I was reading an example of Chung's book "A course in probability theory" 3rd edition. The example says:
Let $\{X_n\}$ be independent random variables with common distribution function $F$ such that $$\mathbb{P}\left(X_1=n\right)=\mathbb{P}\left(X_1=-n\right)=\frac{c}{n^2\log n}$$ where c is the constant $$\frac{1}{2}\left(\sum_{n=1}^{\infty}\frac{1}{n^2\log n}\right)^{-1}$$ If $S_n=X_1+\cdots+X_n$, the idea in this example is to show that $$\limsup\frac{S_n}{n}=+\infty \quad \text{and} \quad \liminf\frac{S_n}{n}=-\infty$$
I understand all of the arguments given by Chung, except the last step. In some point, we can show that $$\mathbb{P}\left(\frac{|S_n|}{n}>\frac{A}{2} \ \text{i.o.}\right)=1 \quad \quad \quad (1)$$
With this in hand, it follows that \begin{equation} \limsup\frac{|S_n|}{n}\ge \frac{A}{2} \quad \text{a.s.} \quad \quad \quad (2) \end{equation} but Chung says that from (1) follows $$\limsup \frac{S_n}{n}\ge \frac{A}{2}$$
So, my question is: why he can drop the absolute value?. I tried to work with (1) in order to arrive to the conclusion given by Chung, writing it in terms of set, using the definition of $\limsup$ for sets and nothing. Moreover, how can I arrive to the conclusion given by Chung without zero one laws?
My question arises in a more general setting, because I've seen that proving $$\limsup \bigg|\frac{S_n}{a_n}\bigg|=1 \quad \text{a.s.}$$ implies $$\limsup \frac{S_n}{a_n}=1=-\liminf \frac{S_n}{a_n}$$ where $\{a_n\}$ is a sequence of no-negative numbers. For example, the law of iterated logarithm .
I appreciate you help.
$\lim \sup \frac {S_n} n$ is measuarble w.r.t. the tail sigma algebra which makes it a.s constant. Similarly, $\lim \inf \frac {S_n} n$ is constant a.s. and $\frac {S_n} n$ is symmetric by independence an symmetry of each $X_n$. It follows that if $\lim \sup \frac {S_n} n=c$ a.s then $\lim \inf \frac {S_n} n=-c$ a.s and $\lim \sup \frac {|S_n|} n=|c|$ a.s.
The arguemnt for the last statement in your question is the same but $a_n \to \infty$ is esential.