Cases where

$sup_{\mu \in E(T)} h_\mu(T) \neq \sup_{\mu \in M(T)} h_\mu(T)$.


For a topological space $X$, let $T: X \to X$ be a continuous application. Then, call the set of $T$-invariant probability measures $M(T)$, and call the set of $T$-ergodic (probability) measures $E(T)$. It is evident that $E(T) \subset M(T)$. But it may happen that $M(T) = \emptyset$. For example, take $X = \mathbb{R}$ and $T(x) = x+1$.

Since an ergodic measure is invariant, it is immediate that

$ \begin{equation*} \sup_{\mu \in E(T)} h_\mu(T) \leq \sup_{\mu \in M(T)} h_\mu(T). \end{equation*} $

The question is whether equality holds or not. When $X$ is compact, it is well known that equality holds. In this case, it is a consequence of Jacobs' Theorem, which states that for any $\mu \in M(T)$, there exists a measure $\tau$, over the set $E(T)$, such that

$ \begin{equation*} h_\mu(T) = \int_{E(T)} h_m(T) d\tau(m). \end{equation*} $

When $X$ is compact (locally compact, in fact), the above equation is a consequence of Choquet Representation Theorem and the Krein-Milman Theorem. (See, for example, Theorem 8.4 from Walters, P. An Introduction to Ergodic Theory)

Now, when $X$ is not necessarily compact, but it is a Borel subset of a compact metrizable set $\widetilde{X}$, Pesin and Pitskel' argue in their Topological Pressure and the Variational Principle for Noncompact Sets, at the end of page 310: (I will rename the spaces and applications in order to conform to this post's notation.)

We may assume that measure $\mu$ is ergodic. In fact, consider the partition $\eta$ of $X$ into ergodic components $X_s,\, s \in S$, of measure $\mu$. Denote by $\mu_s$ the measures on $X_s$ (then $T * \mu_s = \mu_s$), and by $\nu$ the measure on the quotient space $X / \eta$. Then $h_\mu(T) = \int_{Y/\eta} h_m(T) d\nu(m)$.

As far as I understand, $Y/\eta$ is just the same as $E(T)$, since each ergodic component is associated with an ergodic measure. And for the same reason, $\nu$ is just our $\tau$. So, what is being stated is the validity of

$ \begin{equation*} h_\mu(T) = \int_{E(T)} h_m(T) d\tau(m), \end{equation*} $

which in turns implies the equality

$ \begin{equation*} \sup_{\mu \in M(T)} h_\mu(T) = \sup_{\mu \in E(T)} h_\mu(T). \end{equation*} $

In Pitskel' and Pesin's paper, $T$ is not even supposed to be the restriction to $X$ of a continuous transformation $\widetilde{T}: \widetilde{X} \to \widetilde{X}$.


  1. How do I prove that when $X$ is a Borel subset of a compact metrizable space $\widetilde{X}$ and $T$ is a continuous application $T: X \to X$, then for any $\mu \in M(T)$, there exists a measure $\tau$ over $E(T)$ such that $h_\mu(T) = \int_{E(T)} h_m(T) d\tau(m)$?

  2. In case the answer to question "1" is negative, is there a prove for the specific case where $T$ is the restriction of a continuous application $\widetilde{T}: \widetilde{X} \to \widetilde{X}$?

  3. Do you know nice examples of transformations of measurable spaces where $E(T) = \emptyset$ while $M(T) \neq \emptyset$?

PS: This is my first post to MathOverflow. This is really exciting! :-)


3 Answers 3


You make a pretty common mistake here - your question has nothing to do with continuity, compactness etc. and belongs entirely to the measure category. What matters here is that you have a measure preserving transformation of a probability Lebesgue space (sometimes these spaces are also called standard probability spaces - there is a nice wikipedia article about them) and a measurable partition of this space pointwise invariant with respect to the transformation (particular, and, essentially, the only interesting case: the partition into ergodic components of the transformation). Then the entropy of the transformation is the integral of the entropies of its restrictions to the elements of the partition (the formula which appears in Pesin and Pitskel'). In fact, it should not be attributed to them - a much better reference (actually, for the whole entropy theory as well) is Rokhlin's lecture notes on entropy theory in Russian Math Surveys 1967 (the formula we discuss is in section 8.11 there). It takes care of your questions (1) and (2). As for (3), any invariant probability measure has a unique (mod 0) decomposition into ergodic invariant measures (once again, this is an entirely measure category property), so that there are no (nice or not so nice) examples of transformations $T$ with $E(T)=\emptyset$ while $M(T)\neq\emptyset$.


