Let's define $$P(X_n)=\begin{cases}1,&w.p=\; \frac1n \\0,&w.p = 1-\frac1n\end{cases}$$ I want to show that does not exist X such that : ${\displaystyle {\overset {}{X_{n}\,{\xrightarrow {\mathrm {a.s.} }}\,X.}}}$
We know that $$ X_{n} \xrightarrow{\mathrm{a.s.}} X \iff \mathbb{P}\left(\omega:\lim_{n\to \infty}X_n(\omega)=X(\omega)\right)=1. $$
Can you please help me with that ?
This is false. On $(0,1)$ with Lebesgue measure let $X_n =I_{(0,\frac 1 n)}$. Then the hypothesis is satisfied but $X_n$ does tend to $0$ almost surely.
However, if $X_n$'s are independent then the statement is true. $\sum P(X_n=0) =\infty$ and $\sum P(X_n=1) =\infty$. By Borel Cantelli Lemma we see that $X_n$ oscillates with probbaility $1$.