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Consider the circle map $\times d:x\mapsto dx \mod 1$. The lebesgue measure is the only absolutely continuous invariant probability measure, but this map has many other invariant measures. Of course, one can take barycentric combinations of invariant measures to get a new one, so let us restrict to the extremal points, namely ergodic measures.

One can consider a uniform measure on any periodic orbit. There are also singular, atomless invariant measures. For example, the "uniform" measure on the usual middle-third Cantor set is invariant under $\times 3$. All this is pretty explicit and I'm fine with it. But I also heard about a thermodynamical formalism that yield many invariant measures; Bowen's lecture note are on my desk but it does not seem to answer my question, which is the following: what do these measures look like? What is their support? I guess that we cannot answer this for all invariant measures, but maybe for some of them less trivial than the atomic and easy Cantor ones.

Other question: Cantor measures are easily constructed for all $\times d$, $d>2$, but I cannot really get one for $d=2$. Am I clumsy or is there something special to this case?

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Look also at the question mathoverflow.net/questions/37010/…. The map $\times d$ is measurably equivalent to the one sided shift, so some answers there work here (not all since the question is not exactly the same). –  rpotrie Nov 22 '10 at 17:20
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3 Answers 3

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If you do the Cantor measure construction for d=2, you just get Lebesgue measure... so it's a little bit special.

There are lots of fully supported invariant measures for the map $\times d$: the thermodynamic formalism that you mention gives you a whole zoo of them. In particular, if $\phi\colon [0,1]\to \mathbb{R}$ is any Hölder continuous function, then there is a unique "equilibrium state" for $\phi$, which is a probability measure $\mu_\phi$. This measure can be shown to have a certain Gibbs property, which in particular implies that it has full support -- it gives positive measure to every open set in $[0,1]$.

The easiest of these to think about are the Bernoulli measures. Given a sequence $x_1 x_2 \cdots x_n$ where $x_i \in \{0,1,\dots,d-1\}$, consider the interval $$ C(x_1 x_2 \cdots x_n) = [x_1 d^{-1} + x_2 d^{-2} + \cdots + x_n d^{-n}, x_1 d^{-1} + x_2 d^{-2} + \cdots + (x_n+1) d^{-n}]. $$ These generate the $\sigma$-algebra of Borel sets, so given any probability vector $\mathbf{p} = (p_1,\dots,p_d)$, we can define an invariant measure $\mu_{\mathbf{p}}$ by $$ \mu_{\mathbf{p}}(C(x_1 \cdots x_n)) = p_{x_1} p_{x_2} \cdots p_{x_n}. $$ These are equilibrium states for potential functions that are constant on the $d$ intervals $C(x_1)$. Potential functions that are constant on intervals $C(x_1 \cdots x_n)$ for some $n$ yield Markov measures as equilibrium states, and these can also be described quite explicitly.

All these measures are invariant under the $\times d$ map, fully supported on the interval, ergodic, and non-atomic.

See this answer for a little bit more on Markov measures, and this one for some other invariant measures that are ergodic, non-atomic, and fully supported, but have zero entropy. It's worth pointing out that pretty much anything you say about invariant measures for the shift map $\sigma\colon \Sigma_d^+ \to \Sigma_d^+$ (here $\Sigma_d^+ = \{0,\dots,d-1\}^{\mathbb{N}}$) can be translated into a statement about invariant measures for the $\times d$ map, since the two are topologically conjugate on a total probability set -- that is, a set that is given full weight by every invariant measure.

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Great, thanks a lot. –  Benoît Kloeckner Nov 22 '10 at 18:45
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There is a staggering variety of measures invariant under $\times d$; one cannot expect to describe them all in any explicit way. The Gibbs measures described by Vaughn Climenhaga are very special (they have very strong statistical properties which general invariant measures lack), although of course they are very important (they arise naturally for example when trying to compute Hausdorff dimension).

The support of a $\times d$-invariant measure is a compact $\times d$-invariant set; you cannot say anything more than this (conversely, any compact $\times d$-invariant set supports one, and in fact lots, of $d$-invariant measures).

Even though one cannot give any sort of explicit description of a $\times d$-invariant and ergodic measure, there are many properties these measures have which are not shared by general measures. For example, if $\mu$ is $\times d$-invariant and ergodic, then there is $\alpha\in [0,1]$ such that $$ \lim_{r\to 0} \frac{\log(\mu(B(x,r)))}{\log r} = \alpha $$ for $\mu$-almost every $x$ (in particular this is saying the limit on the LHS exists). This is essentially the Shannon-McMillan-Breiman Theorem. Moreover, $\alpha=1$ if and only if $\mu$ is Lebesgue measure.

