Kolmogorov’s 0-1 law

For a probability space $(\Omega ,\mathcal{F},P)$ and a squence of sigma algebras ${\mathcal{F}}_1,{\mathcal{F}}_2,\cdots \subset \mathcal{F}$ are independent and $A\in \mathcal{T}$ where $\mathcal{T}$ is the tail field then $P(A)=0\text{or }1$.

Proof

The idea behind the proof is that we can show that $A$ is independent from itself. We do this by breaking up the series of $\sigma$ -algebras at an index $n$.

Let the event $A\in \mathcal{T}$. Then we know that for the sigma algebra

$$\mathcal{H}=\sigma \left(\bigcup _{j=1}^{\infty }{\mathcal{F}}_j\right)$$

$A\in \mathcal{H}$ because $A$ is in the tail field.

We want to show that $A$ is independent from every event in $\mathcal{H}$.

Note that if we have three collections of events that are independent from each other, then combining two of them into one set will give us 2 independent collections of events. (This is a lemma that should be proved, but not for my notes :) ).

Using this idea, we can combine the first $k$ $\sigma$ -algebras and the next $k-r$ for $r>k$ into two independent ones. In other words,

$$\sigma \left(\bigcup _{n=1}^k{\mathcal{F}}_n\right)\perp \sigma \left(\bigcup _{n=k+1}^r{\mathcal{F}}_n\right)$$

Now let $r\to \infty$ then we have

$$\sigma \left(\bigcup _{n=1}^k{\mathcal{F}}_n\right)\perp \sigma \left(\bigcup _{n=k+1}^{\infty }{\mathcal{F}}_n\right)$$

Note here that $k$ is arbitrary, and $A$ is an element of the second $\sigma$ -algebra, then as $k$ gets very large we see that $A\perp A$. This means that $P(A)=P(A)P(A)$ so $P(A)$ must be 0 or 1.