$\limsup_n$ almost surely constant let $(X_n)_{n \in \mathbb{N}}$ be a sequence of independent real random variables and identically distributed. Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of real numbers such that:
$$\mathbb{P}\left(\limsup_n\left|\frac{1}{n} \sum_{k=1}^{n}{X_k}-x_n\right| <+\infty\right)>0$$
Prove that there exists $\alpha \in \mathbb{R}^+$ such that $\limsup_n\left|\frac{1}{n} \sum_{k=1}^{n}{X_k}-x_n\right|=\alpha \ \ a.s.$
I don't know how to begin, so I am thankful for any ideas.
 A: Although this is already outlined in the comments, I believe the "depth" of the results needed is quite large so I expand and this is an outline of proof.
Consider the tail sigma algebra $\tau = \cap_n \sigma(X_k; k \geq n)$ and by Kolmogorov law of triviality, each event in $\tau$ is trivial, that is the same as saying that the relation $\mathrm{X} \in \tau$ implies $\mathbf{P}(\mathrm{X}) \in \{0,1\}.$
With the previous in mind, one can check that for any random variable $X$ that is measurable with respect to $\tau$ (in fact, all one needs is that $\sigma(X)$ be trivial), the following is true, $X$ is almost surely constant, that is to say, there exists an $\alpha \in \mathbf{R}$ for which $\mathbf{P}(X = \alpha) = 1.$ To see this (rather trivial claim), notice that the sets $\{X < \beta\}$ will have probability $0$ or $1$ and consider $\alpha = \inf\{\beta;
\mathbf{P}(X < \beta) = 1\}.$
Finally apply the previous two observations to $X = \lim\cdot\sup \left| \dfrac{1}{x} \sum\limits_{i=1}^n X_i - x_n \right|,$ the only pesky point is to show this is measurable with respect to $\tau$ but to see this simply notice that in the limit, $\dfrac{1}{n}\sum_{i = 1}^m X_i \to 0$ as $n \to \infty,$ for whatever $m$ fixed beforehand.
Cheers,
