Does uniform recurrence imply uniform convergence of the Birkhoff sums?

Question

Let $(X, T, \mathcal F, \mu)$ be an ergodic measure preserving system with finite measure.

Suppose $T$ is uniformly recurrent, in the following sense:

For every $A \in \mathcal F$, there exists an $M \in \mathbb N$ such that for almost every $x \in A$, we have $T^{n(x)} \, (x) \in A$ for some $n(x) \leq M$.

Question: Does it follow that the convergence of the Birkhoff sums is essentially uniform? That is, for every $f \in L^1 (X)$, there exists a measurable set $E$ of full measure such that

$$\frac{1}{N}\sum_{i = 0}^{N-1} f(T^n (x)) \to \int f \, d\mu$$

uniformly on $E$ as $N \to \infty$.

Answer 1

This is just a comment on your definition of "uniformly recurrent". Can you give an example of a system you'd consider uniformly recurrent?

If this notion is standard, the following must be nonsense, but it seems to me that your notion does not make much sense.

Lemma. If $T$ is ergodic and uniformly recurrent, then it is a single finite cycle.

Proof. Suppose first that the measure of points with an eventual period is $0$. Then by the Rokhlin lemma for noninvertible transformations [1], there exists an $n$-tower for $f$ of measure at least $1 - \epsilon$ for each $\epsilon > 0$, meaning a set $B_{(n)}$ such that for some base $B$, $B_{(n)} = \bigcup_{0 \leq k < n} T^{-k}(B)$ is disjoint and has measure at least $1 - \epsilon$ and the sets $f^{-1}_k(B)$ are disjoint.

From this it is easy to see that there also exist arbitrarily small such sets, just take small subsets of $B$. From this we obtain that we can find for each $n \geq 1$ pick an $n$-tower $B_{(n)}$ such that these towers are disjoint for distinct $n$. Now denoting by $B^n$ is the base of $B_{(n)}$, define $A = \bigcup_{n > 1} B^n \cup T^{-n+1}(B^n)$.

Note that $B^n$ and $T^{-n+1}(B^n)$ have equal measure, so if you pick a random point $x$ in $A$, it will be in in a set of the form $T^{-n+1}(B^n)$ (for some $n$) with probability $0.5$. But then the minimal $n$ of $T$ such that $T^n(x) \in A$ is $n-1$. It follows that $T$ is not uniformly recurrent.

We conclude that there must exist $n, p$ such that the probability that $T^n(x) = T^{n+p}(x)$ is positive. In particular, the shift-invariant set of points satisfying $T^p(x) = x$ has positive measure, so this measure must be $1$. It is well-known that in this case, we can find a cross-section $S$ such that $X = S \cup T(s) \cup \cdots T^{n-1}(S)$ and the union is disjoint (assuming this is on a standard Lebesgue space $[0, 1]$, I think the usual construction is to pick the minimal point on each orbit).

Ergodicity implies that $S$ must be a singleton, as otherwise we can split it in two. Square.

Of course in this case, the answer to your question is "yes".

[1] Avila, Artur; Candela, Pablo, Towers for commuting endomorphisms, and combinatorial applications, Ann. Inst. Fourier 66, No. 4, 1529-1544 (2016). ZBL1360.28012.