Let's define $$P(X_n=1)=\frac1n,\;P(X_n=0)=1-\frac1n$$ $(X_n)$ are independent. I want to show that does not exist X such that : ${\displaystyle {\overset {}{X_{n}\,{\xrightarrow {\mathrm {a.s.} }}\,X.}}}$
We know that $$ X_{n} \xrightarrow{\mathrm{a.s.}} X \iff \mathbb{P}\left(\omega:\lim_{n\to \infty}X_n(\omega)=X(\omega)\right)=1. $$ $(X_n)$ is constant sequence so the limit of $(X_n)$ is $0$ or $1$. Let's consider $\sum_{n=1}^{\infty}P(\omega:X_n(\omega)=1)=\sum_{n=1}^{\infty}\frac1n=\infty.$
$\sum_{n=1}^{\infty}P(\omega:X_n(\omega)=0)=\sum_{n=1}^{\infty}1-\frac1n=\infty.$
So by the second Borell-Cantelli lemma i get :
${\displaystyle P\left(\bigcap _{n=1}^{\infty }\bigcup _{k=n}^{\infty }(\omega:X_n(\omega)=0)\right)=1.}$
${\displaystyle P\left(\bigcap _{n=1}^{\infty }\bigcup _{k=n}^{\infty }(\omega:X_n(\omega)=1)\right)=1.}$.
And now i want to tell that $(X_n)$ cannot has limit because he oscillates between $0$ and $1$. Am i thinking correctly ?