Removing absolute value I have trouble removing the absolute value in the following problem:
Let $\{X_n\}_{n=1}^{\infty}$ be a sequence of real independent identically distributed random variables on some probability space $(\Omega,F,P)$. Assume that they satisfy the following properties:

*

*They are integer valued almost everywhere with $P(X_1=k)=P(X_1=-k)$ for every integer $k$ and $P(X_1=0)=0$.

*We have $\frac{S_n}{n} \to 0$ in probability as $n \to \infty$. Here $S_n=\sum_{i=1}^nX_i$.

*For almost every $\omega \in \Omega$, we have:

\begin{equation*}
\limsup_{n \to \infty} \frac{ |S_n(\omega) |}{n} = \infty
\end{equation*}
Prove that:
\begin{equation*}
P(\{\limsup_{n \to \infty}\frac{S_n}{n}=\infty\})=P(\{\liminf_{n \to \infty}\frac{S_n}{n}=-\infty\})=1
\end{equation*}
How can I remove the absolute value sign? Intuitively, this is obvious because the random variables we have are distributed in a "symmetric" way. However, I have no idea how to write down a fully justified rigorous argument. I tried splitting up $\limsup$ using its definition or writing $S_n=S_n^+-S_n^-$. But none of which seems to be working. A hint will be really appreciated.
 A: Thank you to angryavian for picking up a silly blunder in an earlier attempted proof.  Here's a correction.
$\text{1.}\  \big\{X_n\big\}_{n=1}^\infty\ $ and $\ \big\{{-}X_n\big\}_{n=1}^\infty\ $ have the same distribution. So too, therefore, do $\ \Big\{\frac{S_n}{n}\Big\}_{n=1}^\infty\ $ and $\ \Big\{{-}\frac{S_n}{n}\Big\}_{n=1}^\infty\ $.
$\text{2.}\ \liminf_\limits{n\rightarrow\infty}\frac{S_n}{n}=-\limsup_\limits{n\rightarrow\infty}{-}\frac{S_n}{n}\
    $. Therefore
\begin{align}P\Big(\liminf_\limits{n\rightarrow\infty}\frac{S_n}{n}=-\infty\Big)&=P\Big(\limsup_\limits{n\rightarrow\infty}{-}\frac{S_n}{n}=\infty\Big)\\
&=P\Big(\limsup_\limits{n\rightarrow\infty}\frac{S_n}{n}=\infty\Big)
\end{align}
\begin{align}
\hspace{-1.3em}\text{3.}\ \ \Big\{\limsup_\limits{n\rightarrow\infty}\frac{|S_n|}{n}=\infty\Big\}&\subseteq
   \Big\{\liminf_\limits{n\rightarrow\infty}{}\frac{S_n}{n}={-}\infty\Big\}\\
   &\hspace{1em}\cup\Big\{\limsup_\limits{n\rightarrow\infty}\frac{S_n}{n}=\infty\Big\}\ .\end{align}
$\text{4.}\ \Big\{\limsup_\limits{n\rightarrow\infty}\frac{S_n}{n}=\infty\Big\}\ $ is a tail event of the sequence $\ \big\{X_n\big\}_{n=1}^\infty\ $ (see proof below) and must therefore have probability either $0$ or $1$ by Kolmogorov's zero-one law.
If $\ P\Big(\limsup_\limits{n\rightarrow\infty}\frac{S_n}{n}=\infty\Big)=0\ $, then from item $\text{2}$ above $\ P\Big(\liminf_\limits{n\rightarrow\infty}\frac{S_n}{n}={-}\infty\Big)=0\ $ also, and then it would follow from item $3$ that $\ P\Big(\limsup_\limits{n\rightarrow\infty}\frac{|S_n|}{n}=\infty\Big)=0\ $, contradicting the given condition that $\ P\Big(\limsup_\limits{n\rightarrow\infty}\frac{|S_n|}{n}=\infty\Big)=1\ $.  Therefore
$$
P\Big(\limsup_\limits{n\rightarrow\infty}\frac{S_n}{n}=\infty\Big)=1=P\Big(\liminf_\limits{n\rightarrow\infty}\frac{S_n}{n}={-}\infty\Big)\ .
$$
Proof that $\ \Big\{\limsup_\limits{n\rightarrow\infty}\frac{S_n}{n}=\infty\Big\}\ $ is a tail event:
For any positive integer $\ s\ $
\begin{align}
P\Big(&\limsup_{n\rightarrow\infty}\frac{S_n}{n}=\infty\,\Big|\,X_1,X_2,\dots,X_s\Big)\\
&=P\Bigg(\limsup_{n\rightarrow\infty}\frac{\sum_\limits{i=1}^s X_i+\sum_\limits{i=s+1}^nX_i}{n}=\infty\,\Bigg|\,X_1,X_2,\dots,X_s\Bigg)\\
&=P\Bigg(\limsup_{n\rightarrow\infty}\frac{\sum_\limits{i=s+1}^nX_i}{n}=\infty\,\Bigg|\,X_1,X_2,\dots,X_s\Bigg)\\
&=P\Bigg(\limsup_{n\rightarrow\infty}\frac{\sum_\limits{i=1}^{n-s}X_{i+s}}{n}=\infty\Bigg)\\
&=P\Bigg(\limsup_{n\rightarrow\infty}\Big(1-\frac{s}{n}\Big)\left(\frac{\sum_\limits{i=1}^{n-s}X_{i+s}}{n-s}\right)=\infty\Bigg)\\
&=P\Bigg(\limsup_{n\rightarrow\infty}\frac{\sum_\limits{i=1}^nX_{i+s}}{n}=\infty\Bigg)\\
&=P\Big(\limsup_{n\rightarrow\infty}\frac{S_n}{n}=\infty\Big)
\end{align}
because $\ \big\{X_i\big\}_{i=1}^\infty $ are distributed identically to $\ \big\{X_{i+s}\big\}_{i=1}^\infty\ $. Thus $\ \Big\{\limsup_\limits{n\rightarrow\infty}\frac{S_n}{n}=\infty\Big\}\ $ is independent of $\ X_1,X_2,\dots,X_s\ $ for any finite $\ s\ $, and is therefore a tail event.
