A variation of Borel-Cantelli Lemma 2 This problem is from my midterm. I have not solved it since.
"Given a measure space $(X,M,\mu)$ and measurable sets $A_1,A_2,\dots$ s.t. 
$$\sum \mu(A_i)^2<\infty. $$
Give an example where the above condition is satisfied and $\mu(\cap_{m=1}^\infty \cup_{i=m}^\infty A_i)>0$.
" 
So I think of this example $A_i=(i,i+1/i)$. Clearly it does satisfy the above condition but does not satisfy the condition of Borel-Cantelli Lemma, so this may be the right example. But I cannot calculate $\mu(\cap_{m=1}^\infty \cup_{i=m}^\infty A_i)$ since clearly we cannot use continuity from above with the infinite intersection. Is is the right approach? How do I proceed?
Other examples would be great, too.
Thanks in advance
 A: Let $(X_n)_{n\geq 1}$ be a sequence of independent Bernoulli random variables with $\mathbb{P}\{X_n=0\} = \frac{1}{n}$ for all $n\geq 1$. Consider the events
$$
A_n \stackrel{\rm def}{=} \{X_n =0\}
$$
so that $\mathbb{P}(A_n)=1/n$. Clearly, we have $\sum_{n=1}^\infty \mathbb{P}(A_n)^2 < \infty$. However, by the second Borel-Cantelli lemma,
$$
\mathbb{P}(\limsup_{n\to\infty}A_n) = 1 > 0
$$
(since the events are independent, and $\sum_{n=1}^\infty \mathbb{P}(A_n) = \infty$).
A: It is worth noting that you can do without the second Borel Cantelli lemma, exactly as you intuited.
On $[0,1]$ embedded with the Lebesgue measure measure on the Borel sets (hence a probability measure), consider $H_i$ the $i$-th harmonic number $H_i=\sum_{1 \le j \le i} 1/j$
Set  


*

*$A_i=(\text{fractional part of }H_i, \text{fractional part of} H_{i+1})$ if $H_i$ and $H_{i+1}$ have the same integer part, 

*$A_i =\emptyset$ otherwise.


Then it is not hard to prove that $$\limsup A_i=]0,1[,$$
hence has Lebesgue measure 1.
(Any increasing diverging sequence whose increments vanish in such a way their squares are summable will also do). 
