# Problem with almost sure converges random variable with Bernoulli distribution.

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 $$X_{n} \xrightarrow{\mathrm{a.s.}} X \iff \mathbb{P}\left(\omega:\lim_{n\to \infty}X_n(\omega)=X(\omega)\right)=1.$$

• What does it mean $P(X_n)=1$ with probability $p=\frac1n$? – NCh Mar 1 at 6:26
• $P(X_n=1)=\frac1n$ – Lucian Mar 1 at 9:44
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 $$1$$.