Showing that $ P\left(\lim\limits_{n \to \infty}\frac{S_n}{n} \text{ exists in } \mathbb R\right) = 0. $

Question

I want to show the following:

Let $X_1, X_2,\ldots$ i.i.d. random variables on $(\Omega, \mathfrak A, P)$ with infinite expectation. Let $S_n = X_1+\ldots+X_n$. Then $$ P\left(\lim\limits_{n \to \infty}\frac{S_n}{n} \text{ exists in } \mathbb R\right) = 0. $$

Using the second Borel-Cantelli-Lemma, I already showed that $$ P(|X_n| \ge n \text{ infinitely often}) = 1. $$

Now I'd like to show that $$ P\left(\lim\limits_{n \to \infty}\frac{S_n}{n} \text{ exists in } \mathbb R\right) \le P(|X_n| \ge n \text{ at most finitely often}). \tag{1} $$ Here's what I did:

Suppose that there exists $\omega \in \Omega$ s.t. $X_n(\omega) \ge n$ infinitely often and $\frac{S_{n}(\omega)}{n} \to L \in \mathbb R$.

Note that $$ \lim\limits_{n \to \infty}\frac{S_n(\omega)}{n+1} = \lim\limits_{n \to \infty}\frac{n}{n+1}\frac{S_n(\omega)}{n} = L $$

Then $$ \begin{align} \underset{n\to\infty}{\lim\sup} \left| \frac{S_{n+1}(\omega)}{n+1} - L \right| &= \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} + \frac{S_{n}(\omega)}{n} - L \right| \\ &\ge \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} \right| - \left| \frac{S_{n}(\omega)}{n+1} - L \right| \\ &= \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} \right| - \underset{n\to\infty}{\lim\inf}\left| \frac{S_{n}(\omega)}{n+1} - L \right| \tag{2}\\ &= 1-0 = 1, \end{align} $$ which is a contradiction to $\lim\limits_{n \to \infty}\frac{S_{n}(\omega)}{n} = L$. Therefore $$ \left\{\lim\limits_{n \to \infty}\frac{S_n}{n} \text{ exists in } \mathbb R \right\} \subset \left\{ |X_n| \ge n \text{ at most finitely often}\right\}, $$ which implies $(1)$ and therefore the claim.

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Is this correct? I'm mostly unsure about equation $(2)$. Is it always true that the limsup of a difference of two positive sequences is the limsup of the first sequence minus the liminf of the second one?

Thank you for your help!

Answer 1

As @KaviRamaMurthy correctly pointed out, $(2)$ is incorrect. However, the proof can be fixed since for any real sequences $a_n$ and $b_n$, if $b_n$ is convergent, it holds that $$ \underset{n\to\infty}{\lim\sup}\, (a_n + b_n) = \underset{n\to\infty}{\lim\sup}\, a_n + \lim\limits_{n\to\infty} b_n. $$

So I can write $$ \begin{align} \underset{n\to\infty}{\lim\sup} \left| \frac{S_{n+1}(\omega)}{n+1} - L \right| &= \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} + \frac{S_{n}(\omega)}{n} - L \right| \\ &\ge \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} \right| - \left| \frac{S_{n}(\omega)}{n+1} - L \right| \\ &= \underset{n\to\infty}{\lim\sup} \left| \frac{X_{n+1}(\omega)}{n+1} \right| - \lim\limits_{n\to\infty}\left| \frac{S_{n}(\omega)}{n+1} - L \right| \tag{2´}\\ &= 1-0 = 1, \end{align} $$