Almost sure convergence in strong law of large numbers. Strong Law of Large Numbers is often stated as
$$\overline{X}_n\ \xrightarrow{a.s.}\ \mu \qquad\textrm{when}\ n \to \infty$$
or
$$\Pr\!\left( \lim_{n\to\infty}\overline{X}_n = \mu \right) = 1$$
for $\overline{X}_n$ - average of $n$ i.i.d. random variables with mean $\mu$.
It seems to me from the definition, that order to have a notion of "almost sure convergence" we must have a sequence of random variables $X_i$ on the same probability space; at the same time $\overline{X}_n$ is a random variable on the product of probability spaces of the first $n$ of the $X_i$'s. Of course we can think of all $\overline{X}_k$'s as living on the $n$'th product for $k \leq n$, but to consider all $n$ at the same time, we would have to have an infinite product.  This would make sense, but seems somewhat complicated (infinite product spaces!). So, am I missing something, or is this what's going on and all elementary introductions (and wikipedia) just supress this point?
Edit: Ok, so the original question was a bit misleading - and the clarification is a bit long, see comments below, but I decided to post it also as an answer (it overlaps a bit with the accepted one). 

Why would we need product spaces. $X_i$'s being random variables are measurable functions. Then consider the $\sigma$-field generated by $X_1,\cdots,X_n$. The $\bar{X}_n$ is measurable with respect to this $\sigma$-field. Hence $\bar{X}_n$ is measurable with respect to $\sigma$-field  generated by $X_1,X_2,\cdots$. I think that is all you need.
