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 $$ \phi^N(U)\cap V \neq \emptyset. $$

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    $\begingroup$ 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, 2019 at 15:25
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    $\begingroup$ 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, 2019 at 15:47
  • $\begingroup$ Are you assuming that $\nu$ is finite? at least locally? $\endgroup$
    – YCor
    Oct 30, 2019 at 16:23
  • $\begingroup$ (Deleted a bunch of comments since they have been clarified and now there are no obvious counterexamples.) $\endgroup$ Oct 30, 2019 at 16:37
  • $\begingroup$ Thanks you Willie; I'll do the same. $\endgroup$
    – ABIM
    Oct 30, 2019 at 16:38

1 Answer 1


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? $\endgroup$
    – ABIM
    Oct 31, 2019 at 9:03
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    $\begingroup$ @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, 2019 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, 2019 at 10:46
  • $\begingroup$ It's indeed the exact setting (I was hoping to get something more general) if possible. $\endgroup$
    – ABIM
    Oct 31, 2019 at 10:46
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    $\begingroup$ 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, 2019 at 10:53

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