Durrett’s Probability: Theorem 6.2.6

equidistributionergodic-theoryprobability theorystationary-processes

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.

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Best Answer

$\phi$ 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 $G$.

  1. 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, $\phi^{m}(x)\in[a,b)$.
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