When preparing for my stochastics exam, I tried to show that if $X_1,X_2,\dots$ are independent real-valued random variables, then there exists a $c\in \mathbb{R}\cup \{\pm \infty\}$ such that
\begin{align*} \limsup_n \frac{S_n}{n} = c \end{align*}
almost certainly, where $S_n = X_1+X_2+\dots+X_n$.
I tried to show that by assuming the random variable $\frac{S_n}{n}$ converges almost certainly (taking the pointwise limit) to wasn't constant and leading this to a contradiction to the $X_n$ being independent. But I could not get to the end and also didn't see why the pointwise limit needs to be a random variable as well. Anybody can help me out?
By Kolmogorov's 0-1 Law, we have $\limsup S_n/n = \infty$ with probability $0$ or $1$, same for $-\infty$. If either of these hold with probability 1, then we are done. If not, then the event $\{\limsup S_n/n \leq a\}$ also has probability $0$ or $1$ for every real number $a$. Recall that $\limsup S_n/n$ is a random variable, so the previous observation means the distribution function $F(a) \in \{0,1\}$ for every $a\in \mathbb R$. This can happen if and only if $\limsup S_n/n$ is almost surely constant.