Does mixing automatically imply this seemingly stronger "uniform modulo re-ordering" version of mixing?

Question

THE QUESTION

Let $(X,\mathcal{X})$ be a standard Borel space, $T \colon X \to X$ a measurable map, and $\mu$ a $T$-mixing probability measure.

Is it necessarily the case that for all $A \in \mathcal{X}$ and $\varepsilon>0$, there exists $N_{A,\varepsilon} \in \mathbb{N}$ such that for all $B \in \mathcal{X}$, $$ \#\{n \in \mathbb{N} : |\mu(A \cap T^{-n}(B))-\mu(A)\mu(B)| \geq \varepsilon \} \leq N_{A,\varepsilon} \text{?} $$

Below, I will discuss the meaning and significance of this question in much more detail.

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-- BUT FIRST, A PROGRESS UPDATE --

By adapting Ronnie Pavlov's very nice comments, I can give a result in the direction towards my question, plus a consequent possible strategy for obtaining an affirmative answer to my question.

Let $(X,\mathcal{X})$ be a standard Borel space, $T \colon X \to X$ a measurable map, and $\mu$ a $T$-invariant probability measure. For each $A,B \in \mathcal{X}$ write $$ \rho_n(A,B) = |\mu(A \cap T^{-n}(B))-\mu(A)\mu(B)|. $$

Basic Proposition. To show that $(X,\mathcal{X},\mu,T)$ has the property described in my question, it is sufficient just to consider $A$ belonging to a set $\mathcal{Y} \subset \mathcal{X}$ where $\mathcal{Y}$ is closed under pairwise intersections and $\sigma(\mathcal{Y})=\mathcal{X}$.

Proposition A. If $(X,\mathcal{X},\mu,T)$ is mixing then for every $A \in \mathcal{X}$, every sequence $(B_k)_{k \in \mathbb{N}}$ in $\mathcal{X}$ and every $\varepsilon>0$, $$ \#\{ n \in \mathbb{N} : \#\{k \in \mathbb{N} : \rho_n(A,B_k) < \varepsilon \} < \infty \} < \infty\text{.} $$

(Note that taking constant sequences $B_k=B$ precisely recovers the usual definition of mixing. So this result says that mixing automatically implies a seemingly slightly stronger version of mixing.)

With this, we have the following potential strategy for proving an affirmative answer to the question.

Definition. I will say that $(X,\mathcal{X},\mu,T)$ has re-orderable correlations if there exists a set $\mathcal{Y} \subset \mathcal{X}$ closed under pairwise intersections and with $\sigma(\mathcal{Y})=\mathcal{X}$, such that for every $A \in \mathcal{Y}$ there exists an unbounded sequence $(r_n)_{n \in \mathbb{N}}$ in $\mathbb{N}$, a sequence $(M_N)_{N \in \mathbb{N}}$ in $\mathbb{N}$ and a function $\delta \colon (0,\infty) \to (0,\infty)$ such that for every $\varepsilon>0$ and $N \in \mathbb{N}$, if there exists $B \in \mathcal{X}$ with $$ \#\{n \in \mathbb{N} : \rho_n(A,B) \geq \varepsilon\} \geq M_N $$ then there exists $\tilde{B} \in \mathcal{X}$ with $$ \rho_{r_n}(A,\tilde{B}) \geq \delta(\varepsilon) \quad \forall \, n \in \{1,\ldots,N\}\text{.} $$

Proposition B. If $(X,\mathcal{X},\mu,T)$ is mixing and has re-orderable correlations, then $(X,\mathcal{X},\mu,T)$ has the property described in my question.

So a possible strategy towards proving an affirmative answer is to show that mixing (or perhaps something weaker than mixing) implies re-orderable correlations.

For example, Ronnie Pavlov has shown in the comments that a two-sided Bernoulli shift has re-orderable correlations (with $\mathcal{Y}$ the set of cylinder sets and with $\delta(\varepsilon)=\varepsilon$), and so a two-sided Bernoulli shift fulfils the property described in my question.

Proofs of the above results will be given in the "PROOFS" section at the end of this post.

-- END OF PROGRESS UPDATE --

Now I will return to discussing the significance of the question in more detail.

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Introduction

My question is about whether a mixing measure-preserving dynamical system necessarily exhibits a certain (very weak) kind of "uniformity" of mixing.

The question is particularly motivated by consideration of random dynamical systems (RDS): For a RDS, one can define notions of "almost-sure decay of random correlations" for observables defined over the product of the state space and the underlying probability space; but the result of such a definition is that when we reduce to the "deterministic" case in which the action of the RDS is independent of the noise, since the observables still incorporate the noise, the property of "almost-sure decay of random correlations" does not clearly reduce to classical "decay of correlations" (i.e. mixing): rather, it needs to incorporate at least some level of "extra uniformity" to take account of the randomness in the observable. But perhaps this "extra uniformity" is automatically guaranteed, in which case there is no problem - this is the motivation behind my question.

The structure of the rest of this post is as follows:

• First, I will introduce a relevant notion of convergence of functions that lies strictly between pointwise convergence and uniform convergence.
• Then, I will define (for classical measure-preserving dynamical systems) both classical mixing and the kind of "uniformity" of mixing that I am asking about in my question. (I will then re-pose my question in terms of the terminology introduced.)
• Then, I will describe how my stronger version of mixing is precisely equivalent to saying that the system is guaranteed to have "almost-sure decay of random correlations" when considered as a trivial RDS over a noise space.
• Finally, I will give all the proofs of results stated along the way.

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"Uniform convergence modulo re-ordering"

Suppose we have a set $S$, a function $f \colon S \to \mathbb{R}$ and a sequence $(f_n)_{n \in \mathbb{N}}$ of functions $f_n \colon S \to \mathbb{R}$.

Definition. We say that $f_n \to f$ uniformly modulo re-ordering as $n \to \infty$ if for all $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that every $x \in S$ has $$ \#\{n \in \mathbb{N} : |f_n(x)-f(x)| \geq \varepsilon \} \leq N\text{.} $$

[The name "uniformly modulo re-ordering" is based on characterisation 2 in Proposition 1 below.]

