I am requied to show that $X_n(\omega)=n^{1/p}\mathbb{I}_{0\leq\omega<1/n}$ converges almost surely to $0$ as $n\rightarrow\infty$, where $\mathbb{I}_A$ is the indicator function of the event $A$.
I have already tried to check $\sum_{n=1}^{\infty}\mathbb{P}[|X_n-0|>\epsilon]<\infty$, but this failed, and I'm stuck with showing $\displaystyle\mathbb{P}[\limsup_n\{\omega:X_n>\epsilon\}]=0\:\forall\epsilon>0$. Any help would much be appreciated.
Best Answer
Have you tried drawing what happens as you increase n? Given any $\epsilon > 0$, you can find an $n$ large enough such that $X_n(\omega) = 0$ for any $\omega > \epsilon$. The only problematic point is $\omega = 0$, since $X_n(0) \to \infty$ as $n \to \infty$. However, the Lebesgue measure of a single point is 0.
Now, formally, you should check that the set $\{\omega \in [0, 1]: \lim_{n \to \infty} X_n(\omega) = 0\}$ has measure 1. I hope you can complete the proof from here.