MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is it true that for a given nonnegative number, there exists a measure-theoretical entropy value (supremum of entropies of all partitions under a measure-preserving transformation) that equals this number?

share|cite|improve this question
Take a shift on $G^{\mathbb Z}$ where $G$ is a finite group. The entropy will be $\log |G|$ (let's say logarithm with base $2$). So if you choose $G$ such that $|G|=2^n$ (for $n$ your favourite positive integer) you have the example you are looking for. – Simone Virili Nov 12 '12 at 21:14
In general, using Cantor sets constructions, one can find sub-systems of the shift system of any given entropy between 0 and $\log(n)$. I believe this appears in Furstenberg's disjointness paper for example, although the construction is obviously much older than the paper. This is not a real research question. – Asaf Nov 12 '12 at 22:11
Use the intermediate value theorem for Bernoulli measures on $\lbrace 1,\ldots,n\rbrace^{\mathbb Z}$. – Anthony Quas Nov 12 '12 at 23:58

There is also a paper of Grillenberger, where You can find the construction

share|cite|improve this answer

There are two funny approaches:

  1. Let $S_2$ be a surface of genus 2 with constant curvature $-1$. Then the geodesic flow $\phi_t$ on the unit tangent bundle preserves the Liouville measure $\mu$ and has positive metric entropy: $h_\mu(\phi_t)=|t|\cdot h_\mu(\phi_1)$ with $h_\mu(\phi_1)>0$. So for any given number $r$, just pick $t=r/h_\mu(\phi_1)$. In fact any flow with positive entropy works.

  2. We use the affineness of measure-theoretical entropy $h_{\bullet}(f):\mathcal{M}(f)\to \mathbb{R},\mu\mapsto h_\mu(f)$. Pick a large integer $n\ge e^r$ and consider the full shift $\sigma$ on $\lbrace 1,\cdots,n\rbrace^{\mathbb{Z}}$. Then $\mu=(1/n,\cdots,1/n)^{\oplus\mathbb{Z}}$ has entropy $h_\mu(\sigma)=\log n>r$. Let ${\bf 1}=(1111\cdots)$. Then the enropy of $\mu_p=p\mu+(1-p){\delta_{\mathbf{1}}}$ has entropy $p\log n$. So we set $p=r/\log n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.