Oliver Knill's book proves the famous Kolmogorov's 0-1 law in section 2.2. The statemente is as follows. Let $(\Omega, \mathcal{A},p)$ be a probability space and $\{\mathcal{A}_{\alpha}\}_{\alpha \in I}$ a family of sub-$\sigma$-algebras of $\mathcal{A}$. To each non-empty $J\subset I$, set $\mathcal{A}_{J}$ to be the $\sigma$-algebra generated by the union $\bigcup_{\alpha\in J}\mathcal{A}_{\alpha}$ and, if $J=\emptyset$ then $\mathcal{A}_{\emptyset}=\{\emptyset, \Omega\}$. The tail $\sigma$-algebra $\mathcal{T}$ is defined by $\mathcal{T}:= \bigcap_{J\subset_{f}I}\mathcal{A}_{J^{c}}$, where $\subset_{f}$ indicates the set $J$ is finite. Then:
Kolmogorov's 0-1 Law: If $\{\mathcal{A}_{\alpha}\}_{\alpha\in I}$ is a family of independent sub-$\sigma$-algebras then $\mathcal{T}$ is trivial, in the sense that $p(A) = 0$ or $p(A) = 1$ for every $A \in \mathcal{T}$.
I might be wrong, but I guess a more common version of Kolmogorov's 0-1 law is the following.
Kolmogovov's 0-1 law: [Version 2]: A sequence $\{X_{n}\}_{n\in \mathbb{N}}$ of independent random variables has a trivial tail $\sigma$-algebra.
The tail $\sigma$-algebra, in version two, is usually defined in the following way. If $\sigma(X_{n},X_{n+1},…)$ is the smalles $\sigma$-algebra in which every random variable $X_{n},X_{n+1},…$ is measurable, then the tail $\sigma$-algebra $\mathcal{T}':= \bigcap_{m=n}^{\infty}\sigma(X_{m},X_{m+1},…)$.
Here's my problem. I've been following Knill's book and understoof his version of Kolmogorov's 0-1 law. However, I want to show that this version implies version two (but does it?). Here's the tricky part for me. The two tail-$\sigma$-algebras considered seems a little different to me. For instance, if $I=\mathbb{N}$ the tail-$\sigma$-algebra on Knill's book considers every finite subset of $\mathbb{N}$ while on the second version only those $J$ like $\{1,…,n\}$ are being considered. So, are these two results equivalent (as I'd expect) or is Knill's version weaker than the second version (in the sense that it does not imply the second vesion)?
Best Answer
It is trivially seen that $\bigcap_{J\subseteq_{f}I} A_{J^c}=\mathcal{T}\subseteq \mathcal{T}'.$ To see that the two sigma-algebras are equal, notice that given a finite $J,$ there exists $n$ such that $J\subseteq [n].$ In other words, $A_{[n]^c}\subseteq A_{J^c}$ and hence $A_{[n]^c}\subseteq \bigcap_{J: J\subseteq [n]}A_{J^c}.$ We therefore obtain that $\mathcal{T}'\subseteq\bigcap_{n}A_{[n]^c}\subseteq \bigcap_{J\subseteq_{f}I}A_{J^c}=\mathcal{T}.$