Let $X$ be a separable metric space, $\phi\in C(X,X)$ be a topologically transitive dynamical system, and $V$ be a non-empty open subset of $X$, and $\nu$ be a locally-positive and atomless Borel probability measure on $X$.

Then, for every $\delta \in (0,1)$, does there exist:

  • $\{V_i\}_{i \in \mathbb{N}}$ are open subsets of $V$ satisfying $\nu\left( V-\cup_{n \in \mathbb{N}} V_n \right)=0$

  • A sequence $\{N_i\}_{i \in \mathbb{N}}$, such that the following holds: $$ \nu\left(X - \cup_{i \in \mathbb{N}} \phi^{-N_i}[V_i] \right)<\delta . $$

In intuitive words: there exists an open cover of a non-empty open set be reversed into an almost-everywhere cover of the entire space by appropriately reversing the dynamical sytem?

Auxiliary Definitions:

  • A Borel measure $\nu$ on $X$ is said to be locally-positive iff for every non-empty open subset $U\subseteq X$, $\nu(U)>0$. For example, if $X$ has more than two points then the Dirac is not such a measure.

  • $\phi$ is said to be topologically transitive iff for every two non-empty open subsets $U,V\subseteq X$ there exists some $N\in \mathbb{N}$ such that $$ \phi^N(U)\cap V \neq \emptyset. $$

  • 1
    $\begingroup$ If a solution $(V_i, N_i)_i$ exists, then $(V, N_i)_i$ is also a solution, so $\delta_1$ is unimportant. Is this intentional? $\endgroup$ Oct 30 '19 at 15:25
  • 2
    $\begingroup$ Do you require $V_i$ to be distinct? If not, since the existence of a solution implies $(V, N_i)_i$ is a solution, and if $(V, N_i)_i$ is allowed to be a solution, why not just ignore $V_i$ and work only with $V$? $\endgroup$ Oct 30 '19 at 15:47
  • $\begingroup$ Are you assuming that $\nu$ is finite? at least locally? $\endgroup$
    – YCor
    Oct 30 '19 at 16:23
  • $\begingroup$ (Deleted a bunch of comments since they have been clarified and now there are no obvious counterexamples.) $\endgroup$ Oct 30 '19 at 16:37
  • $\begingroup$ Thanks you Willie; I'll do the same. $\endgroup$
    – BLBA
    Oct 30 '19 at 16:38

No, even if we assume $\nu$ to be invariant under $\phi$.

Let $X = \{0,1\}^\mathbb{Z}$ be the set of two-way infinite binary sequences with the prodiscrete topology, and let $\phi$ be the left shift on $X$. Let $\nu = (\mu_1 + \mu_2)/2$ where $\mu_1$ is the uniform Bernoulli measure on $X$ and $\mu_2$ is an atomless $\phi$-invariant probability measure on some proper subshift of $X$. For simplicity, let's choose $\mu_2$ as the Parry measure on the shift of finite type $Y \subsetneq X$ where $0 0$ is forbidden. Let $V = \{ x \in X : x_0 = x_1 = 0 \}$ be the set of sequences with an occurrence of the forbidden word $0 0$ at the origin. These definitions satisfy your requirements: $\phi$ is well-known to be transitive, $\nu$ gives positive measure to each nonempty clopen set (which form a basis of the topology) and has no atoms, and $V$ is a nonempty open set.

Consider an open cover $(V_i)_{i \in \mathbb{N}}$ of $V$ and a sequence $(N_i)_{i \in \mathbb{N}}$ of integers. For each $i$ the translate $\phi^{-N_i} V_i$ is disjoint from $Y$, so $\nu(X - \bigcup_{i \in \mathbb{N}} \phi^{-N_i} V_i) \geq \nu(Y) = 1/2$.

But yes, if we strengthen the assumptions further.

In my counterexample the ergodic decomposition of $\nu$ features a positive-weight measure $\mu_2$ which is not locally positive. Let's thus assume that $\nu$ has an ergodic decomposition as an integral $\nu = \int_{E(M_\phi)} x \, d\mu(x)$ over the $\phi$-ergodic probability measures on $X$ and $\mu$-a.e. $x \in E(M_\phi)$ is locally positive. Then $x(V) > 0$ holds for those measures $x$. Since they are ergodic, this implies $x(\bigcup_{i \in \mathbb{N}} \phi^{-i} V) = 1$, so that $\nu(\bigcup_{i \in \mathbb{N}} \phi^{-i} V) = \int x(\bigcup_{i \in \mathbb{N}} \phi^{-i} V) \, d\mu(x) = 1$. Then $V_i = V$ and $N_i = i$ give the sequence you're looking for, for every $\delta > 0$. Note that even if I didn't use transitivity in this proof, it's implied by the existence of a locally positive ergodic measure.

Depending on your application, the assumptions of $\phi$-invariance and local positivity of the ergodic decomposition may be too strong. In the context of dynamical systems invariance seems natural, but by itself it's not enough.

  • $\begingroup$ If instead, we strengthen the assumptions on $X$ and $\phi$; namely that it is a separable Banach space and we assume that $\phi$ is linear. Then the counterexample seems to vanish. In this case, it's not clear that the added assumptions are still necessary. Am I right? $\endgroup$
    – BLBA
    Oct 31 '19 at 9:03
  • 2
    $\begingroup$ @MrMMS We can do something similar. Let $X = \ell^1$ and $\phi(x)_n = 2 x_{n+1}$. Define $\mu_1$ by first choosing $c$ from a distribution that is locally positive on $[0, \infty)$, and then choosing each $x_n$ independently and uniformly from $[-c/2^n, c/2^n]$. Define $\mu_2$ similarly but with $[0, c/2^n]$. These are invariant, atomless, and $\mu_1$ is locally positive: For an open ball $B_r(y)$, take $N$ with $\sum_{n \geq N} |y| < r/2$. For a $\mu_1$-random $x$, $\sum_{n \geq N} |x| < r/2$ and $\sum_{n < N} |x-y| < r/2$ with positive probability. Take $V = \{ x \in X : x_0 \in (-1,0) \}$. $\endgroup$ Oct 31 '19 at 10:42
  • $\begingroup$ By the way, it feels like I'm chasing some moving goalposts here. Do you have a specific dynamical system in mind? Or is the case of linear continuous transitive functions on separable Banach spaces the exact setting that you need? $\endgroup$ Oct 31 '19 at 10:46
  • $\begingroup$ It's indeed the exact setting (I was hoping to get something more general) if possible. $\endgroup$
    – BLBA
    Oct 31 '19 at 10:46
  • 2
    $\begingroup$ In both examples, $\nu(X - \bigcup_i \phi^{-i} V)$ can be any number $a \in [0,1]$, since we can choose $\nu = (1-a) \mu_1 + a \mu_2$. $\endgroup$ Oct 31 '19 at 10:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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