random variable takes only rational values with probability one I have found an old exercise that seems very interesting: let $X_{1},X_{2},...$ be i.i.d. bernoullian random variables with $\mathbb{P}(X_{n}=1) = \mathbb{P}(X_{n}=0) = 1/2$. Define $S_{n} := X_{1} + ... + X_{n}$. It is to show that the random variable
$$
M := \sup_\limits{n \in \mathbb{N}}\frac{S_{n}}{n}
$$
with probability $1$ only takes rational values in the intervall $(1/2,1]$. Anyone has an idea, how to prove it?
 A: To add to Arnaud Mortier's comment: the fact that with probability $1$, $S_n/n \to 1/2$ as $n \to \infty$ comes from the Strong Law of Large Numbers.  
The fact that with probability $1$, $S_n/n > 1/2$ for infinitely many $n$ is a consequence of  the Law of the Iterated Logarithm.
A: 
Lemma Let $f: \mathbb{N} \to \mathbb{Q} \cap [0,1]$ be a mapping such that $\lim_{n \to \infty} f(n) = c \in (0,1)$ exists. Then $$M := \sup_{n \in \mathbb{N}} f(n)$$ satisfies $$M \in \{c\} \cup (\mathbb{Q} \cap [c,1]).$$

Proof: Since $\lim_{n \to \infty} f_n = c$ we clearly have $M \geq c$. If $M=c$ we are done, and therefore we will from now assume that $M>c$. Since $\lim_{n \to \infty} f(n)=c<M$ we can choose $N \in \mathbb{N}$ sufficiently large such that $f(n) < M$ for all $n \geq N$. Thus, $$M = \sup_{n \in \mathbb{N}} f(n) = \max_{n=1,\ldots,N} f(n).$$ This shows that the supremum is attained for some $n \in \{1,\ldots,N\}$ and so $M=f(n) \in \mathbb{Q} \cap [0,1]$.

By the strong law of large numbers, we can apply the above lemma to $f(n) := S_n(\omega)/n$ with $c:=1/2$. This proves the assertion.
