Counter example to Borell Cantelli when $\sum \mu(A_n)^2 <\infty$

Question

I am looking for a counter example to the Borell Cantelli Lemma when $$\sum \mu(A_n)^2 <\infty$$

There is a probability solution here:A variation of Borel-Cantelli Lemma 2 but I am looking for one in the Lebesgue measure. It seems clear to me that we have to somehow exploit the fact that the harmonic series diverges and $1/n^2$ converges. However, I could not come up with the intevrals.

Answer 1

One strategy: For $x \in (0,1]$ we can write $$x = \sum_{n=2}^\infty \frac{a_n}{n!}$$ where $a_n$ is a non-negative integer $0 \leq a_n < n$. Further, $a_n$ is uniquely determined by $x$ if we assume there is no $N$ such that $a_n = 0$ for all $n \geq N.$ Therefore, we can write $a_n = a_n(x).$

Now define $A_n = \{x \in (0,1] \mid a_n(x) = 1\}.$

(Compare this to the answer given in your link)