Does this simple random variable converge almost surely? I was trying to find an example of a simple random variable that converges in probability but not almost surely. I came across this example cited in multiple places:
$$X_n = \begin{cases} 1 \ \text{with probability} \ \frac{1}{n} \\ 0 \ \text{with probability} \ \frac{n-1}{n} \end{cases}, \ X = 0$$
It is clear to me that $X_n \overset{p}{\to} X$ but I am having trouble understanding why $X_n$ does not converge to $X$ almost surely. Here is my thought process. I should be able to add some structure to the sample space without changing the problem at all. Suppose the sample space is $[0,1]$ and the probability measure is $P(A) = \int_0^1 \mathbb{1}\{\omega \in A \} d\omega$. Let the underlying structure of the random variables $X_n$ and $X$ be
$$X_n(\omega) = \begin{cases} 1 \ \text{if} \ \omega \leq \frac{1}{n} \\ 0 \ \text{if} \ \omega > \frac{1}{n} \end{cases}, \ X(\omega) = 0 $$
Notice how the random variables are still exactly the same, I have just given some explanation about what the underlying mapping might be between states and outcomes. But now that I have defined the variables this way, I have that
$$P(\{\omega: X_n(\omega) \to X(\omega)\}) = P(\{\omega: X_n(\omega) \to 0\}) = P(\{\omega: \exists N \in \mathbb{N}, \forall n \geq N, \omega > \frac{1}{n} \}) = P((0,1]) = 1$$
So I seem to have proved that $X_n \overset{a.s.}{\to} X$. There must be some mistake in what I have done but I cannot find it.
 A: The two scenarios you give are not necessarily the same.  The first scenario might be the same as the second, but we cannot tell because you did not specify whether or not the $\{X_n\}_{n=1}^{\infty}$ variables are mutually independent. If these are mutually independent then you can invoke Borel-Cantelli to prove that, with probability 1, $X_n=1$ for infinitely many indices $n$. So $X_n$ does not converge to 0 almost surely.
On the other hand the second scenario is completely specified. There it is clear that the variables are not mutually independent, and it is also true, as you say, that the variables do converge to 0 almost surely. The proof you gave for that case is correct.

Note that you can conclude that $X_n\rightarrow 0$ in probability in the first scenario, even though this scenario is incompletely specified, because convergence in probability to 0 only concerns the individual (marginal) distributions of each $X_n$ and does not require information about how the $X_n$ are related to each other through a joint distribution on $(X_1, ..., X_n)$ for each positive integer $n$.
