How should I understand the $\sigma$-algebra in Kolmogorov's zero-one law?

Solution 1:

The usual name for the sigma algebra you have there is the "tail" $\sigma-$algebra. It is all the stuff that does not depend on finitely many of the $\sigma-$ algebras.

I usually think about the Kolmogorov 0-1 law like this: if you have a sequence of independent random variables and an event that is invariant if you ignore finitely many of the variables, then the probability of that event is either 0 or 1.

A typical example of how you can actually use this is to show that the convergence in the classical central limit theorem cannot be almost sure.

Let $X_i$ be a sequence of $iid$ random variables with $EX_i = 0$ and $EX_i^2 = 1$ and let $$S_n = \sum_{i=1}^n X_i$$

It is easy to see that for any fixed $n$ and $M>0$ the event $$\{ \limsup_k \frac{S_k}{\sqrt{k}} >M\}$$ lies in $$\sigma(\cup_{k=n}^\infty F_k)$$ since it only depends on the "tail" of our sequence (this is clear if you write out the definition of $\limsup$). Therefore by independence, the probability of this event is either 0 or 1. Using the central limit theorem, it is not hard to see that this is probability is positive for any fixed $M$ and therefore it is $1$. We then obtain that $$P(\limsup \frac{S_k}{\sqrt{k}} = \infty) = 1$$

Symmetry gives that $$P(\liminf \frac{S_k}{\sqrt{k}} = - \infty) = 1$$ so we cannot have almost sure convergence (or convergence in probability).

Solution 2:

Even though the tail-$\sigma$-algebra is a sub-$\sigma$-algebra of $\mathcal{F}$, it can be in some sense much more complex. In particular, the tail-$\sigma$-algebra may not countably generated even in the most well-behaved cases. This implies that there cannot be any (real-valued) random variable representing the informational content of the tail-$\sigma$-algebra in general.

An atom in a $\sigma$-algebra is a nonempty measurable set with no nonempty measurable proper subset. A measurable space is atomic if every point is contained in some atom (the atoms form a cover).

Lemma: If $(\Omega,\mathcal{F})$ is countably generated, then it is atomic. If there is a $0-1$-valued measure $\mu$ on $(\Omega,\mathcal{F})$, then there exists an atom $A$ such that $\mu(A)=1$.

Proof (Sketch): Let $\mathcal{C}$ be a countable family such that $\sigma(\mathcal{C})=\mathcal{F}$. We can assume without loss of generality that $\mathcal{C}$ is closed under complements. Then for each $\omega\in\Omega$, the set $A(\omega)=\bigcap\{A\ni\omega:A\in\mathcal{C}\}$ is measurable and the atom containing $\omega$. Now if $\mu$ is $0-1$-valued, the for each $A\in\mathcal{C}$ either $\mu (A)=1$ or $\mu (A^C)=1$. The intersection of all measure one events in $\mathcal{C}$ is an atom with measure one.

Corollary: The tail-$\sigma$-algebra on $\Omega=\{0,1\}^\infty$ is not countably generated.

Proof: Endowe the space with the fair coin-flipping measure. Let $\omega=\{\omega_1,\omega_2,\omega_3,\ldots\}$ be any element in $\Omega$. Let $A_\omega^n$ be the set of all sequences in $\Omega$ that coincide with $\omega$ in all coordinates except for, maybe, the first $n$ coordinates. Clearly, $A_\omega^n$ is countable and so is $B_\omega=\bigcup_n A_\omega^n$. So $\mu (B_\omega)=0$. Also, $B_\omega$ is in the tail $\sigma$-algebra. But since $\omega$ was arbitrary, this shows that every atom would have measure zero. By the Lemma, the tail-$\sigma$-algebra is not countably generated.

The argument is taken from Borel Spaces by Rao and Rao (1981).