Borel Cantelli Lemma for non independent random variables

Question

Suppose $Y_n$ is a sequence of random variables, not independent, which converges to $0$ in probability. Define a set $A_n = \{|Y_n|>\epsilon\}$. If $\sum_1^\infty P(A_n) = \infty$, can it be concluded that $A_n$ does not converge to zero almost surely?

Answer 1

No. Borel-Cantelli Lemma states that $P(A_n i.o.) = 1$ only when $\sum_{n=1}^{\infty}{P(A_n)} = \infty$ and $A_n$'s are mutually independent.

Since in your case, $Y_n$'s are not independent, $A_n$'s are also not mutually independent. Hence, you cannot conclude "$A_n$ does not converge to zero almost surely" just using Borel-Cantelli Lemma. You will need additional information.

Edit 1: Adding an example as per suggestion in comments.

Consider $ \Omega = [0,1]$, $\sigma$-algebra to be the boral $\sigma$-algebra

$$Y_n = \begin{cases} 1, \ if \ \ \ \ w \in \left[ 0, \frac{1}{n} \right) \\ 0 \ \ \ otherwise \end{cases} $$

Let the Probability Measure be defined as length of the interval i.e. $$ P(Y_n = 1) = \frac{1}{n} \ , P(Y_n=0) = 1 - \frac{1}{n}$$

Here, you can clearly observe that $Y_n$'s are not independent. But if you define $A_n$ as $$A_n = \{Y_n = 1\}$$ we get $$P(A_n) = P(Y_n = 1) = \frac{1}{n}$$ so you have $$\sum_{n=1}^{\infty}{P(A_n)} = \sum_{n=1}^{\infty}{\frac{1}{n} = +\infty}$$

But for this sequence $Y_n$, you can easily see that $Y_n$'s converge almost surely to $0$ as $ n \to \infty$
So, we have a case where $Y_n$'s are not independent and $P(A_n)$ sums upto $\infty$, but still $Y_n$'s converge almost surely to $0$.
Hence, Independence is must to apply Borel-Cantelli Lemma.

Edit 2: Adding reference
Example taken from this Tutorial Notes. Example 12.3.1