# Robustness of ergodic dynamical systems

Let $$\mathbf X := (X, \mathcal F, \mu)$$ be a standard probability space. For an ergodic measure preserving transformation $$T$$, we define the ergodic robustness $$\mathcal R(T)$$ of $$T$$ as follows:

For $$0 \leq r \leq 1$$, let $$C_r \subset \mathbb N^{\mathbb N}$$ be the subset of monotonically increasing sequences whose natural density exists and is greater than of equal to $$r$$.

Define the quantity $$E(T)$$ by

$$E(T) := \inf \Big\{r \in [0, 1]\;\big| \text{ For all } f \in L^1 (X), (n_k) \in C_r, \lim_{K \to \infty} \frac{1}{K} \sum_{k = 0}^{K-1} f(T^{n_k} (x)) = \int f d\mu \text{ for a.e } x \in X\Big\}.$$

Finally, define $$\mathcal R(T) = 1 - E(T)$$.

Question: Do there exist ergodic measure preserving transformations $$T$$ with $$\mathcal R(T)$$ arbitrarily close to $$1$$? That is, for every $$\varepsilon > 0$$, does there exist an ergodic transformation $$T$$ with $$\mathcal R(T) > 1 - \varepsilon$$?

Remark: The choice to use $$\mathcal R(T)$$ instead of $$E(T)$$ is purely aesthetic to fit the terminology.

Yes. Let $$\nu$$ be a probability measure on $$[0,1]$$. Let $$T$$ be the left shift on a sequence space $$X:=[0,1]^{\mathbb N}$$ equipped with the product $$\sigma$$-field and the product measure $$\mu:=\nu^{\mathbb N}$$. Then for every strictly increasing sequence $$\{n_k\}$$ of positive lower density and $$f \in L^1 (X)$$ we have $$(*) \quad \lim_{K \to \infty} \frac{1}{K} \sum_{k = 0}^{K-1} f(T^{n_k} (x)) = \int f d\mu \text{ for a.e } x \in X \, ,$$ so $$E(T)=0$$. To verify $$(*)$$, it suffices to check it for a dense collection of functions in $$L^1$$. This reduction follows from a Theorem of Garsia (a version of the Banach principle) as stated e.g. in Theorem 4.2 of , see also its application in Theorem 4.3.
If $$f$$ depends only on the first $$q$$ coordinates, then $$(*)$$ follows (for all sequences $$n_k$$) from the law of large numbers if you separate $$n_k$$ into $$q$$ subsequences where each subsequence has gaps at least $$q$$. Since such functions $$f$$ are dense, this completes the proof.
• This system is strong mixing right? I would think that this (or weak mixing maybe) is a necessary condition for the robustness to be $1$. Hmm.. Jun 22 at 3:35