The meaning of almost surely convergence Consider the space of sequences of tosses of a fair coin. In particular let $X_{nH}$ be the number of times that a coin lands on heads in a sequence of $n$ coin flips.
Consider statement $S$ below.
$S$ : In the limit as $n \rightarrow \infty$ almost surely $X_{n}/n \rightarrow 0.5 $.
It seems clear that it is not the case that in the limit as $n \rightarrow \infty$ the
probability $\mathbb{P}(X_{n}/n = 0.5) \rightarrow 1$ since if $n$ is odd it is clear
that $X_{n}/n \neq 0.5$.  So does statement $S$ mean that for each $n$ there exists a
measurable neighbourhood $N_{n}$ about $0.5$ such that with probability 1 each 
$X_{n}/n \in N_n$ and that $N_{n}$ is decreasing with $n$?  
The other subtlety lies in the definition of $X_{nH}$. I am not sure whether this definition indicates that we are to generate all sequences of coin flips then choose a sequence of length $n$ at random and measure how close $X_{nH}$ is to 0.5 for ever increasing $n$.  Or does this definition mean choose  a sequence length $n$ and calculate $X_{nH}$ then flip the coin again and calculate $X_{n+1H}$ for the original sequence plus an extra coin flip.  Of course these two interpretations of the definition could really be the same!
In summary I've been thinking for a while about the meaning of this (and a few other) almost surely statement(s) and I can't progress without understanding these very basic but fundamental questions!  
 A: Almost sure convergence of a sequence of random variables means that P(Limit holds)=1. So, in your case, $\frac{X_n}{n}\rightarrow 0.5\space a.s.$ means $P(\lim\limits_{n\rightarrow\infty}\frac{X_n}{n}= 0.5)=1$ This statement is meant to be interpreted given an entire sequence of $\frac{X_n}{n}$ values (i.e, each realization of an infinite sequence of tosses is 1 sample point/object). What is says is that, except for a coutable set of sequences, each sequence of $\frac{X_n}{n}$ values satistifes the regular/real analysis definition of the limit of a sequence. You state:
"So does statement S  mean that for each n  there exists a measurable neighbourhood $N_n$   about 0.5  such that with probability 1 each $\frac{X_n}{n}\in N_n$   and that $N_n$   is decreasing with n ? 
This statement is more like the weak law of large numbers, or convergence in probability, which states $P(|\frac{X_n}{n}-0.5|<\epsilon)\rightarrow1,\forall \epsilon>0$. For almost sure convergence, what matters is that for almost all sequences ($X_{i,n}$):
$\space\exists N_{i,j}: |\frac{X_{i,n}}{n}-0.5|<N_{i,j},\forall n\geq j$ where $i\in[0,1]$ is a continuous index for the infinite sequence of coin tosses.
Basically, its saying that almost all sequences (i.e, trajectories) approach 0.5 in the same sense that a normal sequence would. This is the key difference between the weak and strong law of large numbers: weak law only says that most of the $\frac{X_n}{n}$ values for a given $n$ will be close to 0.5, while the strong law talks about an entire sample trajectory (single infinite run of tosses), and states that most trajectories get close to 0.5 and stay close, which is something the weak law does not guarantee -- it just says the most of the trajectory values will be close to 0.5, but not that they have to stay close for any given trajectory, (i.e, $\frac{X_{n+1}}{n+1} > N_n$, which the strong law says only happens a countably infinite number of times, so P($\frac{X_{n+1}}{n+1} > N_n)=0$
