The strong law of large numbers states that the sample average converges almost surely to the expected value $\overline{X}_n\ \xrightarrow{\text{a.s.}}\ \mu \qquad\mathrm{when}\ n \to \infty$ .
That is,$$\Pr\left( \lim_{n\to\infty}\overline{X}_n = \mu \right) = 1.$$
I want to ask whats the domain of the random variable $\overline{X}_n$, given that all $X_n$ have same domain $\Omega$?
For the coin tossing problem $\Omega =\{H,T\} $ and $X_n(H)=1 \quad and \quad X_n(T)=0 \quad\forall n $ so, if the domain of $\overline{X}_n$ was $\Omega$ the $\overline{X}_n(H)=1 \quad \overline{X}_n(T)=0 \quad \forall n$, so $\Pr\left({\omega \in \Omega : \overline{X}_n(\omega)=1/2}\right)=0$ , which is not what strong law says.
I want to know whats the domain of this random variable $\overline{X}_n$?
Even if we could define the $\overline{X}_n: \Omega^n\to \{0,1\}$, we would end up with many random variables defined in different probability spaces. In that situation we would not be able to make any sense of the notion "probability that the sequence of $\overline{X}_n$ converges", because we wouldn't have a fixed probability measure to consider all of the outcomes of the $\{\overline X_k\}_{k=1}^\infty$ at once to assign a probability.
What we need is a probability measure defined on sequences of outcomes. More precisely, the sample space is the set of sequences with values in $\Omega = \{H,T\}$, that is, $\Omega^{\mathbb N}$.
In this setting, if $\omega = (\omega_1,\omega_2,\dots)\in \Omega^{\mathbb N}$, then
$$X_k(\omega) = \cases{1 & $\text{if } {\omega_k} = H$\\ 0 & $\text{if } {\omega_k} = T$}$$ and $$\overline{X}_n(\omega) = \frac{X_1(\omega)+\dots+X_n(\omega)}{n}.$$