Let $X_i$ be a sequence of independent random variables, then $P( \sup_n X_n < \infty) \in \{0,1\}$. (Here the random variables take values on the reals, not on the extended reals).
So my solution is this: The event $A = \{\sup_n X_n < \infty\}$ only depends on the behaviour of the sequence at infinity, therefore we have the equality
$$A = \{\sup_n X_n < \infty\} = \{\sup_{n \geq m} X_n < \infty\} = A_m $$
for all positive integers $m$. Hence we have $A = A_m \in \sigma(A_m,A_{m+1},\dots)$ and therefore $A \in \cap_{n=1}^\infty \sigma(A_m,A_{m+1},\dots)$ so that $A$ is a tail event and notice that the events $A_m$ are independent, so by Kolmogorov's 0-1 law it follows that $P(A) \in \{0,1\}$.
I think that the solution is correct, the only problem I have is that i'm not sure how to justify that the events $A_m$ are independent, I think they should be independent because the subsequence $X_{m\geq n}$ are independent and taking sup preserves independence, but i'm not sure abou it. I would appreciate if anyone could help me prove that the events $A_m$ are independent in a more rigorous/formal way.
Best Answer
So I realized that the events $A_m$ are not independent and I fixed the proof, I will post the proof here for the sake of future reference.
Theorem: Kolmogorov's 0-1 law. Let $X_i$ be independent random variables, and define the tail $\sigma$ algebra as $$\mathcal T= \bigcap_{n=1}^\infty \sigma(X_n,X_{n+1},\dots) $$
If $A \in \mathcal T$ then $P(A) \in \{0,1\}$.
Let us now prove the proposition. Define $A = \{sup_n X_n < \infty\}$ We will show that $A^c = \{sup_n X_n = \infty \} \in \mathcal T$ and that will imply that $A \in \mathcal T$. In order to do that notice that
$$\{sup_n X_n = \infty\} = \bigcap_{k=1}^\infty \{X_n \geq k, i.o\} $$
Indeed, if $\omega$ is in the right hand side set, then for all $k \in \mathbb N$ we can find infinitely many $n$ such that $X_n(\omega) \geq k$ consequently for all $k$ we have $sup_n X_n(\omega) \geq k \implies sup_n X_n(\omega)=\infty $.
Now if $\omega$ is such that $sup_n X_n(\omega) = \infty$, then for $k \in \mathbb N$ we can find $N_1$ such that $X_{N_1}(\omega) \geq k$, we can also find $N_2 \neq N_1$ such that $X_{N_2}(\omega) \geq X_{N_1}(w) \geq k$ and so on, therefore we've proven that there are infinitely many $n$ such that $X_n(\omega) \geq k$. Since $k$ was arbitrary, this shows that $\omega \in \cap_{k=1}^\infty \{X_n \geq k, i.o\}$ and hence the equality of sets is proved.
Now we have $$\{X_n \geq k, i.o\} = \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty \{X_m \geq k\} $$
the events $\{X_m \geq k\} \in \sigma(X_m)$ hence $\bigcup_{n=m}^\infty \{X_m \geq k\} \in \sigma(X_n,X_{n+1},\dots)$. Now notice that $\sigma(X_n,X_{n+1},\dots) \supset \sigma(X_{n+1},X_{n+2})$ consequently $$\bigcap_{n=l}^\infty \bigcup_{m=n}^\infty \{X_m \geq k\} \in \sigma(X_l, X_{l+1},\dots) $$ and since $$\bigcap_{n=1}^\infty \bigcup_{m=n}^\infty \{X_m \geq k\} =\bigcap_{n=l}^\infty \bigcup_{m=n}^\infty \{X_m \geq k\} $$ for any integer $l$ this shows that $\{X_n \geq k, i.o\} \in \mathcal T$ for any $k$(basically I proved that the limsup is a tail event but I want this answer to be as detailed as possible) consequently it follows that $$\{sup_n X_n = \infty\} = \bigcap_{k=1}^\infty \{X_n \geq k, i.o\} \in \mathcal T $$ is a tail event and then so is $\{sup_n X_n < \infty\}$ and by Kolmogorov's 0-1 law we obtain $P(sup_n X_n < \infty) \in \{0,1\}$