Find an example of Lebesgue measurable subsets of $[0,1]$ these conditions:

Question

Let $m$ be Lebesgue measure. Find an example of Lebesgue measurable subsets $A_1, A_2, \dots$ of $[0,1]$ such that $m(A_n) > 0$ for each $n$, $m(A_n \Delta A_m) >0$ if $m \neq n$, and $m(A_n \cap A_m) = m(A_n)m(A_m)$ if $m\neq n$.

I've spent some time trying to come up with examples of sets that satisfy these conditions and here's what I've come up with so far:

These sets can't be properly nested since this would not satisfy the multiplicative property.

Each set must intersect every other set in order to satisfy the multiplicative property.

I've tried many variations of shifting shrinking closed intervals through the interval $[0,1]$.

Any nudges in the right direction would be appreciated.

Answer 1

Let $\sum_{k=1}^{\infty}\frac{\theta_k}{2^k}$ be the representation of the real number $\theta \in [0,1]$ in the binary system. For $n=1,2,\cdots$, we set $A_n=\{ \theta: \theta \in [0,1]\& \theta _n=0\}$. The $(A_n)_{n \in N}$ is a required sequence of subsets of $[0,1]$.

Answer 2

Inspired by Michael's comment, here's what I've come up with:

Let $A_n = \cup_{k=1}^{2^{n-1}} \left[\frac{2k-1}{2^n}, \frac{2k}{2^n}\right] $. Since each of these are countable unions of closed intervals, they are certainly Lebesgue measurable, and we claim that this sequence satisfies the given conditions.

If we fix an $A_n$, then this set is comprised of $2^{n-1}$ disjoint unions each of which has length $2^n$. Thus, $m(A_n) = 2^{n-1}\cdot(1/2^n) = 1/2> 0$ for each $n$.

I'm not exactly sure how to $prove$ that these satisfy that the last two conditions are true, but they are certainly clear if you just draw some number lines and see the pattern.