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 $N+\tilde{N}$.)
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**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 $n$.
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**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 https://mathoverflow.net/questions/327938/)

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 https://mathoverflow.net/questions/418835/)

**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.