Kolmogorov zero-one law for lim sup Let $X_1,X_2,\dots$ be a sequence of independent random variables and {$b_n$} a sequence such that $0 \leq b_n \uparrow \infty$, $S_n = \sum_1^nX_i$. How can I show 
$$\mathbb{P}\left(\limsup_{n \rightarrow \infty} \left(\frac{S_n}{b_n}\right) = \mbox{constant}\right) = 1?$$
I have two questions:


*

*I think it is done with an application of Kolmogorov's 0-1 law, but how to show that $\left[\limsup_{n \rightarrow \infty} \left(\frac{S_n}{b_n}\right) \leq c\right]$ is a tail-event, if it is? I think this would be enough to conclude that $F_{\limsup S_n/b_n}(c)$ is 0 or 1 for each $c$ and, therefore, it would be either a finite constant, $+\infty$ or $-\infty$ a.s., Hence why I want to know if infinity counts as constant.

*In this context, does the word "constant" include $+\infty$ or $-\infty$? I mean, if I assume constant cannot be an infinite value, does the conclusion change? I took this question from a past year's exam, so I'm not sure what was agreed here. 


These questions came up because I thought of the following example:


*

*(a) $X_n = n²$ a.s. and $b_n = n$, what makes $\limsup \left( \frac{S_n}{b_n} \right) = +\infty$. If there is nothing wrong with this example and the result holds, then I think the definition of constant includes $+\infty$.

 A: First of all (as pointed out in the comments above), you need to put some assumptions on the sequence $X_n$, such as independence for example. Since otherwise, just taking $X_n = X_1$ for all $n$ and $b_n = n$, we do not get the claim unless $X_1$ is deterministic.
So, henceforth I will assume that we're dealing with independent sequence of random variables. 
Next, you are right that $\pm \infty$ should be treated as a constant. Indeed, assume $X_1$ is integrable with expectation $\mu$, so that we get the strong law of large numbers (SLLN).
Then, 
$$
(*) \lim \frac{S_n}{b_n} = \lim \frac{S_n}{n} \frac{n}{b_n}.
$$
Now, $S_n/n$ converges a.s. to $\mu$, and if you take $b_n$ so that $n/b_n \to \infty$, then (*) converges to infinity a.s. 
Finally, your guess on Kolmogorov's 0-1 is also correct (observe that we need the i.i.d assumption now), as for any constant $c $ (including infinities) the event
$$
E_{c} = \left\{\limsup \frac{S_n}{b_n} \leq c \right\}
$$
is clearly a tail event in view of the fact that $b_n \to \infty$. Hence by Kolmogorov's 0-1 law we have that the distribution function
$$
F(c) =  \mathbb{P} (  \limsup \frac{S_n}{b_n} \leq c )
$$
is either 0 or 1. But $F$ is non-decreasing and $F(+\infty) = 1$ and hence for some $c\in \mathbb{R}$ it has to become 1 unless $F(-\infty) = 1$. Now, taking $c_* = \sup \{ c \in \mathbb{R}: F(c) = 0 \} $ gives us the constant for which
$$
\mathbb{P} (  \limsup \frac{S_n}{b_n} =c_* ) = 1 .
$$
