TITLE: What does a probability of $1$ mean?
QUESTION [12 upvotes]: From a textbook on probability on the Law of Large Numbers:

Theorem 3-19 (Law of Large Numbers): Let $X_1,X_2, \ldots , X_n$ be mutually independent random variables (discrete or continuous), each having finite mean  and variance. Then
  if $S_n = X_1 + X_2 +\dots+ X_n$,

$$ \lim_{n \to\infty} P\left(\left|\frac{S_n}{n} - \mu\right| \geq \varepsilon\right) = 0 $$
Since $S_n$ is the arithmetic mean of $X_1,X_2, \ldots , X_n$, this theorem states that the probability of the arithmetic mean $\frac{S_n}{n}$ differing from its expected value $\mu$ by more than $\varepsilon$ approaches zero as $n \to \infty$. A stronger result, which we might expect to be true, is that $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $  but this is actually false. However, we can prove that $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $
with probability one.
The only difference between the last sentence and the one before that is the phrase 'with probability one'. What does probability one mean here ? The usual definition is that the event occurs with 100% certainty. If that is the case, why is the original assertion $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $ false ?

REPLY [4 votes]: $S_n$ is itself a random variable. Lets say $\mu$ is 0.5. $S_n = 0$ is a possible event iff $X_i=0$ is possible for every i. That means that $lim_{n\rightarrow ∞}\frac{S_n}{n}$ is also a random variable! It is $\mu$ with probability one, but there are events for which it is not $\mu$!