Assume that $X_n^N$, $n=1,\ldots,N$ for all $N>0$ and $Y_n$, for all $n>0$, are 0-1 random variables living in the same probability space. All the $Y_n$'s are independent and Bernoulli with parameter $q$. Furthermore, for all $n$, $Y_n$ and $X_n$ are independent but $X_n$ may depend on $Y_i$ if $i<n$.
Fix $\omega$ such that $\lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N X_n^N(\omega) = p > 0$ and $\lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N Y_n(\omega) = q$.
My question is whether or not the following limit holds true: $$ \lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N X_n^N(\omega) Y_n(\omega) = p q $$
My first difficulty is that I don't see how I can use the independence of $X_n$ and $Y_n$ to reason $\omega$-per-$\omega$.
I assume you are still trying to prove your older conjecture raised in: Almost sure convergence of the average of the product of 0-1 random variables (That was an interesting conjecture and I spent some time thinking about it, but to no avail.) While I believe your older conjecture is true, this current conjecture is false. (I.e. this approach does not work to prove the older conjecture.) Sorry...
Counter example for $p = q = 1/2$:
Define a sample point as the sequence $\{Y_n\}$.
Define $X_1 \equiv 1$ and $X_n \equiv Y_{n-1}$. Clearly $X_n$ and $Y_n$ are independent.
Consider the sample point of alternating $0$s and $1$s, i.e. $\omega = 01010101010...$ Then the successive $X$s are $101010101...$ We have:
$\lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N X_n^N(\omega) = {1\over 2}$
$\lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N Y_n(\omega) = {1\over 2}$
$\forall n: X_n Y_n = 0$ and so $\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N X_n^N(\omega) Y_n(\omega) = 0$
Re: your older conjecture, obviously the hope is to show that such counterexamples have zero measure. Indeed for the specific $X_n \equiv Y_{n-1}$ defined above, your older conjecture should be easily proved in the sense of "almost surely", but just not for EVERY sample point.