application of Kolmogorov 0-1 Law Any random variable measurable with respect to the tail $\sigma$-field
of an infinite independent sequence of $\sigma$-fields is equal to some constant a.s.
My attempt:
Since $\alpha_n \rightarrow \infty$, we have $\limsup \frac{S_n}{\alpha_n} = \limsup \frac{X_j,\ldots,X_n}{\alpha_n}$, thus the variable $\limsup \frac{S_n}{\alpha_n}$ is measurable with respect to the tail σ-algebra of (X_n) and since the (Xn) are all
independent, by the 0 − 1 law, this random variable is almost-surely constant.
Is it correct the attempt ?
Could someone help me pls? Thanks for your time and help
 A: Not sure I get your attempt... the key is that we know that the event $\{X\le x\}$ is a tail event so $F(x)=P(X\le x)$ is zero or one for all $x.$ The intuition is that since $F$ is monotonic in $x,$ and (provided $X$ is not almost surely $-\infty)$ we have $F(-\infty) = 0$ and $F(\infty) = 1,$ there must be some crossover point $c$ at which $F$ jumps from $0$ to $1,$ so that $X=c$ a.s. Naturally, this will be given by $$ c=\sup\{x \mid F(x)=0\}.$$
Then you must show that $P(X=c) =1,$ which is easiest to show here by showing $P(X<c) = 0$ and $P(X\le c) =1.$ For instance, we have $$ 0=P\left(\bigcup_n  \{X<c-1/n\}\right) = P(X<c).$$  
A: Let $\mathcal T$ denote the $\sigma$-algebra which satisfies, by Kolmogorov’s zero–one law, $\mathbb P(A)\in\{0,1\}$ for any $A\in\mathcal T$.
If $X$ is a random variable measurable with respect to $\mathcal T$, then one has, for any $n\in\mathbb Z$, $\{X\in[n,n+1]\}\in\mathcal T$, so that $\mathbb P\{X\in[n,n+1]\}\in\{0,1\}$. But since $\bigcup_{n\in\mathbb Z}\{X\in[n,n+1]\}$ is the whole state space, it must be the case that $\mathbb P\{X\in[n,n+1]\}=1$ for at least one $n\in\mathbb Z$.
Now cut that particular interval $[n,n+1]$ in two halves, $[n,n+1/2]$ and $[n+1/2,n]$. By a similar argument as above, one has either $\mathbb P\{X\in[n,n+1/2]\}=1$ or $\mathbb P\{X\in[n+1/2,n]\}=1$ (or possibly both, but that is not important for the argument). Take the interval for which the probability is $1$, cut it in half again, and choose the half for which the probability of $X$ falling into it is $1$. And so forth, keep cutting in half. Then, use Cantor’s intersection theorem to conclude that the intersection of these ever smaller nested intervals is a singleton: $\{c\}$ for some $c\in\mathbb R$. Complete the proof by showing that $\mathbb P(X=c)=1$ (use that probability measures are continuous from above).
