Are there examples of subshifts (that is, closed shift-invariant subsets of the full shift {$1...n$}${}^{\mathbb{Z}}$) on which the shift is topologically mixing, which admit a shift-invariant probability measure of full support, but no invariant ergodic measure of full support ?

I guess the answer is no, but I can't locate a reference.

Also I would be interested in a topologically mixing subshift that has no invariant probability measure of full support.

  • $\begingroup$ In the title it says "topologically mixing", but in the body it says "topologically transitive". Which would you like? On the basis of work of Weiss and his students I think that mixing is unlikely to make a difference to the result, but a mixing example might not exist in the literature. $\endgroup$
    – Ian Morris
    Jun 2, 2010 at 12:47
  • 1
    $\begingroup$ Oups, that's a typo in the core of the question [edited]. In fact, I just looked at the example in the article of Weiss cited in your answer below, and it is also topologically mixing. Thanks ! $\endgroup$
    – coudy
    Jun 2, 2010 at 13:14
  • $\begingroup$ I've just added the "symbolic dynamics" tag, which I think is appropriate. $\endgroup$
    – Ian Morris
    Jun 8, 2010 at 17:07

2 Answers 2


A construction of a topologically transitive mixing subshift with a fully supported invariant measure, but no fully supported ergodic measure, is given by Benjamin Weiss in the article "Topological transitivity and ergodic measures", Mathematical Systems Theory 1971. It gives a direct combinatorial construction, with $n=2$.

It is easy to give an example of a topologically transitive subshift on two symbols with no fully supported measure: just take the orbit closure under the shift of the sequence with all negative entries equal to one, and all non-negative entries equal to two.


Let me go a little further: I believe that there also exists a topologically mixing subshift on two symbols with no fully-supported invariant measure. I don't know of a reference for this result, but I think that a direct construction should be "not too hard", in the sense that a completely detailed proof would take up perhaps five pages or so.

Here is a very rough sketch. We want to find a sequence $x=(a_n)_{n=-\infty}^\infty \in \{1,2\}^{\mathbb{Z}}$ such that the orbit closure of $x$ is a subshift with the desired properties. Let us denote this orbit closure by $A$, and the shift by $T$. Our criteria mean the following:

1) Topological mixing. This means that for every pair of subwords $u$ and $v$ of $(a_n)_{n=-\infty}^\infty$, there exists an integer $N(u,v)>0$ such that for all $n \geq N(u,v)$, we can find a word $\omega$ of length $n$ such that $u\omega v$ is a subword of $(a_n)_{n=-\infty}^\infty$. Since every nonempty open set contains a cylinder I think it is not too hard to see that this implies topological mixing.

2) There is no fully-supported invariant probability measure on $A$. In particular, every invariant measure on the orbit closure gives zero measure to the cylinder set $\{(b_n)_{n=-\infty}^{\infty} \colon b_0=2\}$. To achieve this we want $(a_n)_{n=-\infty}^\infty$ to have the following property: there is a sequence of positive real numbers $(\varepsilon_n)_{n=1}^\infty$, which decreases to zero, such that for every $n \in \mathbb{N}$, every subword of $(a_n)_{n=-\infty}^\infty$ with length $n$ contains at most $n\varepsilon_n$ instances of the symbol $2$. If this property holds, then it is not difficult to see that \[\lim_{n\to\infty}\sup_{x \in A} \frac{1}{n}\sum_{i=0}^{n-1}\chi_{{x_0=2}}\left(T^ix\right)=0\] and by the Birkhoff ergodic theorem combined with the ergodic decomposition theorem, this forces every invariant measure on $A$ to give zero measure to the aforementioned cylinder set.

I am not really going to say much more, except that I think that these requirements can be seen to be compatible, in the sense that such a sequence $(a_n)_{n=-\infty}^\infty$ exists. To give a very rough idea of how it might be constructed, we could start like this. Begin by defining $a_0:=2$, and $a_{\pm1}:=1$. We will build the sequence outward from the origin in a sequence of inductive stages, roughly as follows. At each stage, we consider the set of all subwords of the finite sequence defined in the previous stage. We construct the new stage by appending copies of these subwords to the sequence defined in the previous stage in a careful manner. To ensure that (1) holds for the limit construction, we will need to include an awful lot of copies of every pair of subwords from the previous stage, separated by big lists of ones, once each for each pair of subwords and each different separating length in a certain range depending on the pair of subwords. On the other hand, to ensure that (2) holds, we need to make sure that every time a subword containing some twos is appended, it is padded out on both sides with big blocks of ones to a sufficient extent that we do not create a subword with too high a proportion of ones in it. The easiest way to achieve this is to ensure that $N(u,v)$ is always very, very, very large relative to the lengths of $u$ and $v$.

I think that this program can be followed, although I haven't gone to the significant effort of trying to write down all of the details. If there is a general moral, then I think it is this: there are an awful lot of subshifts, and if a small collection of criteria on a possible subshift does not lead to an obvious contradiction, then the desired subshift can probably be constructed by this kind of combinatorial procedure.

  • $\begingroup$ You can «go further» in your original answer, too :) $\endgroup$ Jun 8, 2010 at 12:57
  • $\begingroup$ Here is a construction of a mixing binary shift whose only ergodic measure is concentrated on a fixedpoint $0^\infty$: To specify $X$ we describe its language. Let $W$ be the collection of all binary words $w$ such that for any subword $u$ of $w$ if $2^{k-1}+1\le |u| \le 2^{k}$ for some $k$, then the $1$ occurs at less than $k+1$ times in $u$. By Lind&Marcus [Prop. 1.3.4] $X$ is a subshift. Fix cylinders $[u]$ and $[v]$ in $X$. Since $u0^kv0^\infty\in X$ for all sufficiently large $k$, $X$ is mixing. By definition of the language $[1]$ has measure $0$ for all invariant measures in $X$. $\endgroup$ Nov 14, 2014 at 19:09
  • $\begingroup$ (there are no invariant measures other than the Dirac measure on the fixed point $0^\infty$ because $1$ occur with zero density in all points in $X$, on the other hand $X$ is mixing and nontrivial hence uncountable). $\endgroup$ Nov 14, 2014 at 19:12

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