Note that uniform convergence modulo re-ordering is indeed an asymptotic property: if $f_n \to f$ uniformly modulo re-ordering and we have a sequence of functions $\tilde{f}_{\!\!n} \colon S \to \mathbb{R}$ such that $\tilde{f}_{\!\!n}=f_n$ for all sufficiently large $n$, then $\tilde{f}_{\!\!n} \to f$ uniformly modulo re-ordering. (Taking $\tilde{N}$ such that $\tilde{f}_{\!\!n}=f_n$ for all $n \geq \tilde{N}$, just replace $N$ with $N+\tilde{N}$.)

Other characterisations of uniform convergence modulo re-ordering

We now give a couple of alternative characterisations of uniform convergence modulo re-ordering (one of which assumes additional measurable structure).

Given any $\mathcal{N} \subset \mathbb{N}$, we will say that a function $\pi \colon \mathbb{N} \to \mathbb{N}$ is $\mathcal{N}\!$-almost bijective if $\pi$ is injective and $\mathbb{N} \setminus \mathcal{N} \subset \pi(\mathbb{N})$.

For each $x \in S$, let $\mathcal{N}_x:=\{n \in \mathbb{N} : f_n(x)=f(x)\}$.

Proposition 1. (A) The following two statements are equivalent:

1. $f_n \to f$ uniformly modulo re-ordering as $n \to \infty$.
2. There exists an $S$-indexed family $(\pi_x)_{x \in S}$ of functions $\pi_x \colon \mathbb{N} \to \mathbb{N}$ with $\pi_x$ being $\mathcal{N}_x\!$-almost bijective for all $x \in S$, such that defining $g_n(x):=f_{\pi_x(n)}(x)$, we have $g_n \to f$ uniformly as $n \to \infty$.

(B) Furthermore, if $S$ is equipped with a $\sigma$-algebra $\mathcal{S}$ such that $f_n$ and $f$ are $\mathcal{S}$-measurable, then statements 1 and 2 above are equivalent to:

3. For every probability measure $\mathbb{P}$ on $(S,\mathcal{S})$ and every $\varepsilon>0$, $$ \sum_{n \in \mathbb{N}} \mathbb{P}(x \in S : |f_n(x)-f(x)| \geq \varepsilon ) < \infty\text{.} $$

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Mixing, and my question.

Let $(X,\mathcal{X},\mu,T)$ be a measure-preserving dynamical system. (I assume in this terminology that $\mu$ is a probability measure.) For each $n \in \mathbb{N}_0$, we define the function $\rho_n \colon \mathcal{X} \times \mathcal{X} \to [0,1]$ by $$ \rho_n(A,B) = |\mu(A \cap T^{-n}(B)) - \mu(A)\mu(B)|. $$

Definition. We say that $(X,\mathcal{X},\mu,T)$ is mixing if $\rho_n(A,B) \to 0$ as $n \to \infty$ for all $A,B \in \mathcal{X}$.

Remark. If $(X,\mathcal{X},\mu,T)$ is mixing then we have that for all bounded measurable $g_1,g_2 \colon X \to \mathbb{R}$, $$ \int_X g_1 . (g_2 \circ T^n) \, d\mu \to \int_X g_1 \, d\mu \int_X g_2 \, d\mu \ \text{ as } n \to \infty. $$

Now for each $A \in \mathcal{X}$ and $n \in \mathbb{N}$, we write $\rho_n(A,\,\boldsymbol{\cdot}\,) \colon \mathcal{X} \to [0,1]$ for the map $B \mapsto \rho_n(A,B)$. So then, $(X,\mathcal{X},\mu,T)$ is mixing if and only if for each $A \in \mathcal{X}$, $\rho_n(A,\,\boldsymbol{\cdot}\,) \to 0$ pointwise as $n \to \infty$. Hence we can define the following potentially stronger notion of mixing.

Definition. I will say that $(X,\mathcal{X},\mu,T)$ is nicely mixing if for each $A \in \mathcal{X}$, $\rho_n(A,\,\boldsymbol{\cdot}\,) \to 0$ uniformly modulo re-ordering as $n \to \infty$.

Equivalently, $(X,\mathcal{X},\mu,T)$ is nicely mixing if for each $A \in \mathcal{X}$, as $n \to \infty$ the function $B \mapsto \mu(A \cap T^{-n}(B))$ converges to the function $B \mapsto \mu(A)\mu(B)$ uniformly modulo re-ordering.

This notion of "nicely mixing" may at first seem like a somewhat strange definition, but the Theorem further below gives another characterisation of "nicely mixing".

Now my question again:

Assume that $(X,\mathcal{X})$ is a standard Borel space. Is it necessarily the case that if $(X,\mathcal{X},\mu,T)$ is mixing then $(X,\mathcal{X},\mu,T)$ is nicely mixing?

Remark. Regarding the same question with actual uniform convergence in place of the weaker "uniform convergence modulo re-ordering", the answer is fairly clear: If $T$ is invertible and $\mu(A) \in (0,1)$, then we do not have that $\rho_n(A,B) \to 0$ uniformly across $B \in \mathcal{X}$ as $n \to \infty$, since for each $n$ we can just set $B=T^n(A)$, in which case $\rho_n(A,B)=\mu(A)-\mu(A)^2$ for all $n$.

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Motivation: Sample-pathwise mixing in random dynamical systems

For a measurable skew-product map $\Theta \colon \Omega \times X \to \Omega \times X$ whose state space is a standard Borel space $(X,\mathcal{X})$ and whose base is an invertible measure-preserving dynamical system $(\Omega,\mathcal{F},\mathbb{P},\theta)$, given a $\Theta$-invariant measure $\mu$ on $\Omega \times X$ that projects onto $\mathbb{P}$ with disintegration $(\mu_\omega)_{\omega \in \Omega}$, one can define a "random correlation function" $\rho_{g_1,g_2}(n,\omega)$ for a pair of bounded measurable functions $g_1,g_2 \colon \Omega \times X \to \mathbb{R}$, by \begin{align*} &\rho_{g_1,g_2}(n,\omega) = \\ &\ \left| \int_X g_1\!(\omega,x) \, (g_2 \circ \Theta^n)(\omega,x) \, \mu_\omega(dx) - \int_X g_1(\omega,x) \, \mu_\omega(dx) \int_X g_2(\theta^n\omega,y) \, \mu_{\theta^n\omega}(dy) \right|\text{.} \end{align*} For each $g_1,g_2$, this is well-defined up to $\mathbb{P}$-a.s. equality.

