it seems like $$\tau$$ should be empty, even though I know it's clearly not as $$\limsup_n A_n\in\tau$$ (and I understand the proof behind why).

As far as I understand, if $$A\in\tau$$, then $$A\in\sigma(A_n,A_{n+1},\cdots)$$ for every $$n\in\mathbb{N}$$. So any $$A\in\tau$$ can't depend on the first $$A_1,\cdots, A_n$$ events for any $$n$$. The problem is that I have no intuition for what kinds of sets behave this way. It seems to me like if you have any formula for $$A$$ involving the sets $$A_n$$, then there has to be a minimum $$n_0$$ for which $$A_{n_0}$$ is involved in the formula for $$A$$, but this means $$A\not\in\sigma(A_{n_0+1},A_{n_0+2},\cdots)$$, meaning $$A\in\tau$$. So, in my head, $$\tau$$ can't contain any set that can be written in terms of the $$A_n$$'s.

As I said above, I know this is false, as $$\limsup_n A_n$$ and $$\liminf_n A_n$$ are in $$\tau$$. To be honest, the more I think about it the less sense the definitions of these two sets seem to make to me as well. I mean, what does $$\limsup_n A_n$$ even look like if it supposedly doesn't depend on any of the $$A_n$$'s? If it doesn't depend on any of the $$A_n$$'s, then $$\limsup_n A_n = \limsup_n B_n$$ for any two sequences of events $$A_n$$ and $$B_n$$, no?

I guess the flaw in my logic has to do with the idea of "dependence". I know there are events in $$\limsup_n A_n$$, and the logical proposition giving the inclusion of an element in $$\limsup_n A_n$$ does depend on the $$A_n$$'s, so what I said above makes no sense (obviously).

But something is still wrong in my head, and I want to figure out how to right it. Can anyone see where the flaw in my logic is?

## 1 Answer

Regarding your first question: one can show that the intersection of $$\sigma$$-algebras is also a $$\sigma$$-algebra by directly checking the definitions.

Regarding your broader questions, it may help to think about the analogy of a convergent sequence of real numbers: the limit does not depend on any individual element (or finite subset) of the sequence, but we definitely do not have $$\lim_n a_n = \lim_n b_n$$ in general for any two arbitrary convergent sequences $$a_n$$ and $$b_n$$.

• Ahhhhh, that limit analogy does help a bit. Let me think about it a bit more before accepting your answer right away, I might have a followup question. – user3002473 Oct 25 '18 at 20:20