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The baker's map can be defined as $E_{2}:S^{1}\rightarrow S^{1}$, $E_{2}x=2x \mod 1$. This map has various properties, for example, denoting Lebesgue measure by $\lambda$, we have $\lambda(E_{2}^{-1}(A))=\lambda(A)$ for any measurable subset $A\subset S^{1}$, and the set of points with dense orbits under $E_{2}$ has Lebesgue measure 1.

Define a map $\Phi=\phi\circ E_{2}$, where $\phi:[0,1]\rightarrow[0,1]$ is continuous and strictly increasing with $\phi(0)=0,\phi(1)=1$. Assume also that $\phi'$ is continuous. Are there properties similar to those above for the map $\Phi$? [I'm more interested in the second property in particular since I think the first isn't valid.]

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If $E_2\colon S^1\to S^1$ and $\phi\colon[0,1]\to[0,1]$, how are you forming the composition $\phi \circ E_2$? –  Jim Belk Apr 20 '12 at 20:56
    
I'm taking $S^{1}$ to be $[0,1]$ with 0 and 1 identified. –  Yeong-Chyuan Chung Apr 20 '12 at 23:35

1 Answer 1

Certainly $\Phi$ won't preserve Lebesgue measure except in the trivial case $\phi(x)=x$.

The second property can also fail. For example, if $\phi'(0) < 1/2$, $0$ is an attracting fixed point of $\Phi$, i.e. there is an interval $B_0 = [0,\alpha)$ with $\alpha > 0$ such that for all $x_0 \in B_0$, $\Phi^n(x_0) \to 0$ as $n \to \infty$.

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Then are there any similar properties between $E_{2}$ and $\Phi$? We can consider elementary ones too. –  Yeong-Chyuan Chung Apr 21 '12 at 0:19

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