(Such "random correlation functions" appear in e.g. the question Two mixing rates of random dynamical system and the paper Quenched decay of correlations for slowly mixing systems.)

It might seem natural to define "almost-sure mixing" as follows: for every pair of bounded measurable functions $g_1,g_2 \colon \Omega \times X \to \mathbb{R}$, we have that for $\mathbb{P}$-almost all $\omega \in \Omega$, $\rho_{g_1,g_2}(n,\omega) \to 0$ as $n \to \infty$.

However, if the answer to my question is no, then this definition does not "reduce to classical mixing" when the skew-product structure is a direct product structure. To be precise, we have the following theorem (which, for simplicity, I will formulate just in terms of indicator-function observables rather than general bounded measurable observables, although I expect the theorem to generalise to bounded measurable observables).

Theorem. A measure-preserving dynamical system $(X,\mathcal{X},\mu_X,T)$ is nicely mixing if and only if for every invertible measure-preserving dynamical system $(\Omega,\mathcal{F},\mathbb{P},\theta)$, taking $\Theta(\omega,x):=(\theta\omega,T(x))$ and $\mu:=\mathbb{P} \otimes \mu_X$, every $A,B \in \mathcal{F} \otimes \mathcal{X}$ has $\rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\,\boldsymbol{\cdot}\,) \overset{\mathbb{P}\text{-a.s.}}{\to} 0$ as $n \to \infty$.

The "if" direction can, more specifically, take the following form:

Proposition 2. Suppose $(\Omega,\mathcal{F},\mathbb{P},\theta)$ is the Bernoulli shift on the sequence space $\Omega=[0,1]^\mathbb{Z}$, where $[0,1]$ is equipped with the Lebesgue measure. Suppose $\Theta = \theta \times T$ and $\mu=\mathbb{P} \otimes \mu_X$, where $(X,\mathcal{X},\mu_X,T)$ is not nicely mixing. Then one can find $A,B \in \mathcal{F} \otimes \mathcal{X}$ such that for $\mathbb{P}$-almost every $\omega \in \Omega$, $\rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) \not\to 0$ as $n \to \infty$.

So the theorem holds even if "every invertible measure-preserving dynamical system" is replaced by "every Bernoulli automorphism".

The proof of the Theorem is based on the Borel-Cantelli Lemmas (used in a similar manner to the answer to my question Does a sequence of coin-tosses a.s. have a subsequence on which the remainder of the sequence can be identified with the position in the sequence?).

Remark. One other approach that I've seen to studying mixing in RDS is simply to consider $\omega$-independent bounded observables $g_1,g_2 \colon X \to \mathbb{R}$; but a potential problem with this as an approach towards defining mixing is that it respects the Cartesian-product structure of $\Omega \times X$ (as opposed to treating $\Omega \times X$ just as a space equipped with a projection onto $\Omega$), which goes somewhat contrary to the general philosophy of random dynamical systems theory.

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PROOFS.

Proof of Proposition 1(A). First assume 1. It is clear that $f_n \to f$ pointwise as $n \to \infty$, and so in particular, for every $x \in X$ and every non-empty subset $A$ of $\mathbb{N}$, the set $\{|f_n(x)-f(x)| : n \in A\}$ has a maximum. Therefore, for each $x \in X$, by recursion we can construct an injective function $\pi_x \colon \mathbb{N} \to \mathbb{N}$ such that for each $n \in \mathbb{N}$, $$ |f_{\pi_x(n)}(x)-f(x)| = \max \{|f_m(x)-f(x)| : m \in \mathbb{N} \!\setminus\! \pi_x(\mathbb{N}_{<n}) \}\text{.} $$ It is clear that $n \mapsto |f_{\pi_x(n)}(x)-f(x)|$ is decreasing and hence convergent; and since $\pi_x$ is injective and $|f_n(x)-f(x)| \to 0$ as $n \to \infty$, it is then clear that the limit $\lim_{n \to \infty} |f_{\pi_x(n)}(x)-f(x)|$ is $0$. Since, again, $n \mapsto |f_{\pi_x(n)}(x)-f(x)|$ is decreasing, it then clearly follows that every element of $\mathbb{N} \setminus \mathcal{N}_x$ is in the range of $\pi_x$. So $\pi_x$ is $\mathcal{N}_x\!$-almost bijective. It remains to show that $g_n \to f$ uniformly. Fix $\varepsilon>0$, and let $N$ be as in the definition of uniform convergence modulo re-ordering. Then for all $x \in S$ and $n > N$, since every $m < n$ has $|f_{\pi_x(m)}(x)-f(x)| \geq |f_{\pi_x(n)}(x)-f(x)|$ and we also have (by injectivity of $\pi_x$) that $$ \#\{m \in \mathbb{N} : |f_{\pi_x(m)}(x)-f(x)| \geq \varepsilon \} \leq N\text{,} $$ it follows that $|f_{\pi_x(n)}(x)-f(x)|<\varepsilon$, as required.

Now suppose 2. Fix $\varepsilon>0$, and let $N$ be such that every $n > N$ and $x \in S$ has $|g_n(x)-f(x)|<\varepsilon$. Then for each $x \in S$, every $n \in \mathbb{N}$ with $|f_n(x)-f(x)| \geq \varepsilon$ has $n \in \pi_x(\mathbb{N})$ and $\pi^{-1}(n) \leq N$. $\quad\square$

Proof of Proposition 1(B). For convenience, write $S_{n,\varepsilon}:=\{x \in S : |f_n(x)-f(x)| \geq \varepsilon\}$.

