Here is a theorem of Furstenberg:

([2], Theorem 6.18 or [1], Theorem 1.3). If $T:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ is a homeomorphism without periodic orbits (i.e. $T^n(x)\neq x$ for any $n,x$) and $\mu$ is $T$-invariant Borel measure ($\mu(T^{-1}A)=\mu(A)$ for all Borel measurable $A$), then the function $F_{\mu}(x)=\mu([0,x])$ satisfies $F_{\mu}T(x)=T_{\alpha}F_{\mu}(x)\bmod 1$, where $T_{\alpha}x=x+\alpha\bmod 1$ is rotation by $\alpha=F_{\mu}T(0)$ (necessarily irrational).

My question is: which distribution functions can arise in this fashion? With $T$ as above, $\mu$ does not have any atoms (if $\mu(\{x\})>0$, then its orbit would have would have infinite measure by the assumption of no periodic orbits). Is this the only restriction? Does every continuous non-decreasing function $F:[0,1)\to[0,1)$ arise as $F_{\mu}$ for some $(T,\mu)$?

A related question: is every such $T$ (homeomorphism of the circle without periodic points) minimal, i.e. is every orbit dense? If so, then $F_{\mu}$ above is a homeomorphism. I assume this isn't the case otherwise the statement of [2], Theorem 6.18 would be different).

I'd also be happy with an example of a $T$ such that $F_{\mu}$ is not a homeomorphism, probably supported on something Cantor-like if it exists (which I think it does).


[1] H. Furstenberg, $\textit{Strict ergodicty and transformations of the torus}$, Amer. J. Math. 83, 1961, $573-601$

[2] Peter Walters, $\textit{An Introduction to Ergodic Theory}$, Springer

  • $\begingroup$ So F is saying that $T$ is conjugate to an irrational shift mod $1$, where $x\to F_\mu(x)$ is the conjugating map? $\endgroup$ – kimchi lover Jul 20 '17 at 1:28
  • $\begingroup$ @kimchilover yes (I would've drawn a commutative diagram, but I didn't know how). $\endgroup$ – yoyo Jul 20 '17 at 21:34

For starters, it looks like any atomless $\mu$ for which $x\to\mu([0,x])$ is a homeomorphism will do. I think F's result does not cover the case where $\mu$ is the Cantor measure, though: even though $F_\mu: K\to \mathbb R /\mathbb Z$ is continuous and surjective as a map on the Cantor set, pushing $\mu$ to Lebesgue measure, I don't see how to extend it to a homeomorphism $\mathbb R /\mathbb Z\to \mathbb R /\mathbb Z.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.