Showing $X_n \rightarrow 0$ a.s.

Question

If $X_n$ is an infinite sequence of r.v.s all definied on the same space such that $P(X_n=1)=p_n$ and $P(X_n=0)=1-p_n$. How do you prove that $X_n \rightarrow 0$ a.s. if an only if $\sum_{n\geq1}p_n<\infty.$

Answer 1

When $\sum_{n=1}^\infty p_n < \infty$ then applying Borel Cantelli lemma gives you $\mathbb P( \limsup \{ X_n =1 \}) = 0$, so $\mathbb P( \liminf \{ X_n = 0 \}) = 1$. By that we have set of measure $1$ such that for every $\omega$ in that set, there exists $N(\omega)$, such that for $n>N(\omega)$ we have $X_n(\omega) = 0$. That means $X_n \to 0$ almost surely.

Second implication isn't true. Let $\Omega=[0,1]$, $\mathcal F = \mathcal P(\Omega)$ and $\mathbb P = \lambda$, where $\lambda -$ Lebesgue Measure. Define $X_n(\omega) = \chi_{[0,\frac{1}{n}]}(\omega)$. Then $p_n=\mathbb P(X_n =1) = \lambda([0,\frac{1}{n}]) = \frac{1}{n}$, so $\sum p_n = \infty$. However, for any $\omega \in (0,1]$ we have $X_n(\omega) \to 0$

When you add that random variables $X_n$ are independent, then second implication is true, since assuming contrary that $\sum p_n = \infty$, by Borel Cantelli we have $\mathbb P(\limsup \{X_n=1\})=1$, which means that $X_n(\omega) \not \to 0$ for any $\omega \in \limsup \{X_n=1\}$, so in particular $X_n \not \to 0$ almost surely.