First assume 1. Fix $\varepsilon>0$, and let $N$ be as in the definition of uniform convergence modulo re-ordering. For any $R \subset \mathbb{N}$ with $\#R > N$, we have $\bigcap_{n \in R} S_{n,\varepsilon} = \emptyset$. So $$ \sum_{n \in \mathbb{N}} \mathbb{P}(S_{n,\varepsilon}) = \sum_{i=1}^N \mathbb{P}\!\left( \bigcup_{R \subset \mathbb{N}, \, \#R=i} \ \bigcap_{n \in R} S_{n,\varepsilon} \right) \leq N\text{.} $$ Now assume that 1 fails (i.e. $f_n$ does not converge to $f$ uniformly modulo re-ordering), and take a counterexemplary $\varepsilon>0$. For each $N \in \mathbb{N}$, choose $x_N \in S$ such that $$ \#\{n \in \mathbb{N} : x \in S_{n,\varepsilon} \} \geq 3^N\text{.} $$ Take $$ \mathbb{P} = \sum_{N \in \mathbb{N}} 2^{-N}\delta_{x_N}. $$ Then $\sum_{n \in \mathbb{N}} \mathbb{P}(S_{n,\varepsilon})$ is at least $(\frac{3}{2})^N$ for all $N$, and hence is infinite. $\quad\square$

Proof of "only if" direction in Theorem. Suppose that $(X,\mathcal{X},\mu_X,T)$ is nicely mixing, and take any invertible measure-preserving dynamical system $(\Omega,\mathcal{F},\mathbb{P},\theta)$. Let $\mathcal{A}$ be the set of all $A \in \mathcal{F} \otimes \mathcal{X}$ with the property that every $B \in \mathcal{F} \otimes \mathcal{X}$ has $\rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\,\boldsymbol{\cdot}\,) \overset{\mathbb{P}\text{-a.s.}}{\to} 0$ as $n \to \infty$. We will show that

• $\mathcal{A}$ is closed under complements and countable disjoint unions;
• $\mathcal{A}$ includes all rectangles $A=E \times Y$ with $E \in \mathcal{F}$ and $Y \in \mathcal{X}$.

Dynkin's $\pi$-$\lambda$ theorem then gives the desired result.

Given a set $A \subset \Omega \times X$ and a point $\omega \in \Omega$, we write $A_\omega \subset X$ for the $\omega$-section of $A$. Note that $$ \rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) = |\mu_X(A_\omega \cap T^{-n}(B_{\theta^n\omega}))-\mu_X(A_\omega)\mu_X(B_{\theta^n\omega})|. $$ It is easy to check that $\rho_{\mathbf{1}_{(\Omega \times X) \setminus A},\mathbf{1}_B}(n,\omega)=\rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega)$, and so $\mathcal{A}$ is closed under complements. Now let $(A(k))_{k \in \mathbb{N}}$ be a mutually disjoint sequence in $\mathcal{A}$, take any $B \in \mathcal{F} \otimes \mathcal{X}$, and let $\Omega' \subset \Omega$ be a $\mathbb{P}$-full measure set such that for all $\omega \in \Omega$ and $k \in \mathbb{N}$, $\rho_{\mathbf{1}_{A(k)},\mathbf{1}_B}(n,\omega) \to 0$ as $n \to \infty$. Letting $A=\bigcup_{k \in \mathbb{N}} A(k)$, the triangle inequality gives $$ \rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) \leq \sum_{k \in \mathbb{N}} \rho_{\mathbf{1}_{A(k)},\mathbf{1}_B}(n,\omega)\text{.} $$ Now for each $\omega$, $\sum_{k \in \mathbb{N}} \mu_X(A(k)_\omega) \leq 1$ and $\rho_{\mathbf{1}_{A(k)},\mathbf{1}_B}(n,\omega) \leq \mu_X(A(k)_\omega)$. Hence the dominated convergence theorem for discrete sums can be applied to give that for all $\omega \in \Omega'$, $$ \sum_{k \in \mathbb{N}} \rho_{\mathbf{1}_{A(k)},\mathbf{1}_B}(n,\omega) \to 0 \ \text{ as } n \to \infty $$ and so $\rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) \to 0$ as $n \to \infty$. Thus $\mathcal{A}$ is closed under countable disjoint unions. It now remains to show that $\mathcal{A}$ includes all measurable rectangles. First note that for a measurable rectangle $A=E \times Y$, for any $B \in \mathcal{F} \otimes \mathcal{X}$, we have $$ \rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) = \mathbf{1}_E(\omega)\rho_{\mathbf{1}_{\Omega \times Y},\mathbf{1}_B}(n,\omega). $$ Hence it is sufficient just to consider rectangles of the form $A=\Omega \times Y$ with $Y \in \mathcal{X}$. Let us define a $\sigma$-algebra $\Sigma_{\mu_X,T}$ on the $\sigma$-algebra $\mathcal{X}$: namely, let $\Sigma_{\mu_X,T}$ be the smallest $\sigma$-algebra on $\mathcal{X}$ with respect to which the map $Z \mapsto \mu_X(W \cap T^{-n}(Z))$ is measurable for every $W \in \mathcal{X}$ and $n \in \mathbb{N}_0$. Now fix $Y \in \mathcal{X}$ and $B \in \mathcal{F} \otimes \mathcal{X}$, and let $A=\Omega \times Y$. The map $$ \omega \ \ \mapsto \ \ \mu_X(W \cap T^{-n}(B_\omega)) = \! \int_X \mathbf{1}_W(x)\mathbf{1}_B(\omega,T^n(x)) \, \mu_X(dx) $$ is $\mathcal{F}$-measurable for all $W \in \mathcal{X}$ and $n \in \mathbb{N}_0$, and so the map $\omega \mapsto B_\omega$ is $(\mathcal{F},\Sigma_{\mu_X,T})$-measurable. So define the probability measure $\tilde{\mathbb{P}}$ on the measurable space $(\mathcal{X},\Sigma_{\mu_X,T})$ to be the pushforward measure of $\mathbb{P}$ under the map $\omega \mapsto B_\omega$. For each $\varepsilon>0$ and $n \in \mathbb{N}$, let $$ \mathcal{E}_{\varepsilon,n} = \{\tilde{B} \in \mathcal{X} : |\mu_X(Y \cap T^{-n}(\tilde{B}))-\mu_X(Y)\mu_X(\tilde{B})| \geq \varepsilon \}. $$ Note that $\mathcal{E}_{\varepsilon,n} \in \Sigma_{\mu_X,T}$ and, since $\mathbb{P}$ is $\theta$-invariant, $$ \mathbb{P}(\omega : \rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) \geq \varepsilon ) = \tilde{\mathbb{P}}(\mathcal{E}_{\varepsilon,n}). $$ Since $(X,\mathcal{X},\mu_X,T)$ is nicely mixing, Proposition 1(B) gives that for all $\varepsilon>0$, $\sum_{n \in \mathbb{N}} \tilde{\mathbb{P}}(\mathcal{E}_{\varepsilon,n}) < \infty$, and so $$ \sum_{n \in \mathbb{N}} \mathbb{P}(\omega : \rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) \geq \varepsilon ) < \infty. $$ Hence, by the First Borel-Cantelli Lemma, for each $\varepsilon>0$, $\mathbb{P}$-almost every $\omega \in \Omega$ has that $\rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega)<\varepsilon$ for all sufficiently large $n$; and taking a sequence of $\varepsilon$-values tending to $0$ then gives that for $\mathbb{P}$-almost every $\omega \in \Omega$, $\rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) \to 0$ as $n \to \infty$. So $A \in \mathcal{A}$. $\quad\square$

