Let $A_n$ be events with $\mathbb{P}(A_n)=\frac{1}{n^2}$ and define the random variables $X_n=n^2\mathbb{1}_{A_n}$. I want to prove that $X_n \rightarrow 0$ almost surely.
My attempt:
$\mathbb{P}(X_n\neq0)=\mathbb{P}(A_n)=\frac{1}{n^2}\rightarrow 0$. Is this enough to justify that $X_n \rightarrow 0$ almost surely?
Thank you!
No. It is not. Consider the random variables on $[0,1]$ given by $X_k=1_{[\frac{k}{n},\frac{k+1}{n}]}$ for $\sum_{j=1}^{n-1} j\leq k< \sum_{j=1}^n j$. Then, $X_k$ doesn't converge pointwise anywhere, but it does hold that, with the previous notation $\mathbb{P}(X_k\neq 0)=\frac{1}{n}\to 0$ as $k\to \infty$. In general, for this sort of question you need the Borel Cantelli lemma.
Using this, $\sum_{n=1}^{\infty} \mathbb{P}(X_n\neq 0)=\sum_{n=1}^{\infty}\mathbb{P}(A_n)=\sum_{n=1}^{\infty} \frac{1}{n^2}<\infty,$ implying that $\mathbb{P}(X_n\neq 0\; \textrm{i.o.})=0.$ However, that implies that $\mathbb{P}(X_n=0\; \textrm{evt.})=1,$ and $(X_n=0\;\textrm{evt.})\subseteq (X_n\to 0)$. Thus, we get almost sure convergence.
If you are not using this notation, $(A_n\; \textrm{i.o.})=\limsup_n A_n$ and $(A_n\; \textrm{evt.})=\liminf_n A_n$.