# Is any invariant, ergodic measure with full support on an irreducible Markov shift a Markov measure?

I have this question I have been struggling with for a while. It seems rather intuitive, however, I was not able to proof it yet:

Let $\Omega = \{1,2,\cdots,N\}$ a finite alphabet, $\Sigma \subset \Omega^{\mathbb Z}$ be an irreducible Markov shift (i.e. an irreducible 1-step subshift of finite type). Denote by $(\Sigma,\mathcal B)$ the measure space with the usual product $\sigma$-algebra. Further, let $\mu$ be an ergodic, shift-invariant probability measure with full support on $(\Sigma,\mathcal B)$.

Question: Is $\mu$ necessarily a Markov measure? More precisely: Is $\mu$ a measure that takes the value $p_{\omega_0} \prod_{t=1}^{\tau}\Pi_{\omega_{t},\omega_{t+1}}$ on the cylinder sets specified by the finite string $(\omega_0,\omega_1,\dots,\omega_\tau)$, where $\Pi_{i,j}$ are the elements of a compatible stochastic matrix and $p$ is its unique unity stochastic eigenvector?

It is quite easy to construct counter-examples if I drop certain assumptions. However, in the described set-up I could not find one. It feels like that question should have an easy answer, but somehow I don't get it.

Any help would be really appreciated!

Cheers, Bernhard

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Is a Markov measure what I think it is, namely a kind of Fubini product measure derived from eigenvectors of the transition matrix? –  Lee Mosher Apr 24 '14 at 11:26
I think we mean the same, so I edited the OP to be more precise. However, it is not really a product measure, or the appearance of subsequent symbols would be independent of each other. –  Bernhard Altaner Apr 24 '14 at 16:21

Not at all. The simplest example is provided by the so-called $d$-Markov measures. These are Markov measures for the associated shift whose alphabet is the subset of $\Omega^d$ which consists of all $d$-tuples of symbols that occur in $\Sigma$. It is easy to see that for $d>1$ there are more $d$-Markov measures than plain Markov ones.
Thank you so much. Now I really feel embarrassed for even asking. Sometimes the answer is so close... On a related issue: Under the assumptions stated, can you think of a condition on $\mu$ such that it is a Markov measure than specifying its values on a $\pi$-system (that is a system which is closed under finite intersections)? –  Bernhard Altaner Apr 24 '14 at 16:29
Have never thought about $\pi$-systems. However one can formulate such a condition in terms of cylinder sets. Namely, an invariant measure $\mu$ is Markov if and only if the ratios of measures of cylinder sets $\mu(C_{\omega_0,\omega_1,\dots,\omega_t})/\mu(C_{\omega_1,\dots,\omega_t})$ only depend on $\omega_0$ and $\omega_1$. –  R W Apr 24 '14 at 17:34
The cylinder sets themselves form a $\pi$-system, but a rather large one. The question is more if there would be some smaller $\pi$-system, which allows for that...in any case, thank you very much already! –  Bernhard Altaner Apr 24 '14 at 19:31