Is an infinite product probability space compatible with an almost surely statement? My question pertains to the following Theorem which can be found here. 
Theorem (Existence of product measures): 
Let $A$ be an arbitrary set and for each $\alpha \in A$ let $(X_\alpha,B_\alpha,\mu_\alpha)$ be a probability space in which each $X_\alpha$ is a locally compact, $\sigma$-compact metric space.  Then there exists a unique probability measure $\mu_A = \prod_{\alpha \in A}\mu_\alpha$ on $(X_A,B_A)$ with the property that:
$$
\mu_A\left(\prod_{\alpha \in A}E_\alpha  \right) = \prod_{\alpha \in A} \mu_\alpha(E_\alpha)
$$ 
whenever $E_\alpha \in B_\alpha$ for each $\alpha \in A$ and one has $E_\alpha = X_\alpha$ for all but finitely many $\alpha$.
Consider the following example of a product measure (also to be found here):
Example
Let $A = \mathbb{N}$ and for each $\alpha \in A$ let $(X_\alpha,B_\alpha,\mu_\alpha)$ be the two element set $X_\alpha = \{0,1\}$. together with the discrete metric; hence the discrete $\sigma$-algebra.  By the Theorem above there exists a unique probability measure such that $(X_A,B_A,\mu_A)$, together with the above properties, is a probability space. In particular we can think of this as a space of infinite sequences of flips of a fair coin.
Let $Y_n$ be the fraction of heads after tossing a fair coin $n$ times.  Then almost surely,
$$
\lim_{n \rightarrow \infty} Y_n  = \frac{1}{2}
$$ 
By the definition of almost sure convergence we can rephrase this as follows:
$$
\mu_A\left( \lim_{n \rightarrow \infty}Y_n  = \frac{1}{2} \right) = 1
$$
where $\mu_A$ is the product measure discussed above.  Let us rewrite this once again by defining the set $S = \{x \in X_A : \lim_{n \rightarrow \infty} Y_n(x) = \frac{1}{2} \}$ thus we have;
$$
\mu_A(S) = 1
$$ 
My question is: Why does $\mu_A(S)$ make sense? To be slightly less vague, according to the Theorem above $\mu_A$ is defined such that:
$$
\mu_A\left(\prod_{\alpha \in A}E_\alpha  \right) = \prod_{\alpha \in A} \mu_\alpha(E_\alpha)
$$ 
whenever $E_\alpha \in B_\alpha$ for each $\alpha \in A$ and one has $E_\alpha = X_\alpha$ for all but finitely many $\alpha$. It is this finitely with which I have a problem since any $s \in S$ has the infinite property that for all $\epsilon>0 $ there exists an $N$ such that for all $n >N$
$$
\lvert X_n  - \frac{1}{2} \rvert < \epsilon
$$ 
Thus $S$ appears to be made up of a union of sequences which each occur with probability 0.  Any help pointing out my misconception would be greatly appreciated.
 A: There are a few subtleties here. 
The first is the formulation of the existence statement. The statement is that there is a unique measure $\mu$ on a suitable $\sigma$-algebra $B_A$ (see below) on the product space $X_A$ such that
$$
\mu (\prod E_\alpha) = \prod \mu_\alpha (E_\alpha)
$$
under the conditions you stated ($E_\alpha =X_\alpha$ for all but finitely many $\alpha$, ...). 
This is like saying that for a suitable map $f : \Bbb{Q} \to \Bbb{R}$, there is a unique continuous extension $g : \Bbb{R} \to \Bbb{R}$. 
Here, the extension $g$ is not only defined on $\Bbb{Q}$, but on all of $\Bbb{R}$. Likewise, the measure $\mu$ is not only defined on all sets of the form $\prod E_\alpha$ as above, but on a $\sigma$-algebra $B_A$, which contains all these sets. 
Now, the exact choice of this $\sigma$-algebra plays some role, see below. To complete the analogy, there is no unique continuius extension of $f$ above to $h :\Bbb{C}\to \Bbb{R}$, simply because $\Bbb{Q}$ is not dense in $\Bbb{C}$. 
The simplest choice for the $\sigma$-algebra $B_A$ would be the product-sigma-algebra. This is simply the $\sigma$-algebra generated by all products $\prod E_\alpha$ as above. For this construction, you do not need all the assumptions on the spaces $X_\alpha$ (like local compactness, sigma-compactness, ...) if I remember correctly (this could be wrong, but is not the main point of your question). 
The other possibility would be the Borel-$\sigma$-algebra on the product space $X_A$. Here, you use that each of the $X_\alpha$ is a (locally compact) topological group. But then one has to impose some additional condition on $\mu$ to get uniqueness; ordinarily, you will assume that $\mu$ is regular. I only know about this approach in the car that the product $X_A$ is locally compact. This is the case if each $X_\alpha$ is compact for all but finitely many $\alpha$ or if $A$ is finite. 
Finally, each of the random variables $Y_n$ (here, you seem to assume that $A =\Bbb{N}$) is (probably?) simply given by the projection $\pi_n :\prod_m X_m \to X_n, (x_m)_m \mapsto x_n$. 
You can check that this is measurable for the product sigma algebra (and hence also for the (larger) Borel sigma algebra). 
Hence, also $\liminf Y_n$ and $\limsup Y_n$ are measurable, so that the same is true of
$$
S = \{\limsup Y_n =\liminf Y_n =1/2\}. 
$$
Hence, $\mu(S)$ makes sense. 
