Limit superior of $\sum_{j=1}^n X_j$ with $\mathbf{P}[X_j = 1] = \mathbf{P}[X_j = -1] = 0.5$ This is Exercise 2.3.1 from Achim Klenke: »Probability Theory — A Comprehensive Course«.

Exercise:  Let $(X_n)_{n\in N}$ be an independent family of $\mathrm{Rad}_{1/2}$ random variables (i.e., $\mathbf{P}[X_n = −1] = \mathbf{P}[X_n = +1] = \frac{1}{2}$) and let $S_n = X_1 + \dots + X_n$ for any $n \in \mathbb{N}$.
  Show that $\limsup_{n\rightarrow\infty} S_n = \infty$ almost surely.

I can't solve that one.
My thoughts so far: Ok, $A:=\bigl\{\limsup_{n\rightarrow\infty} S_n = \infty \bigr\} = \bigl\{\inf_{n\geq 1}\sup_{\nu\geq n} S_\nu = \infty\bigr\} $ is in the tail field
$A\in\mathcal{T}\bigl((X_n)_{n\in\mathbb{N}}\bigr)$.
Because $\sup_{\nu\geq n} S_\nu$ is monotonically decreasing, $A$ happens exactly when $A_n := \bigl\{\sup_{\nu\geq n} S_\nu = \infty\bigr\}$ happen for all $n\in\mathbb{N}$. But since $A_{n+1}$ implies $A_n$ (the supremum is monotonically decreasing) and $A_{n+1}^c$ implies $A_{n}^c$ (if $\sup_{\nu\geq n+1} S_\nu$ is finite, $\sup_{\nu\geq n} S_\nu$ will still be finite), we get $A_n = A_{n+1}$ and inductively that $A_1 = A_{n}$.
Now, By Kolmogorov’s $0-1$ Law, $\mathbf{P}[A]\in\{0, 1\}$. So we must show that $\mathbf{P}[A^c] = \mathbf{P}[ A_1^c] < 1$. $$A_1^c = \biguplus_{m\in\mathbb{Z}} \Bigl\{\sup_{\nu\geq 1} S_\nu = m \Bigr\} \, .$$
It seems very plausible that all $\mathbf{P}\biggl[\Bigl\{\sup_{\nu\geq 1} S_\nu = m \Bigr\}\biggr]=0$ for all $m\in\mathbb{Z}$, but how do I prove that?
Thank you!

Ok, I don't know if what I've written is helpful or even right (you can disregard it if you want). The question is:

How do I prove the exercise above? 

 A: You are on a good start. We already know that $\mathbb{P} (A) \in \{0, 1\}$, but how do we decide which one?
First remark: a sequence $(X_n)$ is in $A^c$ if and only if its sum is bounded from above, i.e., if and only if there exists $N \in \mathbb{N}$ such that $S_n \leq N$ for all $n$.
Now, let's use the symmetry of the random walk. The map $(X_n)_{n \in \mathbb{N}} \mapsto (-X_n)_{n \in \mathbb{N}}$ preserve the distribution of the sequence of random variables (they are still independent, and have the same distribution for all $n$). Hence,
$$\mathbb{P} (\limsup_n S_n = + \infty) = \mathbb{P} (\limsup_n -S_n = + \infty) = \mathbb{P} (\liminf_n S_n = - \infty).$$
If I write $B$ the event $\{\liminf_n S_n = - \infty\}$, then $\mathbb{P} (A) = \mathbb{P} (B) \in \{0, 1\}$. Let $C$ be the complement of $A \cup B$. Then $C$ is exactly the set of sequences which are bounded from above and from below, so the set of bounded sequences. Hence, 
$$C = \bigcup_{N \in \mathbb{N}} C_N,$$
where $C_N = \{|S_n| \leq N \ \forall n \in \mathbb{N} \}.$
I sum up. If $\mathbb{P} (C) = 0$, then $\mathbb{P} (A \cup B) = 1$. But $A$ and $B$ have the same probability, so $\mathbb{P} (A) \geq 1/2$. Since $\mathbb{P} (A) \in \{0, 1\}$, this forces $\mathbb{P} (A) = 1$. Hence, I only need to prove that $\mathbb{P} (C) = 0$, that is, that each $C_N$ has measure $0$.
Fix $N$. Remark that if there is a sequence of $2N+1$ elements of $(X_n)$ which have the same sign, then the sequence cannot belong to $C_N$.
For $k \geq 0$, let $D_k$ be the event $\{(X_{(2N+1)k}, X_{(2N+1)k+1}, \cdots, X_{(2N+1)k+2N} \text{ all have the same sign} \}$. Then all the $(D_k)$ are independent, and $\mathbb{P} (D_k) = 2^{-2N - 1}$ for all $k$. By Borel-Cantelli's lemma, almost surely, a sequence will lie in an infinity of $D_k$'s, and thus in at least one of them. Since being in any $D_k$ precludes being in $C_N$, this yields $\mathbb{P} (C_N) = 0$.
Since this is true for all $N$, we get $\mathbb{P} (C) = 0$, and from there, $\mathbb{P} (A) = 1$.
A: By the CLT, we have the convergence in distribution
$$
\frac{S_n}{\sqrt{n}}\Longrightarrow N(0,1).
$$
Thus since $\sqrt{n}\nearrow \infty$, it follows that $\limsup_n S_n=\infty$ a.s.

More details: Since $\lim_{n\to\infty}\mathbb P(S_n>\sqrt{n})=\int_1^{\infty}e^{-x^2/2}>0$, it follows that $\mathbb P(\limsup_n S_n=\infty)>0$. Thus by the Kolmogorov 0-1 Law, $\limsup_n S_n=\infty$ a.s.
A: I find a different way to solve this exercise.
Let $S^*:= \limsup_{n\to \infty }S_n$ and $S_n:= \sum_{k=1}^nX_k$. Then its easy to see that $\{ \omega :S^*=\infty \}$ is a tail event so $\Pr [S^*=\infty  ]\in\{0,1\}$. Then the sequence $(X_k)$ induces the product measure $\mu:= \left(\frac12 \delta _1+\frac12 \delta_{-1}\right)^{\otimes {\mathbb N}}$ in $\{-1,1\}^{{\mathbb N}}$ and setting $A:=\{x\in \{-1,1\}^{{\mathbb N}}:\limsup_{n\to \infty } \sum_{k=1}^n x_k\geqslant  0\}$ then we have that $\Pr [S^*\geqslant  0]=\mu(A)$.
Now note that for every $x\in \{-1,1\}^{{\mathbb N}}$ such that $\limsup_{n\to \infty }\sum_{k=1}^n x_k<0$ (where $x_k$ is the $k$-th coordinate of $x$) then the sequence $y$ defined by $y_k:=-x_k$ have the property that $\limsup_{n\to \infty }\sum_{k=1}^n y_k> 0$, so $\mu(A^\complement )=\mu(-A^\complement )$ by the definition of $\mu$ (where $-A^\complement $ is the set of sequences in $A^\complement $ with the signs changed), so $-A^\complement \subset A$, hence it follows that $\mu(A^\complement )\leqslant \mu(A)$ and consequently $\Pr [S^*\geqslant 0]>0$.
By last note that if $\limsup_{n\to \infty }\sum_{k=1}^n x_k=m$ then the sequence $y$ defined by $y_k:=x_{k+1}$ and $y_1:=1$ have the property that $\limsup_{n\to \infty }\sum_{k=1}^n y_k=m+1$ so $\Pr [S^*=m+1]\geqslant  \Pr [S^*=m]$ following a similar argument as the above one. Then the conclusion follows.