Proof of Proposition 2. Choose an $\tilde{A} \in \mathcal{X}$ and $\varepsilon>0$ such that for all $N \in \mathbb{N}$ one can find $\tilde{B} \in \mathcal{X}$ with $$ \#\{n \in \mathbb{N} : |\mu_X(\tilde{A} \cap T^{-n}(\tilde{B}))-\mu_X(\tilde{A})\mu_X(\tilde{B})| \geq \varepsilon\} > N\text{.} $$ Set $A=\Omega \times \tilde{A}$. In analogy to the proof of 3$\Rightarrow$1 in Proposition 1(B), for each $N \in \mathbb{N}$ let $B_N \in \mathcal{X}$ be such that the set $$ R_N := \{n \in \mathbb{N} : |\mu_X(\tilde{A} \cap T^{-n}(B_N))-\mu_X(\tilde{A})\mu_X(B_N)| \geq \varepsilon\} $$ is of cardinality at least $3^N$. Let $$ O = \bigcup_{N=1}^\infty ([2^{-N},2^{1-N}] \times B_N) \, \subset [0,1] \times X \text{,} $$ and let $B$ be the pre-image of $O$ under the projection $((\omega_i)_{i \in \mathbb{Z}},x) \mapsto (\omega_0,x)$ from $\Omega \times X$ to $[0,1] \times X$. So $$ \rho_{\mathbf{1}_A,\mathbf{1}_B}(n,(\omega_i)_{i \in \mathbb{Z}}) = \sum_{N=1}^\infty \mathbf{1}_{[2^{-N},2^{1-N}]}(\omega_n)|\mu_X(\tilde{A} \cap T^{-n}(B_N))-\mu_X(\tilde{A})\mu_X(B_N)|. $$ Note in particular that the random variables $\rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\,\boldsymbol{\cdot}\,)$ are mutually $\mathbb{P}$-independent. Furthermore, writing $Q_n:=\{N \in \mathbb{N} : n \in R_N \}$, we have $$ \mathbb{P}(\omega : \rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) \geq \varepsilon ) = \sum_{N \in Q_n} 2^{-N} $$ for each $n$, and so $$ \sum_{n \in \mathbb{N}} \mathbb{P}(\omega : \rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) \geq \varepsilon ) = \sum_{N,n \text{ with } N \in Q_n} 2^{-N} = \sum_{N \in \mathbb{N}} 2^{-N}\#R_N = \infty. $$ Hence the Second Borel-Cantelli Lemma gives that for $\mathbb{P}$-almost all $\omega$, there are infinitely many $n$ for which $\rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) \geq \varepsilon$. $\quad\square$

Proofs of results in progress update

Proof of Basic Proposition. Let $\mathcal{A}$ be the collection of all sets $A \in \mathcal{X}$ with the property described in the question. By Dynkin's $\pi$-$\lambda$ theorem, it is sufficient to show that $\mathcal{A}$ is closed under complements and countable disjoint unions. Firstly, $\rho_n(A,B)=\rho_n(X \setminus A,B)$, and so $\mathcal{A}$ is closed under complements. Now take a sequence $(A_i)_{i \geq 1}$ in $\mathcal{A}$ that is mutually disjoint, and let $A=\bigcup_{i \geq 1} A_i$. For each $\varepsilon>0$, let $k(\varepsilon) \in \mathbb{N}$ be such that $\mu\!\left( \bigcup_{i > k(\varepsilon)} A_i \right)<\frac{\varepsilon}{2}$. For all $B \in \mathcal{X}$ and $n \in \mathbb{N}$, the triangle inequality gives $$ \rho_n(A,B) \leq \sum_{i=1}^\infty \rho_n(A_i,B) < \tfrac{\varepsilon}{2} + \sum_{i=1}^{k(\varepsilon)} \rho_n(A_i,B)\text{.} $$ Hence we may take $$ N_{A,\varepsilon} = \sum_{i=1}^{k(\varepsilon)} N_{A_i,\frac{\varepsilon}{2k(\varepsilon)}}\text{.} \quad\square $$

To prove Proposition A, we first need the following lemma (whose proof is adapted from a comment in the question https://math.stackexchange.com/questions/4005728/.)

Lemma. For any countable collection $\{A_n\}_{n \in \mathbb{N}} \subset \mathcal{X}$ there exists a compact metrisable topology on $X$ generating $\mathcal{X}$ such that for every $n$, $A_n$ is clopen.

Proof. For each $n$, let $\mathcal{P}_n$ be the coarsest partition of $X$ for which $A_1,\ldots,A_n$ can all be expressed as unions of members of $\mathcal{P}_n$. Let $(\tau_n)_{n \in \mathbb{N}}$ be an increasingly fine sequence of compact metrisable topologies on $X$ such that for each $n$, $\mathcal{X}=\sigma(\tau_n)$ and every member of $\mathcal{P}_n$ is $\tau_n$-clopen. The increasing fineness implies that for integers $n \geq m$, if $x_i \overset{\tau_n}{\to} x$ as $i \to \infty$ then $x_i \overset{\tau_m}{\to} x$ as $i \to \infty$. Equip $X^\mathbb{N}$ with the product topology, i.e. the compact metrisable topology whose convergence corresponds to saying that for each $n \in \mathbb{N}$, the $n$-th coordinate converges in $\tau_n$. It is then easy to see that the diagonal $\{(x,x,x,\ldots) : x \in X \}$ is closed, and so compact. Let $\tau$ be the topology on $X$ that is identified by the bijection $x \mapsto (x,x,\ldots)$ with the induced topology on $\{(x,x,\ldots) : x \in X \}$ from our topology on $X^\mathbb{N}$. Given $n \in \mathbb{N}$ and $\tilde{A} \in \mathcal{P}_n$, if $(x_i)_{i \in \mathbb{N}}$ is a sequence in $\tilde{A}$ such that $(x_i,x_i,\ldots)$ converges to a point $(x,x,\ldots) \in X^\mathbb{N}$ as $i \to \infty$, then in particular $x_i \overset{\tau_n}{\to} x$, and so $x \in \tilde{A}$. Hence for all $n$, every member of $\mathcal{P}_n$ is $\tau$-closed, and so in particular, $A_n$ is $\tau$-clopen. $\quad\square$

