Having received two downvotes, I realised that I needed to considerably improve the layout and clarity of this post. I hope that it is now better-structured and easier to follow!
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**THE QUESTION IN SUMMARY**

>> 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 $B \in \mathcal{X}$,
$$ \#\{n \in \mathbb{N} : |\mu(A \cap T^{-n}(B))-\mu(A)\mu(B)| \geq \varepsilon \} \leq N_{A,\varepsilon} \text{?} $$

I will now discuss this question in much 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 mixing 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 https://mathoverflow.net/questions/327938/ and the paper [Quenched decay of correlations for slowly mixing systems](https://doi.org/10.1090/tran/7811).)

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 mixing 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 $n \to \infty$.*

The proof of the Theorem is based on the Borel-Cantelli Lemmas (used in a similar manner to 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.
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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 mixing 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 $\rho_{\mathbf{1}_A,\mathbf{1}_B}(n,\omega) \geq \varepsilon$. $\quad\square$