Measure theory: Counterexample for a limit of Lebesgue measurable sets Let $(A_k)_{k \in \Bbb {N}}$ be a sequence of Lebesgue measurable subsets of $\Bbb {R}^n$. 
$A_k \in \mathscr {A}_n {} ,(A_k) \subset \Bbb{R}^n \ \forall k \in \Bbb {N}$
Let $ A^* = \{ x \in \Bbb {R}^n \mid x \in A_k \text{ for infinitely many } k \in \Bbb {N} \} $
I'm trying to find a sequence $A_k$ to disprove the following statement 
if $ \lim_{k \to \infty } (\mathscr{L}^n(A_k)) = 0 $ then $ \mathscr{L}^n(A^*) = 0 $
with $ \mathscr{A}_n = \{ A \in \Bbb {R}^n \mid \text{A is Lebesgue-measurable}\}$ and $ \mathscr{L}^n $ is the n-dimensional Lebesgue-measure.
Thanks for any help
 A: I'll give a one dimensional counterexample. For $k \geq 1$ let $s_k = \sum_{i=1}^k \frac{1}{i}$, i.e. the $k$-th partial sum of the harmonic series, and let $B_k$ be the closed interval $[s_k, s_{k+1}]$. Since the harmonic series diverges, the sequence $B_k$ covers $[1, \infty)$. The idea is to wrap these intervals around the unit interval; formally we take $A_k = \{x - \lfloor x\rfloor : x \in B_k\} $ for each $B_k$. Here $x - \lfloor x \rfloor $ is the fractional part of $x$.
Any element of the unit interval is in infinitely many of the $A_k$-s, so $A^* = [0,1)$, which has measure one. On the other hand the measure of $A_k$ obviously tends to zero as $k$ goes to infinity.
A: One dimensional case. Each real number $a \in (0,1)$ admits the expansion in binary  system as follows:
$a=\sum_{k=1}^{\infty}\frac{\theta_k}{2^k}$ where $\theta_k=0$ or $\theta_k=1$;
We define subsets of $(0,1)$ as follows:
Let $A_1=\{\sum_{k=1}^{\infty}\frac{\theta_k}{2^k}:\theta_1=1\},$
    $A_2=\{\sum_{k=1}^{\infty}\frac{\theta_k}{2^k}:\theta_2=1\},$
    $A_3= \{\sum_{k=1}^{\infty}\frac{\theta_k}{2^k}:\theta_3=1,\theta_4=1\},$
    $A_4= \{\sum_{k=1}^{\infty}\frac{\theta_k}{2^k}:\theta_5=1,\theta_6=1\},$
    $A_5= \{\sum_{k=1}^{\infty}\frac{\theta_k}{2^k}:\theta_7=1,\theta_8=1\},$
    $A_7= \{\sum_{k=1}^{\infty}\frac{\theta_k}{2^k}:\theta_9=1,\theta_{10}=1\},$
   and so on.
Let equip $(0,1)$ with standard Lebesgue measure $\mathscr{L}^1$. It is obvious that $(A_k)$ is the sequence of independent Lebesgue measurable subsets of $(0,1)$ and
$$\sum_{k=1}^{\infty}\mathscr{L}^1(A_k)=(1/2+1/2)+(1/4+1/4+1/4+1/4)+\cdots=+\infty.$$
By Boreli-Canteli Lemma we have that $\mathscr{L}^1(A^*)=1$, where $A^*=\{A_k ~o.i.m.\}$.
It is obvious also that $\lim_{k \to \infty}\mathscr{L}^1(A_k)=0$.
$n$-dimensional case. Now we put $B_k=A_k\times [0,1]^{n-1}$. Then $(B_k)$ is the sequence of independent Lebesgue measurable subsets of $(0,1)^n$ and $\sum_{k=1}^{\infty}\mathscr{L}^n(B_k)=\sum_{k=1}^{\infty}\mathscr{L}^1(A_k)=  +\infty$ which by Borel-Canteli Lemma implies that $\mathscr{L}^n(B^*)=1$, where $B^*=\{B_k ~o.i.m.\}$. On the other case we have  $\lim_{n \to \infty}\mathscr{L}^n(B_k)=\lim_{n \to \infty}\mathscr{L}^1(A_k)=0$.
