Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 26K (Probability and Measure)

A measure-preserving transformation $T$ on a probability space $(X,\mathcal{A},m)$ is ergodic if every invariant event $A\in \mathcal{A}$, meaningup to an $m$-null set, satisfies $m(A)\in \{0,1\}$. Thus the invariant sigma-algebra is trivial modulo null sets.

The Kolmogorov zero-one law says that for independent random variables $(Y_n{)}_{n\ge 0}$, every event in the tail sigma-algebrahas 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 $N$,so $A\in \mathcal{T}$. The zero-one law therefore says that the canonical one-sided i.i.d. shift is ergodic.

For a Borel set $A\subseteq [0,1]$, the two inverse branches of the tent map are $y\mapsto y/2$ and $y\mapsto 1-y/2$. Apart from their immaterial common endpoint,Each branch scales Lebesgue measure by $1/2$, and their images lie in opposite half-intervals. HenceAlso $m([0,1])=1$, so $m$ is an invariant probability measure.

The event $\{X_n=1\}$ is $T^{-n}(0,1/2)$. Since $T^n$ has $2^n$ affine branches of slope magnitude $2^n$, this preimage is, up to endpoints, a disjoint union of $2^n$ intervals of length $2^{-(n+1)}$ and has measure $1/2$.

More generally, for every binary word $(a_0,\dots ,a_{N-1})$, its itinerary cylinderis, up to finitely many endpoints, one monotonicity interval of $T^N$ and has measure $2^{-N}$. Thereforeso the $X_n$ are independent and identically distributed random variables, each with the Bernoulli distribution of parameter $1/2$.

The finite itinerary cylinders have diameters at most $2^{-N}$. Their endpoints form a dense set, so their unions generate the Borel sigma-algebra after completion by null sets. Thusin the standard probability-space convention that σ-algebras are identified modulo null sets.

There is a small literal endpoint defect in the uncompleted formulation of the question: $0$, $1$, and $2/3$ all have the all-zero itinerary, so $\sigma (X_0,X_1,\dots )$ cannot separate these Borel singletons. Consequently the displayed equality is false as an equality of raw Borel σ-algebras with the stated open interval, but it is true after completion and modulo null sets, which is the version used in the ergodicity argument.

Yes. Let $A$ be $T$-invariant. Since the itinerary variables generate $\mathcal{A}$ modulo null sets, $A$ can be regarded as an event in the canonical sequence model. For every $N$, invariance and $X_n\circ T=X_{n+1}$ implyThus $A$ belongs to the tail sigma-algebra of the i.i.d. sequence $(X_n)$. By the Kolmogorov zero-one law, $m(A)$ is zero or one. Hence the tent map is an ergodic measure-preserving transformation.