On a finitary version of mixing 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 $1-\varepsilon$
$$
\big|\mathbb{P}[A,X_i\in S] - \mathbb{P}[A] \cdot \mathbb{P}[X_i \in S]\big| < \varepsilon.
$$
 A: 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. 
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 $A$. 
