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Given a measure preserving system $(X, mu, T)$ where $\mu(X) = 1$, we say $T$ is uniformly weak mixing if for all measurable $A$, $B$ and every $\varepsilon > 0$, there exists some $N$ independent of $A$ and $B$ such that

$|\sum_{k=1}^n (1/k)\mu(T^{-k} A \cap B) - mu(A) mu(B)| < \varepsilon \mu(B)$ for all $n > N$.

If $T$ is uniformly weak mixing, then does it hold that for all bounded measurable functions $f$,

$1/n \sum_{k=1}^n T^k f \to \int f d\mu$ uniformly a.e.?

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  • $\begingroup$ I'm just curious if you have any example of a uniformly weak mixing system in which the measure is not a finite sum of atoms... $\endgroup$
    – fedja
    May 17 '19 at 12:19
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I assume the uniformly weakly mixing condition is modified to read

\begin{equation} \left \vert \frac{1}{n} \sum_{k=1}^n \mu(T^{-k}A \cap B) -\mu(A)\mu(B) \right \vert \leq \epsilon \mu(B) \end{equation}

The condition as written in the question does not make sense since if we apply that condition to $A = B = X$ we obtain the partial sums of the harmonic series for the 'averages' on the left.

I claim that if $X$ is nonatomic then there are no uniformly weakly mixing systems on $X$. This can be seen as follows. First observe that a uniformly weakly mixing system must be ergodic, as a nontrivial invariant set clearly violates the condition. This means in particular that $T$ must be aperiodic.

Let $\epsilon = \frac{1}{100}$ and assume toward a contradiction that there exists $N$ as in the uniformly weakly mixing condition. Assume without loss of generality that $N \geq 10^{10}$.

By Rokhlin's lemma there exists a measurable set $C \subseteq X$ such that the sets $C,T^{-1}C,\ldots,T^{-N^2}C$ are pairwise disjoint and such that \begin{equation} \mu(C\cup T^{-1}C \cup \cdots \cup T^{-N^2}C) \geq \frac{1}{2} \end{equation} Since $X$ is nonatomic we can find a measurable set $D \subseteq C$ such that \begin{equation} \mu(D \cup T^{-1}D \cup \cdots \cup T^{-N^2}D) = \frac{1}{2} \end{equation}

For $k \leq N$ we have

\begin{align}&(D \cup T^{-1}D \cup \cdots \cup T^{-N^2}D) \triangle T^{-k}(D \cup T^{-1}D \cup \cdots \cup T^{-N^2}D)\\ &\subseteq C \cup T^{-1}C \cup \cdots \cup T^{-N+1}C \cup T^{-N^2}C \cup \cdots \cup T^{-N^2-N}C\end{align}

Since $\mu(C) \leq \frac{1}{N^2}$ we have

\begin{align} \mu(C \cup T^{-1}C \cup \cdots \cup T^{-N+1}C \cup T^{-N^2}C \cup \cdots \cup T^{-N^2-N}C) & \leq 2N \cdot \frac{1}{N^2} \\ & \leq \frac{2}{10^{10}} \end{align}

Therefore

\begin{align} &\mu((D \cup T^{-1}D \cup \cdots T^{-N^2}D) \cap T^{-k} (D \cup T^{-1}D \cup \cdots T^{-N^2}D)) \\ &= \mu(D \cup T^{-1}D \cup \cdots T^{-N^2-k}D)\\& \hspace{1 cm} - \mu((D \cup T^{-1}D \cup \cdots \cup T^{-N^2}D) \triangle T^{-k}(D \cup T^{-1}D \cup \cdots \cup T^{-N^2}D) ) \\ & \geq \frac{1}{2}-\frac{2}{10^{10}} \geq \frac{49}{100} \end{align}

It follows that if we let $A = B = D \cup T^{-1}D \cup \cdots T^{-N^2}D$ then for $n=N$ we have

\begin{equation} \frac{1}{n} \sum_{k=1}^n \mu(T^{-k}A \cap B) \geq \frac{49}{100} \end{equation}

Thus the uniformly weakly mixing inequality becomes

\begin{equation} \frac{1}{200} \geq \frac{49}{100} - \frac{1}{4} \end{equation}

which is the desired contradiction.

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  • $\begingroup$ The intuition is that a measure-preserving transformation always admits sets which are 'almost invariant'. $\endgroup$ May 17 '19 at 13:29

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