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Here is the question I was reading its solution here Ergodic T invariant measures that can not be Lebesgue measures.:

Let $T: S^1 \to S^1$ be the rotation of the circle by $90$ degrees. I want to find an ergodic $T$-invariant measure.

Here is the solution given there:

Call $\mu_x$ the measure in question. Then for $\phi:X\to\mathbb{R}$ a function one has

\begin{align*} \int \phi\circ f(y) d\mu_x(y) = \dfrac{\phi\circ f(x)+\phi\circ f(f(x))+\cdots+\phi\circ f(f^{p-2}(x))+\phi\circ f(f^{p-1}(x))}{p} = \dfrac{\phi\circ f(x)+\phi\circ f^2(x)+\cdots+\phi\circ f^{p-1}(x)+\phi\circ f^{p}(x)}{p} = \dfrac{\phi\circ f(x)+\phi\circ f^2(x)+\cdots+\phi\circ f^{p-1}(x)+\phi(x)}{p} =\int\phi d\mu_x(y), \end{align*}

so that $\mu_x$ is $f$-invariant.

My questions are:

Can we just do it without integration, since $\mu$ is $f$-invariant iff $$\mu(f^{-1}(U)) = \mu(U)$$ for all $U$?

Also, from where the last equality in the integration above came?where has the $f$'s gone?

Could anyone clarify these points to me please?

John B
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Emptymind
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1 Answers1

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There is no need for integrals:

Take a set $U$ and let $n_1,\dots,n_k$ be the integers in $\{0,\dots,p-1\}$ such that $$ f^{n_i}(x)\in U\quad\text{for $i=1,\dots,k$}. $$ Then $\mu(U)=k/p$. Now note that $$ f^n(x)\in f^{-1}U\quad\text{if and only if}\quad f^{n+1}(x)\in U. $$ Therefore, letting $m_1,\dots,m_l$ be the integers in $\{1,\dots,p\}$ such that $$ f^{m_i}(x)\in U\quad\text{for $i=1,\dots,l$}, $$ we have $l=k$ and $m_i=n_i+1$ for $i=1,\dots,k$. This gives $$ \mu(f^{-1}U)=\frac lp=\frac kp=\mu(U). $$

As for the last equality with the integral, recall that $f^p(x)=x$ and so \begin{align*} &\dfrac{(\phi\circ f)(x)+\cdots+(\phi\circ f^{p-1})(x)+(\phi\circ f^{p})(x)}{p}\\ &= \dfrac{(\phi\circ f)(x)+\cdots+(\phi\circ f^{p-1})(x)+\phi(x)}{p}\\ &= \dfrac{\phi(x)+(\phi\circ f)(x)+\cdots+(\phi\circ f^{p-1})(x)}{p}. \end{align*}

John B
  • 17,726