I was wondering if it's possible to use the Borel Cantelli Theorem in order to ensure that the almost sure convergence DOESN'T exist.
We know that:
Let $A_n$ be a sequence of events in a probability space. If $\sum_{n=1} P(A_n) < \infty$, then $P(A_n i.o.) = 0$, implying that there is the convergence almost surely.
But what If I can find that $\sum_{n=1} \mathcal{P}(A_n) = \infty$?
I'll explain with an example:
Let ${X_n}$ be a sequence of independent random variables such that:
$$
\
X_n =
\begin{cases}
1 \text{with probability} = \frac{1}{n} \\
0 \text{with probability} = 1 – \frac{1}{n}
\end{cases}\
$$
In this case, since the E($X_n) = \frac{1}{n}$ and V($X_n) = \frac{1}{n}$, by using the chebyshev inequality:
$\mathcal{P}(X_n > \epsilon) \leq \frac{1}{n\epsilon^2}$. But $\sum \frac{1}{n\epsilon^2}$ diverges to $\infty$.
Question: Is it possibile to assert that the convergence almost surely it's not possibile or the fact that $\mathcal{P}(A_n)$ is not bounded has no meaning?
Best Answer
Using your example: we know that $\sum_{n=1}^{\infty}\mathbb{P}(X_n=1) = \infty$ and that $\{X_n\}$ are independent. We can use second Borel-Cantelli lemma to get that $\mathbb{P}(\limsup_{n\in\mathbb{N}}\{\omega:X_n(\omega)=1\})=1$, so there is no convergence to $0$ almost surely.