Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 27G (Probability and Measure)

Let $(X_n)$ be independent random variables and letbe their tail sigma-algebra. The Kolmogorov zero-one law states that every $A\in \mathcal{T}$ has probability zero or one.

Indeed, $A$ is independent of $\sigma (X_1,\dots ,X_n)$ for every $n$. The union of these finite-coordinate sigma-algebras generates $\sigma (X_1,X_2,\dots )$, and independence extended from generating pi-systems shows that $A$ is independent of that entire sigma-algebra. Since $A$ itself belongs to it,so $\mathbb{P}(A)\in \{0,1\}$.

On a probability space carrying an infinite IID sequence $(X_n)$ with law $m$, map $\omega$ to $(X_1(\omega ),X_2(\omega ),\dots )$. Its pushforward measure $\mu$ satisfiesfor every cylinder set. Cylinder sets form a pi-system generating $\sigma (\mathcal{C})$, so the sigma-finite uniqueness theorem for measures proves uniqueness. This is the countable product measure $m^{\otimes \mathbb{N}}$.

For a cylinder $A=A_1\times \cdots \times A_N\times \mathbb{R}\times \cdots$,The same generating-class argument extends this equality to every measurable set, so $\theta$ is a measure-preserving transformation.

If $A={\theta }^{-1}A$, then $A={\theta }^{-n}A$ for every $n$, so membership in $A$ is independent of the first $n$ coordinates. Thus $A$ belongs to the tail sigma-algebra. The zero-one law gives $\mu (A)\in \{0,1\}$, proving that the shift is ergodic.