# Question about a particular example of $X_n \to_p X$ but $X_n \nrightarrow_{as} X$

Let's suppose that we have an independent sequence of random variables $X_n \sim Bern(\frac 1 n )$. Then for $\varepsilon$ sufficiently small, $P(\vert X_n - 0\vert > \varepsilon) = P(X_n = 1) = \frac 1n \to 0$ hence $X_n \to_p 0$. But according to this wikipedia article, an appeal to the Borel-Cantelli lemma is sufficient to show that $X_n \nrightarrow_{as} 0$.

My question: suppose our probability space is $([0,1], \mathscr F, \lambda)$ where $\mathscr F$ is the Borel $\sigma$-algebra generated by $[0,1]$ and $\lambda$ is the standard Lebesgue measure, which I believe is a probability measure in this case. Let $A_n = [0, \frac 1 n ]$ and define $X_n(\omega) = 1_{A_n}(\omega)$, so that $\lambda(X_n = 1) = \frac 1 n$ and therefore my sequence of $X_n$ are independent Bernoulli RVs with success probabilities $\frac 1 n$. Now consider $\lim_{n \to \infty} X_n(\omega)$. $\forall \omega \in (0, 1]$, eventually $X_n(\omega) = 0$ so doesn't this imply that $\forall \omega \notin \{0\}$, which is a null set w.r.t. $\lambda$, we have pointwise convergence?

What's going on here?

Update

@Did pointed out that my $X_n$ are not independent, which in retrospect is painfully obvious. Now I would like to know how to tweak them so that I do have independence and thus can see how the pointwise limits fail in this instance. This means that I need to come up with a sequence of sets $A_n \subset [0,1]$ such that $\lambda(A_n \cap A_m) = \frac 1{nm}$ for each $n \neq m \in \mathbb N$. I'm trying to come up with something but I haven't been able to get anything with independence in general, rather than just independence between $X_n$ and $X_{n+1}$. Any help would be very appreciated as I continue to work on this.

• What happens is that your sequence is not independent. – Did Nov 14 '16 at 15:51
• @Did Ah, that makes a lot of sense! Thank you very much. How could I modify my construction so that the $X_n$ are independent so I can see how the pointwise limits fail with positive probability? My guess is that in the course of it, the sets $A_n$ will necessarily bounce around enough? – alfalfa Nov 14 '16 at 16:05