Welcome to mathoverflow!

I believe I can answer question 3. My reference here is Walters, P. An introduction to Ergodic Theory. Chapter 6.2.

When $T: X \mapsto X$ is a continuous transformation of a compact metrisable space $X$, then there will always be a measure for which $T$ is measure preserving. Hence $M(T)$ is never empty. (Corollary 6.9.1)

The space $M(T)$ is compact, convex and nonempty. Hence it has an extreme point by this argument.

The extreme points are the ergodic measures (Walters theorem 6.10(iii))

Hence $E(T) \neq \emptyset$.

  • $\begingroup$ I want $E(T)$ to be empty. $\endgroup$ Commented Oct 3, 2011 at 3:28
  • $\begingroup$ Perhaps I misunderstood. Are you interested in the case where $X$ is not compact? For if the space is compact, $E(T)$ will not be empty. $\endgroup$ Commented Oct 3, 2011 at 6:15
  • $\begingroup$ Yes, I am interested in the non-locally-compact case. I've made a specific question on another post: mathoverflow.net/questions/77036/… $\endgroup$ Commented Oct 3, 2011 at 11:33

I think the answer is positive for the special case (2).

In this case every $(X,T)$-invariant measure $\mu$ is also $(\widetilde{X},T)$-invariant and hence admits an ergodic decompostion, say $\tau=\tau_\mu$. Since $\mu(X)=1$, $m(X)=1$ for $\tau$-a.e. $m\in E(\widetilde{X},T)$. In other words, the ergodic decomposition of $\mu$ is 'localized' on $(X,T)$. Note that $h_m(X,T)=h_m(\widetilde{X},T)$ for each $m$ with $m(X)=1$. Therefore


After André Caldas: I take a snapshot (link) of Theorem 6.4 in Chapter II of Mane's book:

alt text http://www.freeimagehosting.net/newuploads/e2770.jpg

  • $\begingroup$ How do you prove that $\mu(X) = 1$ implies $m(X) = 1$ for $\tau$-a.e. $m$? Remember that having an "ergodic decomposition" means that for every continuous $\phi: \widetilde{X} \to \mathbb{R}$, $\int \phi d\mu = \int_{E(T)} \int \phi dm d\tau$. Also, notice that it might happen that $X \neq T^{-1}X$. $\endgroup$ Commented Jan 3, 2012 at 13:41
  • $\begingroup$ Another characterization of ergodic decomposition is <br> $\mu(E)=\int_{E(T)}m(E)d\tau(m)$ for every Borel subset $E\subset \widetilde{X}$. See the remark between Theorem 6.4 and Corollary 6.5 in the book "Ergodic Theory and Differentiable Dynamics" by Mane. Indeed Corollary 6.5 is exactly the case we used here. $\endgroup$
    – Pengfei
    Commented Jan 4, 2012 at 2:48
  • $\begingroup$ Then I find the following statement in the beginning of Section II.6 of Mane's book: Let $X$ be a compact metric space and $T: X\to X$ a measurable map. $\cdots$ When $\mathcal{M}_f(X)\neq\emptyset$ it is reasonable to ask whether it contains any ergodic maps (typo, should be measures). The answer is yes, and it follows from the results in this section. -----Theorem 6.4 there is about the ergodic decomposition of invariant measure (if exist). Assuming this theorem, then your quesion (1) is true for the even general map $f$. $\endgroup$
    – Pengfei
    Commented Jan 4, 2012 at 2:55

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