Proof of Proposition A if $T$ is invertible. Arguing by contrapositive, suppose we have a counterexemplary $A$, $(B_k)$ and $\varepsilon$, and write $$ R = \{ n \in \mathbb{N} : \#\{k \in \mathbb{N} : \rho_n(A,B_k) < \varepsilon \} < \infty \}\text{.} $$ Fix a compact metrisable topology on $X$ whose Borel $\sigma$-algebra is precisely $\mathcal{X}$, such that the sets $T^n(A)$, $n \in \mathbb{N}_0$, are all clopen. Since $X$ is compact, any bounded subset of $L^\infty(\mu)$ is relatively compact in the metrisable topology on $L^\infty(\mu)$ generated by the set of mappings $\{g \mapsto \int_X f.g \, d\mu : f \in C(X) \}$. Hence we can find $g \in L^\infty(\mu)$ and a subsequence $(B_{m_k})$ of $(B_k)$ such that $\int_{B_{m_k}} f \, d\mu \to \int_X f.g \, d\mu$ for all $f \in C(X)$. For each $n \in \mathbb{N}$, we have \begin{align*} \rho_n(A,B_{m_k}) = & \ |\mu(T^n(A) \cap B_{m_k})-\mu(A)\mu(B_{m_k})| \\ \overset{k \to \infty}{\to} & \ \left| \int_{T^n(A)} g \, d\mu - \mu(A) \int_X g \, d\mu \right| \\ =& \ \left| \int_A g \circ T^n \, d\mu - \mu(A) \int_X g \, d\mu \right| =: \rho_n(A,g)\text{,} \end{align*} and so for each $n$ in the unbounded set $R$, we have $\rho_n(A,g) \geq \varepsilon$, and so $\mu$ is not $T$-mixing. $\quad\square$

Proof of Proposition A when $T$ is non-invertible. We use the "natural extension" to reduce to the invertible case. It is well-known that, since $(X,\mathcal{X})$ is a standard Borel space and $\mu$ is a $T$-invariant probability measure, the set $X_{\text{ext}} := \{(x_i)_{i \in \mathbb{Z}} \in X^\mathbb{Z} : T(x_i)=x_{i+1} \forall \, i \in \mathbb{Z} \}$ is non-empty and on $X_{\text{ext}}$ (equipped with the induced $\sigma$-algebra from $\mathcal{X}^{\otimes \mathbb{Z}}$) there is a unique probability measure $\mu_{\text{ext}}$ all of whose coordinate-projections coincide with $\mu$. Write $T_{\text{ext}} \colon X_{\text{ext}} \to X_{\text{ext}}$ for the left-shift map $$ T_{\text{ext}}((x_i)_{i \in \mathbb{Z}}) = (x_{i+1})_{i \in \mathbb{Z}} = (T(x_i))_{i \in \mathbb{Z}}\text{.} $$ Obviously $T_{\text{ext}}$ is invertible, and one can show that if $\mu$ is $T$-mixing then $\mu_{\text{ext}}$ is $T_{\text{ext}}$-mixing, in which case Proposition A applies to $(\mu_{\text{ext}},T_{\text{ext}})$. We then recover Proposition A for $(\mu,T)$ by taking sets $\pi_0^{-1}(A)$ and $\pi_0^{-1}(B_k)$. $\quad\square$

Proof of Proposition B. Assume that $(X,\mathcal{X},\mu,T)$ has re-orderable correlations but fails the property described in my question. Take $\mathcal{Y}$ as in the definition of re-orderable correlations, and (on the basis of the Basic Proposition) take a counterexemplary $A \in \mathcal{Y}$ and $\varepsilon>0$ for the property described in my question. Let $(r_n)$ be the sequence corresponding to $A$ as in the definition of re-orderable correlations. So we can find $\delta>0$ and a sequence $(\tilde{B}_N)_{N \in \mathbb{N}}$ in $\mathcal{X}$ such that for all $n,N \in \mathbb{N}$ with $N \geq n$ we have $\rho_{r_n}(A,\tilde{B}_N) \geq \delta$. Hence in particular, since the set $\{r_n\}_{n \in \mathbb{N}}$ is obviously an infinite set, the conclusion of Proposition A fails by taking $\delta$ in place of $\varepsilon$. So $(X,\mathcal{X},\mu,T)$ is not mixing. $\quad\square$

Answer 1

Reposting some proofs I gave in the comments as a (partial) answer, demonstrating that the answer to Julian's question is "yes" for Bernoulli/i.i.d. systems, and so that it is not resolved by some usual "there's no way to guarantee a rate of mixing" examples. The question actually seems pretty subtle, although my honest guess is that the answer is "no" in general.

Say $\mu$ is i.i.d. on $S^{\mathbb{Z}}$ for some finite set $S$ (and that $T = \sigma$ is the left shift on $S^{\mathbb{Z}}$), that $A$ depends only on coordinates $-k$ through $k-1$, and $\epsilon > 0$. Assume for a contradiction that Julian's property fails. Then, for all $n$, there exist $B_n$ and $i_0,...,i_{n-1}$ s.t. $|\mu(A\cap \sigma^{i_j}B_n)-\mu(A)\mu(B_n)| \geq \epsilon$ for $0 \leq j < n$. (I've removed the negative power for simplicity, since $\sigma$ is invertible here.)

We can assume without loss of generality that $i_0 > k$ and $i_j - i_{j-1} > 2k$ for all $0 < j < n$ (since this decreases the size of $\{i_0, \ldots, i_{n-1}\}$ by at most a factor of $2k$).

