Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 25J (Probability and Measure)

For a measurable map, each preimage ${\theta }^{-1}(A)$ lies in $\mathcal{E}$, so ${\mu }_{\theta }$ is well-defined and nonnegative. The preimage of the empty set is empty, and preimages commute with disjoint countable unions, givingfor pairwise disjoint measurable $A_j$. Thus ${\mu }_{\theta }$ is a measure, the pushforward measure ${\theta }_{\ast }\mu$. Its total mass is $\mu (E)$, because ${\theta }^{-1}(E)=E$. Finiteness is not needed for this construction, but is used in the uniqueness argument next.

If $\theta$ is measure-preserving, equality holds on every measurable set and hence on the specified $\pi$-system. Conversely, let $\mathcal{D}=\{A\in \mathcal{E}:{\mu }_{\theta }(A)=\mu (A)\}$. Both finite measures have equal total mass. Therefore $E\in \mathcal{D}$, complements of members remain in $\mathcal{D}$ by subtracting from the finite total mass, and disjoint countable unions remain in $\mathcal{D}$ by countable additivity. Thus $\mathcal{D}$ is a Dynkin system.

It contains the generating pi-system $\mathcal{A}$, so Dynkin lemma gives $\mathcal{E}=\sigma (\mathcal{A})\subseteq \mathcal{D}$. ConsequentlyThis is exactly the measure-preserving property.

The Birkhoff ergodic theorem states: for a measure-preserving transformation $\theta$ of a finite measure space and $f\in L^1$, the averages $A_{n}f=n^{-1}{\sum }_{j=0}^{n-1}f\circ {\theta }^j$ converge almost everywhere and in $L^1$ to an invariant function $f^{\ast }$. On a probability space,where $\mathcal{I}$ is the invariant sigma-algebra modulo null sets. In particular the integral is preserved, and if $\theta$ is ergodic then $f^{\ast }=\int fd\mu$ almost everywhere; for a finite space of nonunit mass, this constant is $\mu (E{)}^{-1}\int fd\mu$ when $\mu (E)>0$.

The maximal ergodic lemma says that for real integrable $f$, with $S_{k}f={\sum }_{j=0}^{k-1}f\circ {\theta }^j$ and $E_N=\{{\max }_{1\le k\le N}{S}_{k}f>0\}$,The same inequality holds on $E_{\infty }=\{{\sup }_{k\ge 1}{S}_{k}f>0\}$, by taking the increasing limit and dominated convergence. These statements do not require the transformation to be invertible or the space to be ergodic.

The map swaps each adjacent pair of binary digits: $({\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,\dots )\mapsto ({\omega }_2,{\omega }_1,{\omega }_4,{\omega }_3,\dots )$. Canonical expansions with no eventual all-ones tail remain canonical under this swap, and ${\theta }^2=\mathrm{id}$.

Under Lebesgue measure, any specified values of $m$ distinct binary digits have probability $2^{-m}$. Indeed they describe a union of dyadic intervals at the largest of the specified positions, with that total length. The preimage of a cylinder prescribing its first $m$ output digits prescribes $m$ distinct input digits and therefore has the same measure. The half-open dyadic intervals, with the empty set, form a pi-system generating the Borel sigma-algebra. Apply the preceding uniqueness result to obtainThe choice of binary expansion matters at dyadic endpoints for the map's pointwise definition; the chosen convention removes that ambiguity and those endpoints are null in the measure calculation.

Consider $A=\{x:{\omega }_1={\omega }_2\}=[0,1/4)\cup [3/4,1)$. The first digit pair is swapped by the transformation, so the equality of its digits is unchanged: ${\theta }^{-1}(A)=A$. Yet $\mu (A)=1/2$. An ergodic measure-preserving transformation has only invariant events of measure zero or one. ThusIts order-two property also explains why averaging along orbits cannot in general sample the entire interval.