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.?



A sufficient condition for uniform convergence of ergodic averages?
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.


