The strong law of large numbers tells us that for $X_n$ independently identically distributed with expectation $\mu$ we have almost surely,
$$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}X_k=\mu$$
If the $X_n$ are Bernoulli distributions with probability $p$ then there is a natural probability space on $2^\Bbb{N}$ representing the outcomes of these distributions (It is assumed that the associated probability measure is complete). In this context we have via the strong law of large numbers,
$$\Pr(\mathcal{K})=1 \text{ where }\mathcal{K}=\{A\in2^\Bbb{N}|\lim_{n\rightarrow\infty}\frac{|A\cap[n]|}{n}=p\}$$
Now consider the class $\mathcal{S^*}=\{S\in2^\Bbb{N}|\lim_{n\rightarrow\infty}\frac{|S\cap[n]|}{n}=1\}$.
It seems reasonable to conclude (in analogy to the strong law of large numbers) that for any $S\in\mathcal{S^*}$ we have,
$$\Pr(\mathcal{K}_S)=1 \text{ where }\mathcal{K}_S=\{A\in2^\Bbb{N}|\lim_{n\rightarrow\infty}\frac{|A\cap S\cap[n]|}{|S\cap[n]|}=p\}$$
It is clear from this via standard measure theory that for any countable collection $\mathcal{U}\subseteq\mathcal{S^*}$ we have,
$$\Pr(\bigcap_{S\in\mathcal{U}}\mathcal{K}_S)=1$$
The question I have is, does $\Pr(\bigcap\limits_{S\in\mathcal{S^*}}\mathcal{K}_S)=1$?
This would be in a certain sense a strengthening of the strong law of large numbers, however it seems like it might be too strong and so no longer hold. As far as I can tell in the very special cases of $p=0,1$ it does hold, but other than that I am not sure how to proceed. Any help or references would be appreciated.