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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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