Let $\sigma$ be the shift function in the space of two-sided infinite sequences of $\{0,1\}$, $X=\{0,1\}^\mathbb{Z}$ equipped with product topology. We know that there is some point $x\in X$ with dense orbit under $\sigma$. Now the sequence:


of $\sigma$-invariant probability measures on $X$, has a convergent subsequence in weak* topology which tends to some invariant measure $\mu$.

Is $\mu$ ergodic for $\sigma$?

Is $\mu$ the usual product measure on $X$?


1 Answer 1


Not necessarily.

Why $\mu$ is not necessarily the uniform product measure:

Even if the orbit of $x$ is dense, there could be arbitrary bias in the frequency it visits different regions. For instance, let $w^{(1)}, w^{(2)}, \ldots$ be an enumeration of $\{0,1\}^*$ (i.e., all finite binary words). Construct $x$ as

$$x = \cdots 0\,0\,0\,\underline{0}\,0^{n_1}\,w^{(1)}\,0^{n_2}\,w^{(2)}\,0^{n_3}\,w^{(3)}\cdots\;.$$

(Underline denotes the symbol at the origin.) The orbit of $x$ is clearly dense, but if you choose $n_1, n_2,\ldots$ appropriately, the measures $\mu_n$ converge to the Dirac measure concentrated at the all-$0$ sequence. (More specifically, the frequency of indices $i$ such that the block $x_{[i,i+k)}$ is anything but $0^k$ converges to $0$.)

Why $\mu$ is not necessarily ergodic:

Repeat a similar construction this time with

$$x = \cdots 0\,0\,0\,\underline{0}\,0^{n_1}\,w^{(1)}\,1^{n_2}\,w^{(2)}\,0^{n_3}\,w^{(3)}\,1^{(n_4)}\cdots\;.$$

(That is, the blocks separating $w^{(i)}$s are alternately all-$0$s and all-$1$s.) Again, choosing $n_1,n_2,\ldots$ suitably, the measures $\mu_n$ converge to the uniform convex combination of the two Dirac measures concentrated at the all-$0$ and all-$1$ sequences.

In fact, there is a theorem of Kakutani stating that every probability measure on $\{0,1\}^{\mathbb{Z}}$ can be obtained as the orbit statistics of some sequence in $\{0,1\}^{\mathbb{Z}}$. Modifying such a sequence slightly, you can make sure that it has dense orbit.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .