I'm studying for a final exam I have tomorrow and have a question about the following problem.
Let $\Omega = [0,1]$ and let $P([a, b]) = b-a$ for all $0\le a\le b\le 1$. Let $X(\omega) = 0 $ for all $\omega \in \Omega$ an let $X_n (\omega) = 2^n \mathbf I_{[0,\frac{1}{n})} $ for $ n\ge 1$ Analyze whether $X_n (\omega) $ converges almost surely or only in probability to $X(\omega)$.
I believe $\mathbf I $ is an indicator function here.
We didn't cover this topic in too much detail, and I haven't done many problems involving the central limit theorem. So, I'd like to discuss whether my result is correct, if my reasoning for my answer is correct, and/or why I am wrong.
My answer to this problem is the following.
$$ \lim_{n \rightarrow \infty} X_n (\omega) = \lim_{n \rightarrow \infty} 2^n \mathbf I_{[0,\frac{1}{n})} (\omega) = \infty $$
Therefore, $X_n (\omega)$ does not converge. But, $X(\omega)$ converges to $0 \ \ \forall \ \ \omega \in \Omega $. Hence, $X_n (\omega) $ only in probability to $X(\omega)$.
I'm not sure how to apply the indicator function here, so this might be a source of error if my answer is incorrect. Are there more details I neglected to mention, how would a proper solution to this problem be described?
Thank you.
For any $\omega,$ we have $$ \lim_{n\to \infty} I_{[0,1/n)}(\omega)2^n =0$$ since eventually $1/n<\omega.$ Thus the sequence $X_n$ is almost surely eventually zero. Thus it also converges in probability to zero.
However, it doesn't converge in $L^1$ since $$ E(X_n) = \frac{2^n}{n} \to \infty.$$