The <Kolmogorov zero-one law> says that for <independent random variables> $(Y_n)_{n\geq0}$, every event in the <tail sigma-algebra>
$$
\mathcal T=\bigcap_{N\geq0}\sigma(Y_N,Y_{N+1},\ldots)
$$
has <probability measure>[probability] zero or one.

In the <canonical model of an independent and identically distributed sequence>, the sample space is a sequence space $E^{\mathbb N}$ with the product law, $Y_n$ is the $n$th coordinate, and the left <Bernoulli shift> $S$ satisfies $Y_n\circ S=Y_{n+1}$. If $A$ is shift-invariant, then for every $N$,
$$
A=S^{-N}A\in\sigma(Y_N,Y_{N+1},\ldots),
$$
so $A\in\mathcal T$. The zero-one law therefore says that the canonical one-sided i.i.d. shift is <ergodic measure-preserving transformation>[ergodic].

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 26K / b / Solution

= Canonical model of an iid sequence
{synonym}

For a probability law $\nu$ on $E$, take the sequence space $E^{\mathbb N}$ with <product measure> $\nu^{\otimes\mathbb N}$. The coordinate maps are an <independent and identically distributed random variables>[independent and identically distributed sequence] with common law $\nu$, and they generate the product <sigma-algebra>.

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