How is almost sure convergence defined without reference to the underlying probability measure space? Suppose I have some "leisurely defined" (real-valued) random variables $X_n, X$ - by this I mean no reference is made to the underlying probability space $(\Omega, \mathcal{F}, P)$ acting as the domain space of our random variables. Now suppose I'd like to prove that $X_n$ converges to $X$ almost surely. The definition, as I understand it, is
$$
\text{Pr}(\{\omega\in\Omega\ :\ \lim_n X_n(\omega) = X(\omega)\}) = 1
$$
which we write more succinctly as $\text{Pr}(\lim_n X_n = X) = 1$.
I don't really see how to do this without reference to the formal definition of our random variables. The expression $\lim_n X_n = X$ doesn't really mean anything (at least to me) when we just view our random variables as "things which can take on any number of values with certain probabilities."
For instance if $X^i$ were iid standard normal variables, and $X_n := \frac1n\sum_{i=1}^n X^i$, what would it mean to say $X_n\to 0$ almost surely? It's not as if the values $X_n$ takes on somehow "converge" to zero. We can't appeal to $X_n$'s distribution ($N(0, 1/n)$) and say this is what converges to zero, as this would be convergence in distribution. Nor can we appeal to $X_n$'s tail probabilities, $\text{Pr}(|X_n| > \epsilon)\to0$, as this would be convergence in probability. How could we check $X_n\overset{a}{\to} 0$, without writing down $(\Omega, \mathcal{F}, P)$?
It occurred to me that we can check whether
$$
\text{Pr}\left(\bigcap_n\bigcup_m\bigcap_{k\ge m}\left\{|X_k - X| < \frac1n\right\}\right) = 1
$$
which makes no reference to $(\Omega, \mathcal{F}, P)$, and perhaps we could use the Borel-Cantelli lemmas to simplify this further (though we might need to know that the events are independent), but this is still rather cumbersome in my opinion compared to just knowing the probability space beforehand.
 A: This answer summarizes my comments: There is always some probability space $(\Omega, \mathcal{F}, P)$, even if the problem does not explicitly say so.
Here is a useful sufficient condition for convergence almost surely that works for a sequence of arbitrary random variables (not necessarily independent).  The lemma works more generally for convergence to a random variable $Y$, but I consider convergence to a constant $c$ for simplicity.
Lemma: Let $\{Y_n\}_{n=1}^{\infty}$ be a sequence of random variables (all on the same probability space). Suppose there is a real number $c$ such that
$$\sum_{n=1}^{\infty} P[|Y_n-c|>\epsilon] < \infty \quad \forall \epsilon>0$$
Then $Y_n\rightarrow c$ almost surely.

The above lemma is so useful that it is the standard way of proving the law of large numbers. Indeed, suppose $\{X_n\}_{n=1}^{\infty}$ are i.i.d. with finite mean and variance.  Define $m=E[X_n]$ and $\sigma^2=Var(X_n)$ for all $n \in \{1, 2, 3, ...\}$. Define
$$M_n=\frac{1}{n}\sum_{i=1}^n X_i \quad \forall n \in \{1, 2, 3, ...\}$$

*

*You can use Chebyshev to show that $P[|M_n-c|>\epsilon]\leq \frac{\sigma^2}{\epsilon^2n}$ for all $n \in \{1, 2, 3,...\}$ and so the above lemma directly implies $M_{n^2}\rightarrow c$ almost surely. Some more details are needed to show that $M_n\rightarrow c$ almost surely (you would need to show $M_n$ does not significantly change between perfect squares $n=i^2$ and $n=(i+1)^2$).


*Assuming a finite fourth moment, so that $E[(Y_n-c)^4]=\mu<\infty$ for all $n$,  you can show $P[|M_n-c|>\epsilon]\leq \frac{1}{\epsilon^4}\cdot O(1/n^2)$ and so the above lemma directly implies $M_n\rightarrow c$ almost surely.
Both of the bullets above use the fact
$$ \sum_{n=1}^{\infty} \frac{1}{n^2}<\infty$$
