Let $S_n = \sum_{i =1}^n X_i $. Then the event $[\lim_{n \to \infty} \frac{S_n}{n} = 0]$ is an event found in the tail sigma algebra generated by the $\{X_n\}$.
The proof is straightforward I believe : the trick is to realize that $\sum_{n=1}^\infty X_n < \infty$ iff $\sum_{n=m}^\infty X_n < \infty$. Therefore, the event
$$ [\lim_{n \to \infty}S_n/n = 0] \text{ is equivalent to } [\lim_{n \to \infty}(\sum_{i =m}^n X_i)/n = 0] \text{ for any fixed }m$$
and the result holds.
Question: Is it possible to extend this result to another limit (other than zero) ? For example, if the $\{X_n\}$ are i.i.d then by SLLN we know that the sample average converges to $\mu$. In this case, would the event $[\lim_{n \to \infty} S_n/n = \mu]$ also be measurable wrt to the tail sigma field?
As long as the $X_n$ are surely finite, we have $\lim_{n\to\infty}S_m/n = 0$, for any fixed $m$.
Therefore, under the same assumption, $$\lim_{n\to\infty}S_n/n = \lim_{n\to\infty}(S_n-S_m)/n$$
Since for $n > m$, $S_n-S_m \in \sigma(X_k : k > m)$, we can see $$\{\lim_{n\to\infty}S_n/n = \mu\} = \{\lim_{n\to\infty}(S_n-S_m)/n = \mu\} \in \sigma(X_k : k > m)$$ for all $m$, so $\{\lim_{n\to\infty}S_n/n = \mu\}$ is in the tail $\sigma$-field.