Let $X_n$ be random independent variables, $S_n = \sum_{i=1}^n X_i$. I want to show that $\lim\sup_n X_n$ and $\lim \sup_n \frac{S_n}{n}$ are degenerate variables. We call a random variable $X$ degenerate if $X = c $ a.s for some $c \in \mathbb{R}\cup\{\pm\infty\}$. Note that the variables are not assumed to be iid., just independent.
My idea for the first claim: We consider the tail-$\sigma$-algebra, which is definied as $T = \cap_{n\in\mathbb{N}} T_n$, where $T_n = \sigma(X_k : k\geq n)$. We know that $\lim\sup_{n\geq m} X_n$ is $T_m$-measurable, however it's easy to see that $\lim\sup_{n\geq m} X_n = \lim\sup_n X_n =: X$, hence $X$ is $T_m$-measurable for all $m$, so it's $T$-measurable. By Kolmogorov's $0,1$ law, we know that for any $F\in T$, we have either $\mathbb{P}(F) = 0$ or $\mathbb{P}(F) = 1$. Trivially, $\mathbb{P}(X^{-1}([-\infty, \infty])) = 1$. If we already have $X=-\infty$ a.s, then $X$ is degenerate, otherwise there must exist an $c\in \mathbb{R} \cup \{\infty\}$ s.t. $\mathbb{P}(X^{-1}([-\infty, c))) = 0$ but $\mathbb{P}(X^{-1}([-\infty, c])) = 1$, hence $X=c$ a.s., so $X$ is degenerate again.
Does this work so far? Now I do not know how to show the second claim. It looks like the law of big numbers, but we don't have identical distributions. What can I try here?
Best Answer
Your proof is fine for the first one. For the second one simply observe that $\lim \sup_n \frac 1 n \sum\limits_{i=1}^{n}X_i =\lim \sup_n \frac 1 n \sum\limits_{i=k}^{n}X_i$ for any $k$ since $\lim_n \frac 1 {n} \sum\limits_{i=1}^{k-1}X_i=0$. Once again we see that $\lim \sup_n \frac 1 n \sum\limits_{i=1}^{n}X_i$ is tail-mesurable so Kolomogorov's $0-1$ law can be used to finish the proof.