Note that $\mu$ is i.i.d., so exchangeable. Take a self-map $\pi_n$ of $S^{\mathbb{Z}}$ sending coordinate $i_j + m$ to $2kj + m$ for $0 \leq j < n$ and $-k \leq m < k$, and consider $C_n=\pi_n(B_n)$; clearly $\mu(C_n) = \mu(B_n)$. Then, since $A$ depends only on coordinates in $[-k, k)$, $\mu(A \cap \sigma^{i_j} B_n) = \mu(A \cap \sigma^{2kj} C_n)$ for $0 \leq j < n$. So, $\mu(A\cap \sigma^{2kj} C_n)-\mu(A)\mu(C_n))| \geq \epsilon$ for $0 \leq j < n$.

Take a weakly convergent sequence $C_{m_n} \rightarrow C$. (Here, I mean that we identify a set $D$ with the measure $\mu_D$ defined by $\mu_D(E) = \mu(D \cap E)$, and take a weak(star) convergent subsequence.) Then, since $A$ is clopen (i.e. $\chi_A$ is continuous), $\mu(A\cap \sigma^{2kj}C)-\mu(A)\mu(C)| = \lim_{n \rightarrow \infty} \mu(A\cap \sigma^{2kj}C_{m_n})-\mu(A)\mu(C_{m_n})| \geq \epsilon$ for all $j$. And this contradicts mixing of $\mu$.

Finally, we note that for arbitrary $A$ and $\epsilon$ (i.e. $A$ doesn't depend on only finitely many coordinates), there exists $A'$ where $\mu(A \triangle A') < \epsilon/3$. Then, for any set $B$, $|\mu(A)\mu(B) - \mu(A')\mu(B)| < \epsilon/3$, and for any $j$, $\mu(A\cap \sigma^j B) - \mu(A' \cap \sigma^j B)| < \epsilon/3$. By the triangle inequality, $|\mu(A' \cap \sigma^j B) - \mu(A')\mu(B)| < \epsilon/3 \Longrightarrow |\mu(A \cap \sigma^j B) - \mu(A)\mu(B)| < \epsilon$. This means that, using terminology from the original question, $N_{A, \epsilon} \leq N_{A', \epsilon/3}$. The second quantity has been proved finite, so the first is as well, completing the proof.

Answer 2

Theorem: Suppose $(X,\mu,\mathcal{B})$ is a standard probability space. Let $T:X\to X$ be a mixing transformation. Given any measurable set $A$ and $\epsilon > 0$, there exists $N_{A,\epsilon}\in \mathbb{N}$ such that for all $B \in \mathcal{B}$, $$\#\{ n\in \mathbb{N} : |\mu(B\cap T^{-n}A)-\mu(A)\mu(B)|\geq \epsilon \} \leq N_{A,\epsilon} .$$ Proof: Suppose the theorem is not true. Then there exists a measurable set $A$ and $\epsilon >0$ such that no finite $N_{A,\epsilon}$ exists. Since $T$ is mixing, there exists $N_1\in \mathbb{N}$ such that for $m\geq N_1$, $|\mu(A\cap T^{-m}A)-\mu(A)^2| < \frac{\epsilon^2}{2}$.
Choose $N_2$ such that $\frac{1}{N_2} < \frac{\epsilon^2}{2}$.
Let $N = 2N_1N_2$. Since we are assuming the theorem does not hold, there exists a measurable set $B$ such that $$\#\{ n\in \mathbb{N} : |\mu(B\cap T^{-n}A)-\mu(A)\mu(B)|\geq \epsilon \} \geq N .$$ Thus, there exists a sequence $n_1<n_2<\ldots < n_{N_2}$ such that $|n_j-n_i| \geq N_1$ for $i\neq j$ and either for all $i\neq j$, $$\mu(B\cap T^{-n_i}A)-\mu(A)\mu(B) \geq \epsilon ,\mbox{(1)}$$ or for all $i\neq j$, $$\mu(A)\mu(B)-\mu(B\cap T^{-n_i}A) \geq \epsilon .\mbox{(2)}$$ Assume the first claim above holds. A proof for the second claim is similar.
Note for $i\neq j$, $|n_j-n_i|\geq N_1$, either $n_j>n_i$ or $n_i>n_j$. If $n_j > n_i$, then $n_j \geq n_i + N_1$ and we have \begin{align} |\mu(T^{-n_i}A\cap T^{-n_j}A) - \mu(A)^2| &= |\mu(T^{-n_i}(A\cap T^{-n_j+n_i}A)) - \mu(A)^2| \\ &= |\mu(A\cap T^{-n_j+n_i}A) - \mu(A)^2| < \frac{\epsilon^2}{2} . \end{align} If $n_i > n_j$, then $n_i\geq n_j+N_1$ and we still have \begin{align} |\mu(T^{-n_i}A\cap T^{-n_j}A) - \mu(A)^2| &= |\mu(T^{-n_j}(T^{-n_i+n_j}A\cap A)) - \mu(A)^2| \\ &= |\mu(T^{-n_i+n_j}A\cap A) - \mu(A)^2| < \frac{\epsilon^2}{2} . \end{align} Thus, \begin{align*} \int_X \big( \frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x)-\mu(A)\big)^2 d\mu &= \frac{1}{N_2^2}\sum_{i,j=1}^{N_2} \big( \mu(T^{-n_i}A\cap T^{-n_j}A) - \mu(A)^2 \big) \\ &= \frac{1}{N_2^2}\sum_{i=1}^{N_2} \big( \mu(T^{-n_i}A\cap T^{-n_i}A) - \mu(A)^2 \big) \\ &+ \frac{1}{N_2^2}\sum_{i\neq j} \big( \mu(T^{-n_i}A\cap T^{-n_j}A) - \mu(A)^2 \big) \\ &< \frac{1}{N_2} + \frac{\epsilon^2}{2} < \epsilon^2 . \end{align*} Hence, by the Cauchy-Schwarz Inequality, \begin{align*} \int_X |\frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x)&-\mu(A)| I_B(x) d\mu \\ &\leq \sqrt{\int_X |\frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x)-\mu(A)|^2 d\mu} \sqrt{\int_X I_B^2(x) d\mu} \\ &< \epsilon . \end{align*} Therefore, \begin{align*} \frac{1}{N_2}\sum_{i=1}^{N_2} \big( \mu(B\cap T^{-n_i}A) &-\mu(A)\mu(B)\big) \\ &= \int_X \Big( \frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x)-\mu(A)\Big) I_B(x) d\mu \\ &\leq \int_X |\frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x) - \mu(A)| I_B(x) d\mu \\ &< \epsilon . \end{align*} This contradicts (1). If (2) holds, we still get \begin{align*} \mu(A)\mu(B) &- \frac{1}{N_2}\sum_{i=1}^{N_2} \mu(B\cap T^{-n_i}A) \\ &= \int_X \Big( \mu(A) - \frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x)\Big) I_B(x) d\mu \\ &\leq \int_X |\frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x) - \mu(A)| I_B(x) d\mu \\ &< \epsilon ,\ \mbox{which is a contradiction.} \end{align*} $\Box$ This theorem is inspired by results in Blum-Hanson (1960): On the mean ergodic theorem for subsequences.

