Durrett's Probability: Theorem 6.2.6

Question

I am having some difficulty understanding the concept of measure preserving, invariance, and ergodic.

Here is a proof from Theorem 6.2.6 in Durrett's Probability: Theory and Examples, 5e (p.338) (available at https://services.math.duke.edu/~rtd/PTE/pte.html), which states that If A = [a, b), then the exceptional set is $\emptyset$. Here the $\varphi$ is a measure-preserving transformation, and $A_k = [a + 1/k, b-1/k)$.

I am trying to understand two points:

1. Why do we need to show that $G$ is dense in $[0, 1)$?
2. Why does $\varphi^m \omega_k \in A_k$ imply $\varphi^m x \in A$ if $x \in [0, 1)$, $\omega_k \in G$, with $|\omega_k - x| < 1/k$? My guess is that this has something to do with the measure-preserving property of $\varphi$ or the invariance of $A$ or $A_k$ (if they are), but I am not sure where and how exactly these concepts were used in this context.

Thank you very much.

Answer 1

In this case $\varphi:[0,1[\to[0,1[$ is an irrational rotation.

1. Birkhoff Ergodic Theorem says that the time averages with almost any initial condition can be switched with space averages. Here the statement is that time time averages with any initial condition can be switched with space averages, that is, the exceptional set is truly empty. This is needed because Birkhoff Ergodic Theorem doesn't guarantee a priori that for the initial condition $0$ the statement holds, which is the content of the "number theoretic result" Weyl Equidistribution Theorem. (The fact that irrational rotations are uniquely ergodic is also relevant to this discussion; see Topological Invariance of Unique Ergodicity .)

2. This has to do with $\varphi$ being length preserving, rather than measure preserving. If $\omega_k$ is within a $1/k$ distance to $x$, then for any $n$ $\varphi^n(\omega_k)$ will be within a $1/k$ distance to $\varphi^n(x)$.

Answer 2

$\phi$ is not any invariant function with respect to the Lebesgue measure on $[0,1)$, but a translation: $$\phi(x)=x+\theta\mod1$$ for some $\theta\in[0,1)$.

1. It is not that one has to prove that $G$ is dense. The strategy that Rick Durrett set out is to first construct a set $G$ that is large (full measure) in $[0,1)$ where the result of the ergodic theorem holds for each open interval $A_k=(a+1/k,b-1/k)$ at any point of $G$ $(k>2/(b-a), k\in\mathbb{N}$. The fact that $P[G]=1$ implies that $G$ is dense in $[0,\infty)$ (otherwise there is a point $x\in [0,1)$ and a neighborhood $V$ of $x$ in $[0,1)$ of radius $\delta>0$ such that $V\cap G=\emptyset$. This would imply that $P[G]=P[G\cap V]+P[G\setminus V]\leq 1-\varepsilon<1$, contradicting the fact that $P[G]=1$.

The remaining of the proof consists on letting the ergodic averages operate on points in $[0,1)\setminus G$. As $G$ is dense, you get almost the same as operating on points in $G$.

2. For all $k>\frac{2}{b-a}$ set $A_k=(a+1/k,b-1/k)$. If $x\in [0,1)\setminus G$, by density, there are points $\omega_k\in G$ such that $|x-\omega_k|<\frac1k$. Since $S_n\mathbb{1}_{A_k}(\omega_k)\xrightarrow{n\rightarrow\infty}b-a-\frac1k$ there are infinitely many $m$'s such that $\mathbb{1}_{A_k}(\phi^m(\omega_k))=1$, i.e. $\phi^{m}(\omega_k)\in A_k$. Notice that $|\phi^m(x)-\phi^m(\omega_k)|=|x+m\theta-\omega_k-m\theta|=|x-\omega_k|<\frac1k$ and so, $\phi^{m}(x)\in[a,b)$.