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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?

Remark: This is a potential sharpening of an earlier result.

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1 Answer 1

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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.

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    $\begingroup$ Very nice answer! So I guess this condition turns out to be much stronger than expected… $\endgroup$
    – Nate River
    Commented Jul 23, 2021 at 2:09
  • $\begingroup$ Actually, is there an even simpler proof? I claim that the entire system is ergodic, otherwise we can find invariant sets $E$ and $E^c$ of non null measure - then taking $A = E$ and $B = E^c$ in the criterion for $\varepsilon$-almost weak mixing, we find that the system is not $\varepsilon$-almost weak mixing for any $\varepsilon$. If this is right, we can possibly ask for much stronger conclusions than merely ergodic. $\endgroup$
    – Nate River
    Commented Jul 23, 2021 at 6:51
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    $\begingroup$ If $X$ is the union of two mixing systems of measure $\frac 12$ each, say $X_1\cup X_2$, then take sets $A$ and $B$ with $\mu(A\cap X_i)=a_i$ and $\mu(B\cap X_i)=b_i$. Then for large $k$, $|\mu(T^{-k}A\cap B)-\mu(A)\mu(B)|\approx |(a_1b_1/\mu(X_1)+a_2b_2/\mu(X_2))-(a_1+a_2)(b_1+b_2)|$ $=|a_1b_1 + a_2b_2 - a_1b_2 - a_2b_1|=|(a_1-a_2)(b_1-b_2)|\le \mu(A)\mu(B)$, so that $T$ is 1-almost weakly mixing. $\endgroup$
    – Anthony Quas
    Commented Jul 23, 2021 at 6:58

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