**Problem set up:**


Let $\mathbf X := (X, \mathcal A, \mu)$ be a standard probability space.


We say that a measure preserving transformation $T$ on $\mathbf X$ is *$\varepsilon$-almost weakly mixing* if for every $\delta > \varepsilon$, and every pair of non-null measurable sets $A, B \in \mathcal A$, there exists an $N > 0$ such that for all $n > N$, we have $$\frac{1}{n} \sum\_{k = 1}^{n} |\mu(T^{-k}A \cap B) - \mu(A)\mu(B)| < \delta \mu(A)\mu(B).$$


We say a measure preserving transformation $G$ on $X$ *admits an ergodic component* if there exists some non-null measurable subset $E$ of $X$ such that $G(E) \subset E$, and the “restricted system” ($\mathbf E, G\_{|E})$, with $\mathbf E := (E, \mathcal A\_{|E}, \mu\_{|E})$ is ergodic. Here $\mathcal A\_{|E}$ is the restricted sigma algebra, and $\mu\_{|E}$ is defined by $\mu\_{|E}(A) := \mu(A \cap E)/\mu(E)$.



> 
> **Question:** Does there exist some $\varepsilon > 0$ such that any $\varepsilon$-almost weakly mixing transformation $T$ on $\mathbf X$ admits an ergodic component?
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**Remark:** This is a potential sharpening of an [earlier result](https://mathoverflow.net/questions/392667/does-an-almost-mixing-transformation-admit-a-non-null-ergodic-component).



Does an “almost weakly mixing” transformation admit a non-null ergodic component?
Your question is equivalent to asking whether $T$ being $\epsilon$-almost weak mixing implies that the invariant $\sigma$-algebra, $\mathcal I$ contains atoms. The answer is yes.


I will prove the contrapositive. I claim if $\mathcal I$ contains no atoms, then $T$ is not $\epsilon$-almost weak mixing for any $\epsilon>0$. To see this, notice that there are invariant sets of arbitrarily small measure. Let $A$ be an invariant set and let $B=A$. Then for each $k$, $|\mu(T^{-k}A\cap B)-\mu(A)\mu(B)|=\mu(A)-\mu(A)^2=\mu(A)(1-\mu(A))=\frac {1-\mu(A)}{\mu(A)}\mu(A)$.
In particular, $T$ is not $\epsilon$-almost weak mixing.


