$A_k\subset\mathbb{R}$ such that $\lim\sup A_k=\mathbb{R}$ but $\lambda(A_k)=1$ (Lebesgue measure) for all $k$.

Question

Construct a sequence of measurable sets $A_k\subset\mathbb{R}$ such that $\lim\sup A_k=\mathbb{R}$ but $\lambda(A_k)=1$ (Lebesgue measure) for all $k$.

My thoughts: Since $\lim\sup A_k=\cap_{n=1}^{\infty}\cup_{k\ge n}A_k$, we want for any sufficiently large $n$ (i.e., for all $n>N$ for some natural number $N$), the union of $A_{n},A_{n+1},...$ to cover $\mathbb{R}$, and the measure of each $A_k$ is 1. So I thought maybe we can make the sequence periodic, and in each period, the union of the $A_k$ covers $\mathbb{R}$. Thank you.

Answer 1

Let
$A_1=[0,1]$, $A_2=[-1,0]$, $A_3=[1,2]$,
$A_4=A_1$, $A_5=A_2$, $A_6=A_3$, $A_7=[-2,-1]$, $A_8=[2,3]$,
$A_9=A_4$, $A_{10}=A_5$, $\ldots$, $A_{14}=[-3,-2]$, $A_{15}=[3,4]$
and so on. That is in each step add two new sets by repeating all of the previous sets...

Answer 2

Pick an enumeration of $\mathbb Q$ and let $A_k=[q_k-\frac12,q_k+\frac12]$. Then for any $x\in \mathbb R$ there exist infinitely many rationals with $|q-x|<\frac12$, hence for each $n\in\mathbb N$ there exists $m>n$ with $|q_m-x|<\frac12$. That amounts to $x\in\bigcap_n\bigcup_{m>n} A_m$, as was to be shown.