The <Rudolph measure rigidity theorem> has an essential ergodicity hypothesis. If a Borel probability measure $\mu$ on the <circle group> is invariant under both $T_2$ and $T_3$, is ergodic for the semigroup generated jointly by these maps, and either map has positive <Kolmogorov-Sinai entropy>, then
$$
\boxed{\mu=m.}
$$
Equivalently, a jointly ergodic common invariant measure other than <Lebesgue measure> has zero entropy for both maps. Joint ergodicity means that every set invariant modulo $\mu$ under both maps has measure zero or one. Positive entropy without this hypothesis is insufficient: $\tfrac12m+\tfrac12\delta_0$, with $\delta_0$ a <Dirac measure>, is a common invariant measure of positive entropy and is not $m$.

The <Host equidistribution theorem> states that if $p,q\geq2$ are relatively prime integers and $\nu$ is invariant and ergodic under $T_p$, with $h_\nu(T_p)>0$, then for $\nu$-almost every $x$ the sequence $(T_q^nx)$ is an <equidistributed sequence> for <Lebesgue measure>. Explicitly, for every continuous $g$ on the circle,
$$
\boxed{\frac1N\sum_{n=0}^{N-1}g(T_q^nx)\longrightarrow\int g\,dm
\quad\text{for }\nu\text{-almost every }x.}
$$
The non-ergodic form assumes $T_p$ invariance and positive entropy for almost every component in the <ergodic decomposition> $\nu=\int\nu_\omega\,d\lambda(\omega)$. Applying the ergodic theorem of Host on each such component gives the same almost-everywhere conclusion for $\nu$. More generally, its conclusion holds on the part supported on positive-entropy components. A positive value of $h_\nu(T_p)$ alone does not eliminate zero-entropy components.

To deduce the joint version of the <Rudolph measure rigidity theorem>, suppose $h_\mu(T_2)>0$; if only $T_3$ has positive entropy, interchange the roles. Write the $T_2$ <ergodic decomposition> as $\mu=\int\nu\,d\tau(\nu)$. Since $T_3$ commutes with $T_2$, its <pushforward measure> sends a $T_2$ ergodic component $\nu$ to a $T_2$ ergodic component $(T_3)_*\nu$. On each component, $T_3$ is a <factor of a measure-preserving system> with fibres of size at most three. We use the standard <entropy preservation under a finite-to-one factor>:
$$
h_{(T_3)_*\nu}(T_2)=h_\nu(T_2).
$$
The reason for this standard entropy fact is that, conditional on a complete factor point, every finite orbit name has at most three possibilities; its conditional entropy is bounded by $\log3$, and division by the orbit length gives zero relative entropy.

The component at $T_3x$ is $(T_3)_*\nu_x$ <almost everywhere>; this follows from commutation and the componentwise ergodic averages. The component entropy function $r(x)=h_{\nu_x}(T_2)$ is therefore invariant under both $T_2$ and $T_3$. Joint ergodicity makes it constant <almost everywhere>, and <affinity of entropy under ergodic decomposition> identifies the constant as $h_\mu(T_2)>0$. Thus almost every $T_2$ component has positive entropy, exactly the condition required in the non-ergodic <Host equidistribution theorem>. It follows that $\mu$-almost every point equidistributes for $m$ under $T_3$.

For any continuous $g$, invariance under $T_3$ and the <dominated convergence theorem> now give
$$
\int g\,d\mu
=\int\frac1N\sum_{n=0}^{N-1}g(T_3^nx)\,d\mu(x)
\longrightarrow\int g\,dm.
$$
Continuous functions determine Borel probability measures on the circle, so $\mu=m$, proving the deduction.

For the normal-number example, let $\omega_1,\omega_2,\ldots$ be independent fair binary digits and define
$$
\pi(\omega)=\sum_{j=1}^\infty\frac{2\omega_j}{3^j},\qquad
\nu=\pi_*\left(\left(\tfrac12,\tfrac12\right)^{\mathbb N}\right).
$$
This <Cantor Bernoulli measure> is supported on the middle-third <Cantor set> $C$. If $S$ is the <Bernoulli shift>, then $T_3\circ\pi=\pi\circ S$ on the circle. Consequently $\nu$ is $T_3$ invariant and ergodic: a $T_3$ invariant event pulls back to an $S$ invariant event, which has probability zero or one.

Take the ternary digit <measurable partition> $\eta=\{[0,1/3),[1/3,2/3),[2/3,1)\}$. Its block partition of length $n$ has, up to null endpoints, $2^n$ positive-measure atoms under $\nu$, each of measure $2^{-n}$, and all other atoms have measure zero. Hence
$$
H_\nu(\eta_0^{n-1})=n\log2,\qquad
h_\nu(T_3)\geq h_\nu(T_3,\eta)=\log2>0.
$$
Apply the <Host equidistribution theorem> with $p=3$, $q=2$. For $\nu$-almost every $x\in C$, the sequence $(T_2^nx)$ equidistributes for <Lebesgue measure>, so $x$ is a <normal number> in base $2$ by <normality and equidistribution under integer multiplication>.

The ternary expansion of $\nu$-almost every such $x$ contains only $0$ and $2$, so the frequency of digit $1$ is zero rather than $1/3$. The ambiguous ternary endpoints form a countable $\nu$ null set and can be removed. Therefore $x$ is not a <normal number> in base $3$. We have proved the stronger almost-everywhere existence statement
$$
\boxed{\nu\{x\in C\cap[0,1):x\text{ is normal in base }2\text{ and not normal in base }3\}=1.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 108 / 4 / Solution

Let $C_n$ be the union of the $2^n$ closed intervals remaining after $n$ steps of the <Cantor set> construction. Their total length is $(2/3)^n$. Any point outside $C_n$ has a ternary expansion with a $1$ among its first $n$ digits, so $A^c\subseteq C_n$. Endpoints of deleted intervals also belong to $A$ because they have an alternative ternary expansion containing a $1$, as in $2/3=0.1222\ldots{}_3$.

A partition using all endpoints of $C_n$ has lower Darboux sum at least
$$
1-\left(\frac23\right)^n.
$$
Since $A$ is dense, every upper Darboux sum is one. Letting $n\to\infty$ gives equal upper and lower integrals, and therefore
$$
\boxed{f\text{ is Riemann integrable and }\int_0^1f(x)\,dx=1}.
$$

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 1 / 12F / ii / Solution

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The Cantor set is obtained from $[0,1]$ by repeatedly deleting the open middle third of every remaining interval. Equivalently, it consists of the numbers that admit a base-three expansion using only the digits zero and two.


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