When working with Dominik Kwietniak and Jakub Konieczny, the question appeared:

Let $X$ and $Y$ be two subshifts on the same alphabet, $M(X)$, $M(Y)$ the sets of shift-invariant measures on $X$ and $Y$, respectively, $M_e(X)$ and $M_e(Y)$ be the set of the ergodic ones.

The Besicovitch measure $\bar{d}$ considered on the set of all shift-invariant measures can be lifted in a way the Hausdorff distance is established and we can define the distance between the sets of ergodic measures, or all invariant measures, as follows:

$$\bar{d}(M_e(X),M_e(Y))=\max\left(\sup_{\mu\in M_e(X)}\inf_{\nu\in M_e(Y)}\bar{d}(\mu,\nu),\sup_{\nu\in M_e(Y)}\inf_{\mu\in M_e(X)}\bar{d}(\mu,\nu)\right),$$

$$\bar{d}(M(X),M(Y))=\max\left(\sup_{\mu\in M(X)}\inf_{\nu\in M(Y)}\bar{d}(\mu,\nu),\sup_{\nu\in M(Y)}\inf_{\mu\in M(X)}\bar{d}(\mu,\nu)\right).$$

If we are not wrong (if someone disagree I can add the details), it is true that $$\bar{d}(M_e(X),M_e(Y))\le \bar{d}(M(X),M(Y)).$$

The question is whether $$\bar{d}(M(X),M(Y))\le \bar{d}(M_e(X),M_e(Y)).$$

The reason why we "claim" it is the observation that it is true when on the left side are the convex hulls of the ergodic measures instead of all invariant measures.

The idea of the proof is to take for given $\mu\in M(X)$ its ergodic decomposition $\mu=\int_{M_e(X)}\mu'd\lambda(\mu')$. Every ergodic $\mu'$ assign with an ergodic measure $\nu'$ on $Y$ that satisfies $$\bar{d}(\mu',\nu')<\bar{d}(M_e(X),M_e(Y))+\varepsilon.$$ It defines a map $f$ from $M_e(X)$ to $M_e(Y)$, $f(\mu')=\nu'$. Now integrate $$\nu=\int_{M_e(X)}f(\mu') d\lambda(\mu')$$ and prove that $\bar{d}(\mu,\nu)<\bar{d}(M_e(X),M_e(Y))+\varepsilon.$

But I do not know how to show measurability of the map $f$. The definition of the mapping is not very constructive and is not unique at all.

Any comment, suggestion would be appreciated!


2 Proof based on approximations by convex combinations

This proof is much simpler and does not involve the advanced notions and theorems from the descriptive set theory.

Since the role $X$ and $Y$ is interchangeable with respect to the assumptions, it is satisfactory to prove that for every $D>\bar{d}(M_e(X),M_e(Y))$ and $\mu\in M(X)$, there is $\nu\in M(Y)$ such that $\bar{d}(\mu,\nu)\leq D$.

Let $\mu_n=\sum^{k_n}_{i=1}a_{n,i}\mu_{n,i}$, $n\in\mathbb{N}$, be a sequence of convex combinations of ergodic measures $\mu_{n,i}$ on $X$ that converges to $\mu$ in the weak$^\star$ topology. By the assumption on $D$, for every $\mu_{n,i}$ there exists an ergodic measure $\nu_{n,i}$ on $Y$ such that $\bar{d}(\mu_{n,i},\nu_{n,i})<D$. It means that there exists a joining $\xi_{n,i}$ of $\mu_{n,i}$ and $\nu_{n,i}$ such that $$\int_{X\times Y}1_{x_0\neq y_0}d\xi_{n,i}(x,y)<D.$$ Put $\xi_n=\sum^{k_n}_{i=1}a_{n,i}\xi_{n,i}$. If needed, one can pass to a subsequence, to get $\xi_n$ convergent in the weak$^\star$ topology. Denote its limit by $\xi$. It is not diffcult to see that the marginal measure on $X$ is $\mu$. Moreover, $$\int_{X\times Y}1_{x_0\neq y_0}d\xi(x,y)\leq D,$$ because the integral depends on the measure continuously. If we denote by $\nu$ the marginal measure on $Y$, we get $\bar{d}(\mu,\nu)\leq D$.


This is not an answer, but rather a long comment. I also suspect the answer is yes, but my strategy is running into a difficulty that is very similar to yours:

Here is what I am thinking: let $\alpha=\bar d(M(X),M(Y))$. For each ergodic measure, $\mu$ on $X$, there is a joining $\bar\mu$ of $\mu$ and an ergodic measure $\nu_\mu$ on $Y$ such that $\int \mathbf 1_{x_0\ne y_0}\,d\bar\mu\le\alpha$. There is a subset $X_0(\mu)$ of $\mu$-measure 1 such that for each $x\in X_0(\mu)$, there exists a $y\in Y$ such that $(x,y)$ is generic for $\bar\mu$. Now I want to consider $X_0=\bigcup_{\mu\in M_e(X)}X_0(\mu)$. Question: is $\eta(X_0)=1$ for each $\eta\in M(X)$?

