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.?

  • $\begingroup$ 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

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}


\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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.