3

Indeed, induction is a good way: for $n=1$ this can checked directly. Now, suppose that $T^n\left( \left[ \cfrac{k}{2^n},\cfrac{k+1}{2^n}\right]\right) = [0,1] , \forall n\in\mathbb N, 0<\cfrac{k}{2^n}<1$ and let us show it for $n+1$. Let $0\leqslant k\leqslant 2^{n+1}-1$. If $k+1\leqslant 2^n$, then on $\left[ \cfrac{k}{2^{n+1}},\cfrac{k+1}{2^{n+1}}\...


2

Suppose that $f$ take values on $\mathbb{R}$, otherwise the above equality it's not true. In other words, assume that $-\infty<f(x)<\infty$ for every $x\in X$. Fix some $n$ and $x\in X$. Then, $$-\infty<2^nf(x)<\infty.$$ In other words, $k\leq 2^nf(x)<k+1$, where $k=[2^nf(x)]$ is the integer part of $2^nf(x)$. Divide both sides by $2^n$ to get ...


1

The precise statement of this result is as follows: For any stationary process $\{X_n\}_{n= 1}^{\infty}$, there exists a doubly-infinite stationary sequence $\{Y_n\}_{n=-\infty}^{\infty}$ s.t. $X_1,\ldots$ and $Y_1,\ldots$ have the same distribution. Proof. By Kolmogorov's extension theorem, it suffices to define the finite dimensional distributions of $\{...


1

You know that for any $A \cap H \subset X$ there is a measurable set $C \subset X$ such that $A\cap H \subset C$ and $\mu^∗(A\cap H)=\mu(C)$. Note that $A\cap H \subset C\cap H \subset C$ and that $ C\cap H $ is measurable. So we have $$ \mu^∗(A\cap H) \leq \mu(C \cap H) \leq \mu(C)$$ Since $\mu^∗(A\cap H)=\mu(C)$, we have $\mu^∗(A\cap H) = \mu(C \cap H) =...


1

Since $f_n$ is stationary, it means that $$f_n(\omega,x)=f_n(\tau_{-x} \omega,0)$$ almost everywhere. Therefore, the weak limit $f_0(\omega,\cdot)$ of $f_n(\omega,\cdot)$ (in $L^2_{loc}(\mathbb{R}^3)$) verifies $$f_0(\omega,x)=f_0(\tau_{-x} \omega,0).$$ Then I pose $\tilde{f_0}(\omega)=f_0(\omega,0)$ for almost all $\omega$ and we can easily check that $f_0$ ...


1

Here is a proof that doesn't exactly follow the outline indicated in the question. Let $W$ be a Poincaré sequence, $(X,\mathcal B,\mu,T)$ an m.p.s., and let $A\subset X$ have $\mu(A)>0$. We will find $n\in W\cap m\mathbb Z$ such that $\mu(T^{-n}A\cap A)>0$. Let $(Y,\mathcal D,\nu,S)$ be the rotation on $m$ points: $Y= \{0,\dots,m-1\}$, $Sy = y+1$ mod $...


1

First $X$ is just the set of (bi infinity) sequences of element of $Y$ and $T$ is just the shift in the sequence, you change the indices by $n \mapsto n+1$. Now the measure $m$ is as follow $m(.....i....)=p_i$ that is the measure of the set of sequences having $I$ at a given position is $p_i$. And then $$ m(...i_1 i_2 - i_l....)=p_1p_2-p_l $$ Where "$......


1

It's a change of index. The initial inequality reads $$g_{m+n} \leq g_m + g_n \circ \tau^m \quad \forall n, m \geq 0.$$ Now, let $m' := m$ and $n' := m+n$ I can replace $m$ with $m'$ and $n$ with $n'-m'$. This inequality becomes $$g_{n'} \leq g_{m'} + g_{n'-m'} \circ \tau^{m'} \quad \forall n' \geq m' \geq 0.$$ Note that the quantification on $n$, $m$ ...


1

1 As you said the Theorem proving the existsence of the Haar measure gives regularity. To understand why this holds, you must read the proof of Theorem 0.13. 2 Let $E \subset G$ be any Borel set. Then $$ \mu(E)=m(A^{-1}E)= \sup \{ m(K) : K\subseteq A^{-1}E , K \mbox{ compact } \} $$ Now use the fact that $A$ is onto and $G$ is compact to show that each ...


1

You haven't really given a counterexample because assuming that $f$ has the shadowing property, for every $\epsilon>0$ choose $\delta>0$ such that $\delta<\epsilon$. Now you need to fix $r$ as $0<r<\delta$ for $g(x)$ to be a $\delta$-pseudo orbit. For this $r$ and a starting point $x$, $x_n=g^n(x)$, fix $p=x+r \in \mathbb{R}$, then $$|f^n(p)-...


1

$\mu(\{x_1,\dots,x_q\}) = 1$ implies $\mu = \sum\limits_{i=1}^q\frac{\delta_{\{x_i\}}}{q}$ is correct. But I do not understant how this close the case $\mu (A)=1$ since you can repeat infinitely times your procedure without reaching a atom of the measure. If $A$ is invariant and $\mu(A)=0$ then $A^c$ the complement, is also invariant and $\mu(A^c)=1$. That ...


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