If the answer to this question is yes, then I think there is a positive answer to your question: pick an i.i.d. sequence $(x^{(n)})_n$ of points of $X$ with distribution $\eta$. Now choose $y^{(n)}$ so that $(x^{(n)},y^{(n)})$ is generic for a joining $\bar\mu^{(n)}$ with $\int \mathbf 1_{x_0\ne y_0}\,d\bar\mu^{(n)}\le\alpha$. Now claim that $\lim_{N\to\infty} 1/N\sum_{n=0}^{N-1}\mu^{(n)}$ converges to a joining whose $X$-marginal is $\eta$, and showing that the distance from $\eta$ to $M(Y)$ is at most $\alpha$.

  • $\begingroup$ I posted below two proofs. I should say that you suggest another way how to prove it. I will think about it. $\endgroup$ Aug 16 '17 at 17:21
  • $\begingroup$ A detail in your proof brings me to the fact I overlooked. Namely, that for every ergodic measure $\mu$ on $X$, $\inf_{\nu\in M_e(Y)}\bar{d}(\mu,\nu)$ is attained. It is true, although the space $(M_e(Y),\bar{d})$ is not compact. Nice. $\endgroup$ Aug 16 '17 at 17:58

Finaly, I found two ways of proving the claim myself.

1 - Proof based on a measurable selection

First, there is a way how to finish the ideas sketched in the same post with the question. One can show the measurability of the mapping $f$ using some of measurable selection or measurable uniformization theorems from the descriptive set theory. We use here Jankov-von Neumann uniformization theorem, see Theorem 18.1 in Kechris, Alexander S., Classical descriptive set theory, Graduate Texts in Mathematics. 156. Berlin: Springer-Verlag. xx, 402 p. (1995). ZBL0819.04002.. More precisely, we use the version introduced in Exercise 18.3 in the same book.

There is also a small change with respect to the previous post, namely the codomain of $f$ is different. For technical reason I found easier to look for an appropriate function $f$ from $M_e(X)$ to $M_e(X\times Y)$.

Fix $D>\bar{d}(M_e(X),M_e(Y))$ and put

$$P=\left\{\xi\in M_e(X\times Y)\mid \int_{X\times Y}1_{x_0\neq y_0} d\xi(x,y)\leq D\right\}.$$

It is a closed subset in the Polish space $M_e(X\times Y)$ equipped with the weak$^*$ topology. Hence, $P$ itself with the restriction of Borel structure $M_e(X\times Y)$ is a standard Borel space as well as the space $M_e(X)$ where the weak$^*$ topology is considered. By the assumption on $D$, the standard projection $\pi_X:P\to M_e(X)$ is onto and continuous, in particular Borel. The Jankov-von Neumann theorem and its corrolary in the Exercise 18.3 in the Kechris' book ensures that there exists a function $f:M_e(X)\to P$ such that $\pi_X(f(\mu))=\mu$ and which is measurable with respect to the $\sigma$-algebra on $M_e(X)$ generated by analytic sets. By Lusin theorem, $f$ is universally measurable since $M_e(X)$ is standard Borel (see Theorem 21.1 in Kechris' Book). It implies, that for every $\mu\in M(X)$ and its ergodic decomposition $\mu=\int_{\mu'\in M_e(X)}\mu'\lambda(\mu')$, the Pettis integral

$$\xi=\int_{\mu'\in M_e(X)}f(\mu')\lambda(\mu')$$

is well defined. It is easy to see that $\xi$ is an $\sigma\times\sigma$-invariant measure on $X\times Y$ that projects on $\mu$.

Moreover, $$\int_{X\times Y}1_{x_0\neq y_0}d\xi(x,y)\leq D.$$ It implies, that for the projection $\nu=\pi_Y(\xi)$, $\bar{d}(\mu,\nu)\leq D$. It concludes the proof.

  • $\begingroup$ I should acknowledge Dominik Kwietniak and Tomas Downarowicz for advising me to use a measurable selection theorem. Both suggested to use Kuratowski-Ryll-Nardzewski theorem in the version stated at wikipedia. Let me notice, that in the Kechris' book, the statement of the Kuratowski-Ryll-Nardzewski theorem is different. Since a text book is a better source than wikipedia, I use the Jankov-von Neumann selection theorem. $\endgroup$ Aug 16 '17 at 17:28

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