On The Convergence of Ergodic Measures - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:47:04Z http://mathoverflow.net/feeds/question/23443 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23443/on-the-convergence-of-ergodic-measures On The Convergence of Ergodic Measures Leandro 2010-05-04T15:24:19Z 2010-09-09T01:54:54Z <p>I would like to know an example (not using the Gibbs measure Theory) of a sequence of measures $\mu_n:\mathcal B\to[0,1]$ , where $\mathcal B$ is the $\sigma$-algebra of the borelians of a compact space $X$ such that :</p> <p>1) $\mu_n$ is ergodic, with respect to a fixed continuous function $T:X\to X$, for all $n\in\mathbb N$;</p> <p>2) $\mu_n\to \mu$ in the weak-$*$ topology and $\mu$ is not ergodic.</p> http://mathoverflow.net/questions/23443/on-the-convergence-of-ergodic-measures/23449#23449 Answer by Ian Morris for On The Convergence of Ergodic Measures Ian Morris 2010-05-04T16:06:31Z 2010-05-04T17:40:58Z <p>Let $X=\{0,1\}^{\mathbb{N}}$ with the infinite product topology (which is metrisable). For each $n \geq 1$, define $x_n$ to be the sequence given by $x_i=0$ for $1 \leq i \leq n$, $x_i=1$ for $n+1 \leq i \leq 2n$, and $x_{2n+i}=x_i$ for all $i$. Let $T \colon X \to X$ be the shift transformation $T[(x_n)]= (x_{n+1})$. We have $T^{2n}x_n=x_n$ for every $n \geq 1$, so the measure $\mu_n$ defined by $\mu_n:=(2n)^{-1}\sum_{j=0}^{2n-1}\delta_{T^jx_n}$ is an ergodic invariant Borel probability measure for $T$. Let $\overline{0}$ denote the element of $X$ corresponding to an infinite sequence of zeroes, and similarly let $\overline{1}$ denote the infinite sequence of ones; we have $\lim_{n \to \infty} \mu_n = \frac{1}{2}(\delta_{\overline{0}}+\delta_{\overline{1}})$, and this limit is not ergodic (since the set containing only the point ${\overline{0}}$ has measure 1/2 but is invariant).</p> <p>There is a nice paper by Parthasarathy - called, I think, "On the category of ergodic measures" - which shows that for this particular dynamical system and some of its generalisations, the set of all ergodic measures and the set of all non-ergodic measures are both weak-* dense in the set of all invariant measures, so this phenomenon can actually happen quite a lot.</p> <p>(Hmm, the definition of $X$ above is supposed to have curly set brackets in it, but I can't get them to appear for some reason. Anyway, it's supposed to be the set of all one-sided infinite sequences of zeroes and ones.) </p> http://mathoverflow.net/questions/23443/on-the-convergence-of-ergodic-measures/23463#23463 Answer by fedja for On The Convergence of Ergodic Measures fedja 2010-05-04T17:39:47Z 2010-05-04T17:39:47Z <p>If you just want something elementary to present to students, take $X$ to be the annulus $1\le|z|\le 2$ and $T(z)=ze^{i(|z|-1)}$. Then the Lebesgue measure on every circle of radius $1+\frac 1n$ is ergodic (if the ergodicity of the irrational rotation has not been covered yet, take the counting measures on some orbits on the circles of radii close to $1$ where the rotation is rational instead) but the weak limit, which is the Lebesgue measure on the unit circle, is as far from ergodic as it can possibly be.</p> <p>Of course, Ian's example is far more interesting in many respects. :-)</p> http://mathoverflow.net/questions/23443/on-the-convergence-of-ergodic-measures/25134#25134 Answer by Vaughn Climenhaga for On The Convergence of Ergodic Measures Vaughn Climenhaga 2010-05-18T14:59:54Z 2010-05-18T14:59:54Z <p>There's a property called "entropy density of ergodic measures" (or variations on that terminology), which states that given an invariant measure &mu;, you can find a sequence of ergodic measures &mu;<sub>n</sub> that converges to &mu; in the weak* topology, and furthermore, the lim inf of the entropies h<sub>&mu;<sub>n</sub></sub>(f) is at least h<sub>&mu;</sub>(f). In other words, not only can you approximate an arbitrary invariant measure using ergodic measures, but you can do so without losing any of the entropy of the original measure.</p> <p>Of course not every system has entropy density, but there are many interesting ones that do. Pfister and Sullivan use this property in a couple papers -- see "Large deviations estimates for dynamical systems without the specification property" (Nonlinearity <strong>18</strong>, 2005, 237-261), and "On the topological entropy of saturated sets" (Ergod. Th. &amp; Dynam. Sys. <strong>27</strong>, 2007, 929-956). In particular, they show that entropy density follows from something they call the <em>g-almost product property</em>. This latter property is a weaker form of the classical <em>specification property</em> (it's also been called <em>almost specification</em>). </p> <p>There are many systems known to have specification -- for example, if an Anosov systems, an Axiom A system, a subshift of finite type, or an interval map is topologically mixing, then it has specification, and hence has entropy density. There are also classes of systems that satisfy the g-almost product property (but not specification) and hence have entropy density: &beta;-shifts are one example.</p> http://mathoverflow.net/questions/23443/on-the-convergence-of-ergodic-measures/38132#38132 Answer by Martin for On The Convergence of Ergodic Measures Martin 2010-09-09T01:54:54Z 2010-09-09T01:54:54Z <p>consider the torus map $(x,y) \mapsto (x+y, y)$. For every $y$, this gives a rotation on the circle $S^1 \times y$. That is, Lebesgue measure on any "horizontal" circle is preserved. For $y$ rational it is not ergodig but for $y$ rational it is. Well, that's it. Lebesgue on a circle with rational rotation is weak* approximated by Lebesgue on circles with irrational rotation.</p>