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Let $X$ be a Borel probability space (i.e. equipped with a measure $\mu$ on the Borel $\sigma$-algebra such that $\mu(X) = 1$) with a measure-preserving transformation $T$ such that every point has a dense orbit. Does it follow that the measure-preserving system $(X,\mu,T)$ is ergodic?

I have heard that the answer is no, but I haven't been able to think of any examples.

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Furstenberg constructed smooth skew products that are minimal (every orbit is dense), but have uncountably many ergodic measures. A non-trivial linear combination would be an invariant non-ergodic measure. His paper is "Strict ergodicity and transformation of the torus". I recall it also can be done rather easily in symbolic setting. Also, I think, Boshernitzan has a nice construction of a minimal interval exchange (4 intervals?) that is not uniquely ergodic. –  Andrey Gogolev Sep 1 '11 at 18:48
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Andrey Gogolev also provided an answer, but you want to assume that $X$ is a metric (or at least topological) space, $T$ is continuous and the $\sigma$-algebra is the Borel $\sigma$-algebra. Otherwise the question doesn't really make sense. –  Pablo Shmerkin Sep 1 '11 at 18:57

1 Answer 1

Here is a symbolic construction alluded to by @Andrey Gogolov

Start with two words $W^{(1)}_1=a$ and $W^{(1)}_2=b$.

Then inductively create sequences of pairs of new words by stringing together copies of the words you created at the previous stage. You want to do this in such a way that the dominance of $a$'s in the 1-words and $b$'s in the 2-words does not become diluted.

So for example, try $W^{(n)}_1=[ W^{(n-1)}_1 ] ^ {2^n-1} W^{(n-1)}_2$ and $W^{(n)}_2= [ W^{(n-1)}_2 ]^{2^n-1} W^{(n-1)}_1$ ( so that the $W^{(n)}$ words are $2^n$ times the length of the previous pair of words ). Let $L^{(n)}$ be the length of the blocks at the $n$th stage.

Now let $X$ be the set of all bi-infinite sequences such that each finite block of the sequence occurs in some $W^{(n)}_i$ (and hence all $W^{(m)}_j$ with $m>n$).

Claim that $X$ is minimal (all points have dense orbits): let $B$ be a block appearing in $X$ somewhere. By definition, it's a sub-word of some $W^{(n)}_j$. Now let $x\in X$. Choose a block in $x$ of length exceeding $2L^{(n+1)}$. By definition of $X$ again, this block occurs inside a level $m$ block for some $m>n+1$. Since level $m$ blocks are simply concatenations of level $n+1$ blocks, the original block in $x$ must contain a complete level $n+1$ block. Since the level $n+1$ blocks contain both of the level $n$ blocks, it follows that $x$ contains the block $B$. This establishes the minimality of $X$.

As for lack of unique ergodicity, let $\rho^{(n)}$ denote the density of $a$'s in the level $n$ 1-block (so that $1-\rho^{(n)}$ is the density of $a$'s in the level $n$ 2-block). Recursively, we have $\rho^{(n+1)}=[ (2^{n+1}-1)\rho^{(n)} + (1-\rho^{(n)}) ]/2^{n+1} = (1-2^{-n})\rho^{(n)} + 2^{-(n+1)}$. This is specially chosen to remain bounded away from 1/2. Hence there exists $b>1/2$ such that the density of $a$'s in a level $n$ 1-block is bounded below by $b$, whereas the density in a level $n$ 2-block is bounded above by $1-b$. This contradicts unique ergodicity.

Another very nice family of examples of minimal but non-uniquely ergodic systems was given by Mazur and Smillie in the context of polygonal billiards.

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