Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 16 / 4

Let $(X,\mathcal{B},\mu )$ be a probability space, let $T$ be a measure-preserving transformation, and let $g\in L^1(\mu )$. The Birkhoff ergodic theorem asserts thatfor almost every $x$, where $\mathcal{I}=\{B\in \mathcal{B}:\mu (T^{-1}B△B)=0\}$ is the invariant sigma-algebra. The limit is integrable, satisfies $g^{\ast }\circ T=g^{\ast }$ almost everywhere, and has $\int g^{\ast }d\mu =\int gd\mu$. On a probability space the convergence also holds in $L^1$. No invertibility of $T$ is required.

If $T$ is ergodic, its invariant sigma-algebra is trivial modulo null sets, so its conditional expectation is constant. In that case the concise form is

Identify the circle group with $[0,1)$ modulo its endpoints, and let $\lambda$ be normalized Lebesgue measure. We first check invariance. On the $r$th inverse branch, $x=(y+r)/m$ for $0\le r<m$. For an integrable function $g$, substitution givesIn particular, testing indicator functions shows that the integer multiplication map on the circle preserves $\lambda$.

To prove ergodicity, let $B$ be a measurable set with $\lambda (E_m^{-1}B△B)=0$ and put $h=1_B$. Then $h(E_{m}x)=h(x)$ almost everywhere. Its Fourier coefficients areUsing invariance of $h$ and substituting separately on the same $m$ branches givesIn the last line, the finite geometric series of $m$th roots of unity is zero unless $m$ divides $k$, when it equals $m$. For any nonzero integer $k$, divide repeatedly by $m$ until the resulting integer is no longer divisible by $m$. The recurrence then implies $c_k=0$.

Since the Fourier basis is a complete orthonormal basis of $L^2([0,1),\lambda )$, the absence of every nonconstant Fourier coefficient implies $h=c_0=\lambda (B)$ almost everywhere. An indicator function takes only the values zero and one, so $\lambda (B)$ must be zero or one. This proves

Outside the countable set of dyadic rational numbers, the binary expansion $x={\sum }_{j\ge 1}{ϵ}_j(x)2^{-j}$ is unique. Multiplication by two modulo one deletes the first binary digit, soLet $g=1_{[1/2,1)}$. By part (b), the doubling map is a measure-preserving transformation and is ergodic for Lebesgue measure. Applying the Birkhoff ergodic theorem to this indicator function givesfor almost every $x$. The dyadic rationals have Lebesgue measure zero, so the choice of their two possible binary expansions does not affect the conclusion. Thus the limiting frequency of the digit one is $1/2$ almost everywhere.