To begin my question, I first need to introduce a kind of convergence of functions that lies strictly between pointwise convergence and uniform convergence.

Uniform convergence modulo re-ordering

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 $\mathbb{N} \setminus \mathcal{N} \subset \pi(\mathbb{N})$.

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

Proposition. The following two statements are equivalent:

1. For all $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that every $x \in S$ has $$ \#\{n \in \mathbb{N} : |f_n(x)-f(x)| \geq \varepsilon \} \leq N\text{.} $$
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 $n \to \infty$.

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 $\varepsilon>0$, $$ \sum_{n \in \mathbb{N}} \mathbb{P}(x \in S : |f_n(x)-f(x)| \geq \varepsilon ) < \infty\text{.} $$

Proof. 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 $x \in X$, 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 clear that $\lim_{n \to \infty} |f_{\pi_x(n)}(x)-f(x)|=0$. 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 statement of 1. 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 $\pi^{-1}(n) \leq N$.

So we have shown the equivalence of 1 and 2. Assuming the extra measurability assumptions, we now show the equivalence of 1 and 3. 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 statement of 1. 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, 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. $\quad\square$

Definition. We say that $f_n \to f$ uniformly modulo re-ordering as $n \to \infty$ if the equivalent statements 1 and 2 in the above Proposition hold.

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 the $N$ in the first characterisation in the Proposition with $N+\tilde{N}$.)

MY QUESTION.

Let $(X,\mathcal{X},\mu,T)$ be a measure-preserving dynamical system. 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 $\rho_n \to 0$ pointwise as $n \to \infty$.

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 can consider $\rho_n(A,\,\boldsymbol{\cdot}\,)$ as a function from $\mathcal{X}$ to $[0,1]$.

Assume that $(X,\mathcal{X})$ is a standard Borel space. If $(X,\mathcal{X},\mu,T)$ is mixing, does it necessarily follow that for every $A \in \mathcal{X}$, $\rho_n(A,\,\boldsymbol{\cdot}\,) \to 0$ uniformly modulo re-ordering as $n \to \infty$?

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 $n$.

Motivation: Sample-pathwise mixing in random dynamical systems

For a skew-product map $\Theta \colon \Omega \times X \to \Omega \times X$ on a standard Borel space $(X,\mathcal{X})$ over an invertible mixing dynamical system $(\Omega,\mathcal{F},\mathbb{P},\theta)$, given a $\mathbb{P}$-projecting $\Theta$-invariant measure $\mu$ with disintegration $(\mu_\omega)_{\omega \in \Omega}$, one sometimes defines 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 $$ \left| \int_X g_1\!(\omega,x) \, (g_2 \circ \Theta^n)(\omega,x) \, \mu_\omega(dx) - \int_X g_1(\omega,x) \, \mu_\omega(dx) \int_X g_2(\theta^n\omega,y) \, \mu_{\theta^n\omega}(dy) \right|\text{,} $$ which, for each $g_1,g_2$, is well-defined up to $\mathbb{P}$-a.s. equality.

(See e.g. the question Two mixing rates of random dynamical system)

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 $\rho_{g_1,g_2}(n,\,\boldsymbol{\cdot}\,) \overset{\mathbb{P}\text{-a.s.}}{\to} 0$ 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:

Theorem. 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 mixing and is a counterexample for my question. Then one can find measurable sets $A_1,A_2 \in \mathcal{F} \otimes \mathcal{X}$ such that for $\mathbb{P}$-almost every $\omega \in \Omega$, $\rho_{\mathbf{1}_{A_1},\mathbf{1}_{A_2}}(n,\omega) \not\to 0$ as $n \to \infty$.

Proof. Take a counterexemplary $A \in \mathcal{X}$ and let $A_1=\Omega \times A$. With the counterexemplary $A$, take a counterexemplary $\varepsilon$ to statement 1 in the Proposition. For each $N \in \mathbb{N}$, let $B_N \in \mathcal{X}$ be such that the set $$ R_N := \{n \in \mathbb{N} : |\mu_X(A \cap T^{-n}(B_N))-\mu_X(A)\mu_X(B_N)| \geq \varepsilon\} $$ is of cardinality at least $3^N$. Let $$ B = \bigcup_{N=1}^\infty ([2^{-N},2^{1-N}] \times B_N) \, \subset [0,1] \times X \text{,} $$ and let $A_2$ be the pre-image of $B$ under the projection $((\omega_i)_{i \in \mathbb{Z}},x) \mapsto (\omega_0,x)$ from $\Omega \times X$ to $[0,1] \times X$. So $$ \rho_{\mathbf{1}_{A_1},\mathbf{1}_{A_2}}(n,(\omega_i)_{i \in \mathbb{Z}}) = \sum_{N=1}^\infty \mathbf{1}_{[2^{-N},2^{1-N}]}(\omega_n)|\mu_X(A \cap T^{-n}(B_N))-\mu_X(A)\mu_X(B_N)|. $$ Note in particular that the random variables $\rho_{\mathbf{1}_{A_1},\mathbf{1}_{A_2}}(n,\,\boldsymbol{\cdot}\,)$ are mutually $\mathbb{P}$-independent. Furthermore, writing $Q_n:=\{N \in \mathbb{N} : n \in R_N \}$, we have $$ \mathbb{P}(\omega : \rho_{\mathbf{1}_{A_1},\mathbf{1}_{A_2}}(n,\omega) \geq \varepsilon ) = \sum_{N \in Q_n} 2^{-N} $$ and so $$ \sum_{n \in \mathbb{N}} \mathbb{P}(\omega : \rho_{\mathbf{1}_{A_1},\mathbf{1}_{A_2}}(n,\omega) \geq \varepsilon ) = \sum_{N,n \text{ with } N \in Q_n} 2^{-N} = \sum_{N \in \mathbb{N}} 2^{-n}\#R_N = \infty. $$ Hence the Second Borel-Cantelli Lemma gives that for $\mathbb{P}$-almost all $\omega$, there are infinitely many $n$ for which $\rho_{\mathbf{1}_{A_1},\mathbf{1}_{A_2}}(n,\omega) \geq \varepsilon$. $\quad\square$

(The above proof via a Borel-Cantelli approach is inspired by 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.