# IID Random Variables that are not constant can't converge almost surely

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!

## 1 Answer

$\left\{\omega: \lim X_n(\omega) \le c \right\}$ 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.

• How do you know your second sentence. "But this ... for every n" – Leo Spencer Dec 4 '13 at 5:51
• If $X_1$ is not $c$ almost surely then it must be bounded away from $c$ with positive probability. Then the $X_n$ are i.i.d. so they all enjoy this same positive probability bound away from $c$. – Deven Ware Dec 4 '13 at 5:55
• Forgot about the i.d. part, of course. I been working too much today. Time to go to bed haha. – Leo Spencer Dec 4 '13 at 5:57
• I don't understand the first sentence. You have $\{\omega: \lim X_n(\omega)=c\}$ has probability zero or one for every $c$. What prevents these probabilities from all being zero? As they would be, for instance, if $\lim X_n$ were a continuous random variable. – user940 Dec 4 '13 at 6:07
• @DevenWare Actually I am not convinced that we have the bound. Suppose $X_1$ was $c$ everywhere but on a set of measure zero. Then doesn't that violate your argument? – Leo Spencer Dec 4 '13 at 6:25