Answer 3

The answer to my question is yes!! [The assumption that $(X,\mathcal{X})$ is a standard Borel space is not needed.]

A beautiful proof of a modified version of the statement has been provided by user65023, who says that it is "inspired by results in Blum-Hanson (1960): On the mean ergodic theorem for subsequences." I now give the proof of the answer to my question, in my own words.

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First some generalities:

Fix a set $S$, a sequence $(f_n)_{n \in \mathbb{N}}$ of functions $f_n \colon S \to \mathbb{R}$, and a function $f \colon S \to \mathbb{R}$.

Definition. We say that $f_n$ converges uniformly modulo re-ordering to $f$ if for all $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $x \in S$, $$ \#\{n \in \mathbb{N} : |f_n(x)-f(x)|\geq\varepsilon \} \leq N. $$

Now for any finite $J \subset \mathbb{N}$ with $\#J \geq 2$, let $\Delta J$ denote the smallest distance between distinct elements of $J$.

Proposition. Suppose that, over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence: $$ \sup_{x \in S} \left| f(x) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(x) \right| \to 0 \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$ Then $f_n$ converges uniformly modulo re-ordering to $f$.

Proof. Arguing by contrapositive, suppose we have $\varepsilon>0$ and a sequence $(x_k)_{k \in \mathbb{N}}$ in $S$ such that $\#\{n : |f_n(x_k)-f(x_k)| \geq \varepsilon\} \to \infty$ as $k \to \infty$. Without loss of generality, suppose there is a subsequence $(x_{m_k})$ of $(x_k)$ such that, writing $I_k=\{n : f(x_{m_k}) \geq f_n(x_{m_k}) + \varepsilon\}$, we have $\#I_k \to \infty$ as $k \to \infty$. Take a finite $J_k \subset I_k$ for each $k$, such that $\#J_k$ and $\Delta J_k$ both tend to $\infty$ as $k \to \infty$. We have that $ f(x_{m_k}) - \frac{1}{\#J_k}\!\sum_{n \in J_k} f_n(x_{m_k}) \geq \varepsilon$ for all $k$. $\ \square$

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Now a re-statement and proof of the result:

Let $(X,\mathcal{X},\mu,T)$ be a mixing probability-preserving transformation. Fixing $A \in \mathcal{X}$, define $f_n \colon \mathcal{X} \to \mathbb{R}$ and $f \colon \mathcal{X} \to \mathbb{R}$ by \begin{align*} f_n(B) &= \mu(A \cap T^{-n}(B)) \\ f(B) &= \mu(A)\mu(B). \end{align*}

Theorem. $f_n$ converges uniformly modulo re-ordering to $f$.

We now prove this, using the above Proposition. We will equip the space of $[0,1]$-valued functions on $X$ identified up to $\mu$-almost sure equality with its natural topology, namely the topology of $L^p$ convergence for any $p \in [1,\infty)$ or equivalently the topology of convergence in probability.

Since $\mu$ is $T$-invariant, we have that for all $B \in \mathcal{X}$ and $n,N$ with $0 \leq n \leq N$, \begin{align*} f_n(B) &= \int_{T^{-N}(B)} \mathbf{1}_{T^{n-N}(A)} \, d\mu \\ f(B) &= \int_{T^{-N}(B)} \mu(A) \, d\mu. \end{align*} So for any non-empty finite $J \subset \mathbb{N}$ and any $B \in \mathcal{X}$, applying the above with $N=\max J$ gives $$ f(B) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(B) \ = \ \int_{T^{-\max J}(B)} \mu(A) - \tfrac{1}{\#J}\!\sum_{n \in J} \mathbf{1}_{T^{n-\max J}(A)} \, d\mu. $$ Therefore, to be able to apply the Proposition, it will be sufficient to show the following.

Lemma. Over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence: $$ \tfrac{1}{\#J}\!\sum_{n \in J} \mathbf{1}_{T^{n-\max J}(A)} \to \mu(A) \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$

Proof. We have \begin{align*} \| \mathrm{RHS}-\mathrm{LHS} \|_{L^2(\mu)}^2 &= \left\| \tfrac{1}{\#J}\!\sum_{n \in J} (\mu(A) - \mathbf{1}_{T^{n-\max J}(A)}) \right\|_{L^2(\mu)}^2 \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_X (\mu(A) - \mathbf{1}_{T^{n_1-\max J}(A)})(\mu(A) - \mathbf{1}_{T^{n_2-\max J}(A)}) \, d\mu \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_X \mu(A)^2 - \mu(A)\mathbf{1}_{T^{n_2-\max J}(A)} - \mu(A)\mathbf{1}_{T^{n_1-\max J}(A)} + \mathbf{1}_{T^{n_1-\max J}(A)}\mathbf{1}_{T^{n_2-\max J}(A)} \, d\mu \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \left( \int_X \mathbf{1}_{T^{n_1-\max J}(A)}\mathbf{1}_{T^{n_2-\max J}(A)} \, d\mu - \mu(A)^2 \right) \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n \in J} \big( \mu(A) - \mu(A)^2 \big) \ + \ \left( \tfrac{1}{\#J} \right)^{\!2} \!\!\! \sum_{\substack{\text{distinct} \\ n_1,n_2 \in J}} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\ &\leq \underbrace{\tfrac{1}{\#J}}_{\to 0} \ + \ \underbrace{\sup_{n \geq \Delta J} \big| \mu(A \cap T^{-n}(A)) - \mu(A)^2 \big|}_{\to 0}. \quad \square \end{align*}

[NB: The fact that $T$ is mixing only entered at the very final step.]