For $d=2$ you cannot do a straightforward Cantor construction, but you can look at the set of all points in $[0,1]$ such that the binary expansion does not have two consecutive zeros (for example). This is a topological Cantor set of dimension less than $1$. There are many invariant measures supported on the set (such as Markov measures). This is in some sense the simplest nontrivial $\times 2$-invariant set.

Alternatively, you can first do a Cantor construction for $d=4$ (call the resulting measure $\mu$) and then get a $\times 2$-invariant measure out of it by setting $\nu = \frac{1}{2}(\mu + T_2\mu)$ where $T_2=\times 2$.

Although not directly related to your question, let me recall one of the most famous open problems in ergodic theory. There are lots of measures invariant and ergodic under each of $\times 2$ and $\times 3$. Lebesgue measure, as well as some discrete measures, are invariant under both $\times 2$ and $\times 3$. Are there any others? (The answer is known to be "no" if the $\alpha$ in the displayed equation above is not zero.)

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Excellent points -- certainly the Gibbs measures are a relatively small exhibit in the zoo. I'll make one pedantic comment, which is that you need the word "compact" in your observation on arbitrary invariant sets supporting an invariant measure. Every compact invariant set supports an invariant measure, but for non-compact invariant sets this is not necessarily the case (just fix a function $\phi$ and consider the set of points for which the Birkhoff averages of $\phi$ fail to converge). –  Vaughn Climenhaga Nov 22 '10 at 22:57
    
Thanks! Compactness is of course crucial, so I've edited my post. –  Pablo Shmerkin Nov 22 '10 at 23:17
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To supplement the previous answers (and perhaps further illustrate that there are a lot of invariant measures!) I thought I'd mention what "typical" invariant measures for the $d$-fold expanding map look like. Fix a $d$-fold expanding map $T$, and consider the set $\mathcal{M}$ of all $T$-invariant Borel probability measures on the circle. We can make $\mathcal{M}$ into a compact, metrisable topological space using the weak topology, which is characterised by the fact that a sequence of measures $(\mu_n)$ converges to $\mu$ if and only if the real sequence $(\int f d\mu_n)$ converges to $\int fd\mu$ for every continuous function $f$ from the circle to the reals. We can then ask what typical elements of $\mathcal{M}$ look like in the sense of Baire category.

It turns out that the answer is this: there is a dense $G_\delta$ subset of $\mathcal{M}$ in which every measure is fully supported, weak-mixing for $T$, but not strong-mixing for $T$. All of these measures are non-atomic and pairwise mutually singular with respect to one another. So there are enormously many fully supported invariant measures - so many that it's not easy to say anything about them at all which carries with very much generality.

This result is basically due to K. R. Parthasarathy in the 1961 paper "On the category of ergodic measures". In that paper it's proved for the two-sided full shift on $d$ symbols (or even infinitely many symbols) but the same proofs go through with little change.

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That's a very nice general result. Do I recall correctly that there are two papers from the early days of ergodic theory with titles like "Generic transformations are mixing" and "Generic transformations are not mixing", where the only difference is the type of mixing being considered? Or was it the same kind of mixing, and a different definition of generic? I can't remember the details at the moment... –  Vaughn Climenhaga Nov 22 '10 at 23:00
    
I think that the earlier papers are for generic MPTs of a Lebesgue space, although I don't recall what topology on the space of transformations is used. As I recall the result is that weak-mixing is generically true and strong mixing is generically false, but I don't know the references. I read a summary of the results once, but I forget where - maybe Walters? In the context described in my answer, it's also easy to show (using upper semi-continuity and the density of measures supported on periodic orbits) that generic measures have zero entropy. I forget whether that's in the paper though. –  Ian Morris Nov 22 '10 at 23:05
    
Also, if you can find the references, I'd be interested to have a look at those papers =o] –  Ian Morris Nov 22 '10 at 23:06
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I believe the references are: Halmos, Paul R. In general a measure preserving transformation is mixing. Ann. of Math. (2) 45, (1944). 786–792. Rohlin, V. A "general'' measure-preserving transformation is not mixing. (Russian) Doklady Akad. Nauk SSSR (N.S.) 60, (1948). 349–351. If I am not wrong these results are well treated in the book by Alpern and Prasad which in particular proves the fact that being "generic" as measure preserving transformation is equivalent as being generic as volume preserving homeomorphism of a manifold of dimension $\geq 2$. –  rpotrie Nov 23 '10 at 7:50
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