Let $(X\_1,X\_2,\ldots)$ be a stationary, mixing sequence of real random variables. Then it holds (for example) for any event $A$ that is measurable in $\sigma(X\_1,X\_2,\ldots)$ and any $S \subseteq \mathbb{R}$ that
$$
\lim\_{i \to \infty} \big|\mathbb{P}[A,X\_i\in S] - \mathbb{P}[A] \cdot \mathbb{P}[X\_i \in S]\big|=0.
$$


I am looking for a finitary version of this statement. That is, something of this sort: for every $\varepsilon>0$ there is an $n$ large enough such that, if $A$ is measurable in $\sigma(X\_1,\ldots,X\_n)$, and if $i$ is chosen uniformly in $\{1,\ldots,n\}$, then with probability $1-\varepsilon$
$$
\big|\mathbb{P}[A,X\_i\in S] - \mathbb{P}[A] \cdot \mathbb{P}[X\_i \in S]\big| < \varepsilon.
$$



On a finitary version of mixing
Let $J\_n$ be a random variable independent of $X:=(X\_1,X\_2,\ldots)$ and uniformly distributed in the set $\{1,\ldots,n\}$. For each $i\in\{1,\ldots,n\}$, let 
\begin{equation}
 p(i):=P(A,X\_i\in S). 
\end{equation}
We need to show that 
\begin{equation}
 p(J\_n)\to P(A)P(X\_1\in S)
\end{equation}
in probability uniformly in all $A\in\sigma(X\_1,\ldots,X\_n)$; the convergence here everywhere is as $n\to\infty$. 


Let us show a bit more -- that this convergence is uniform over all $A$ in the underlying sigma-algebra (say $\Sigma$). Suppose then that the weak mixing condition holds (which of course will hold if the strong mixing holds). Then the sequence $X$ is ergodic; see e.g. the Remark on page 13 in [Ergodic Theory](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0ahUKEwjfnfyv59rXAhUJxGMKHXCAC_EQFggwMAI&url=http%3A%2F%2Fwww.maths.manchester.ac.uk%2F~cwalkden%2Fmagic%2Flecture04.pdf&usg=AOvVaw2EtKUTpo6BcFyJVDXEyVHU). 
Now the ergodic theorem implies that  

\begin{equation}
 d\_n(S):=\frac1n\,\sum\_{i=1}^n I\{X\_i\in S\}-P(X\_1\in S)\to0
\end{equation}
almost surely and hence in probability, where $I\{\cdot\}$ denotes the indicator function. So, for every $\varepsilon>0$ there is a natural $n\_\varepsilon$ such that for all natural $n>n\_\varepsilon$ we have $P(|d\_n(S)|>\varepsilon)<\varepsilon$; therefore and because $|d\_n(S)|\le1$, 
\begin{equation}
 E|d\_n(S)I\{A\}|\le E|d\_n(S)| \le\varepsilon P(|d\_n(S)|\le\varepsilon)+P(|d\_n(S)|>\varepsilon)\le2\varepsilon. 
\end{equation}
So, $d\_n(S)I\{A\}\to0$ in $L^1$ uniformly in all $A$. So, 
\begin{equation}
 Ep(J\_n)=
 \frac1n\,\sum\_{i=1}^n P(A,X\_i\in S)
 =Ed\_n(S)I\{A\}+P(A)P(X\_1\in S)\to P(A)P(X\_1\in S)
\end{equation}
uniformly in all $A$. 
Similarly, letting $(\tilde A,\tilde X)$ denote an independent copy of $(A,X)$, we have 
\begin{multline\*}
 Ep(J\_n)^2=
 \frac1n\,\sum\_{i=1}^n P(A,X\_i\in S)^2
 =\frac1n\,\sum\_{i=1}^n P(A,\tilde A,X\_i\in S,\tilde X\_i\in S) \\ 
 \to P(A,\tilde A)P(X\_1\in S,\tilde X\_1\in S)=P(A)^2P(X\_1\in S)^2
\end{multline\*}
uniformly in all $A$;  

here we use the fact that the weak mixing property is preserved under the direct product (see e.g. page 5 in <http://arxiv.org/abs/math/0603575v1> ) and hence the sequence of pairs $((X\_1,\tilde X\_1),(X\_2,\tilde X\_2),\dots)$ is ergodic. 
Thus, $Ep(J\_n)\to P(A)P(X\_1\in S)$ and $Var\,p(J\_n)\to0$ uniformly in all $A$. So, by Chebyshev's inequality, indeed $p(J\_n)\to P(A)P(X\_1\in S)$ in probability uniformly in all $A$. 


