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I'm learning Kolmogorov's zero-one law in probability theory:

Let $(Ω,{\mathcal F},P)$ be a probability space and let $F_n$ be a sequence of mutually independent $\sigma$-algebras contained in $\mathcal{F}$. Let $$G_n=\sigma\bigg(\bigcup_{k=n}^\infty F_k\bigg)$$ be the smallest $\sigma$-algebra containing $F_n, F_{n+1}, \dots$. Then Kolmogorov's zero-one law asserts that for any event $$ F\in \bigcap_{n=1}^\infty G_n$$ one has either $P(F) = 0$ or $1$.

I've no idea how $G_n$ and $\bigcap_{n=1}^{\infty} G_n$ would look like. What's the point of such construction? Could any one come up with some concrete examples of how this theorem works?

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Small question, how do we know that $\sigma(\cup_{k=n}^\infty F_k)$ is "the smallest $\sigma$-algebra containing $F_n,F_{n+1},...$"? For example, $\sigma(A\cup B)$ is not the smallest $\sigma$ algebra containing $A,B$, is it? – BlueBuck Oct 4 '15 at 16:23
@BlueBuck:… – Jack Oct 5 '15 at 0:09
Sorry but I don't see how that article answers my question. I understand what the $\sigma$-algebra generated by a family (such as $F_n,F_{n+1},...$) means, I just don't understand why the $\sigma$-algebra generated by the union of sets in that family gives the $\sigma$-algebra generated by the family. – BlueBuck Oct 5 '15 at 18:21
@BlueBuck I think: 1 The smallest $\sigma$-algebra containing $A,B$ is $\sigma(A,B) := \sigma(\sigma(A) \cup \sigma(B))$, if $A$ and $B$ are events. 2 $\sigma(A,B) = \sigma(A \cup B)$ if $A$ and $B$ are $\sigma$-algebras – BCLC Jan 24 at 13:18
up vote 13 down vote accepted

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).

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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).

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May I know how you define "atomic" here? Thanks – Theta33 Feb 8 '13 at 7:46
@Theta30 It's defined above: "An atom in a $\sigma$-algebra is a nonempty measurable set with no nonempty measurable proper subset." – Michael Greinecker Feb 8 '13 at 8:13
I refer to "atomic" as used in Lemma. – Theta33 Feb 8 '13 at 8:22
@Theta30 Oh, sorry- Iedited it now. Atomic means that every point lies in some atom. – Michael Greinecker Feb 8 '13 at 8:26
Dumb question: How does this answer OP's question? – BCLC Jan 24 at 13:19

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