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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 ?

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  • $\begingroup$ You've asked several questions related to the same problem without any real thoughts of your own. $\endgroup$ – Foobaz John Feb 28 at 22:52
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    $\begingroup$ Proof by contradiction is always a good place to start when you want to prove something doesen't exist. $\endgroup$ – Daniel Gendin Feb 28 at 23:01
  • $\begingroup$ What does it mean $P(X_n)=1$ with probability $p=\frac1n$? $\endgroup$ – NCh Mar 1 at 6:26
  • $\begingroup$ $P(X_n=1)=\frac1n$ $\endgroup$ – Lucian Mar 1 at 9:44
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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 $1$.

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