TITLE: IID Random Variables that are not constant can't converge almost surely
QUESTION [6 upvotes]: I am trying to prove the following.
If $\{ X_n \}$ are iid random variables and not constant, then $R:=P\{ \omega \mid  X_n(\omega)\text{ converges} \}=0$
Using independence I know that by Kolmogorov's 0-1 law, that if $R$ is not $0$ then $R=1$.  So I think the way to do this proof is by contradiction.  So I am trying to show $R=1$ implies the $X_n$ are constant using their identical distribution but sadly it is not working.  Help would  be appreciated.  Thanks!

REPLY [4 votes]: $\left\{\omega: \lim X_n(\omega)  \le c \right\}$ is in the tail $\sigma$-algebra for every $c$. So by Kolmogorov's $0-1$ law they have to converge to a constant if they converge.
But this can't happen if $X_n$ are not constant. Because in this case for any constant $c$ we have some $\epsilon > 0$ so that  
$P(X_n(\omega) < c - \epsilon) \ge \delta > 0$ for every $n$ 
or $P(X_n(\omega) > c + \epsilon) \ge \delta > 0$ for every $n$. 
In either case, we see that $X_n(\omega)$ cannot converge to $c$ almost surely.